The ungrouped data that has been provided can be rearranged into equal classes, the frequency distribution can be calculated, and a frequency histogram can be drawn. The data that has been given is:(4.6 4.5 4.7 4.6 4.5 4.6 4.3 4.6 4.8 4.2 4.6 4.5 4.7 4.5 4.5 4.6 4.6 4.6 4.8 4.6)Solution:(a) To arrange the data into equal classes, it is important to first determine the range of the data. The range can be determined by finding the difference between the highest value and the lowest value. Range = Highest value - Lowest value Range = 4.8 - 4.2Range = 0.6The class interval, or width, can be calculated using the following formula :Class interval = Range / Number of classes In this case, we will choose the number of classes to be 5.Class interval = 0.6 / 5Class interval = 0.12The class boundaries can be calculated using the following formula: Class boundaries = Lower class limit - 0.5 to Upper class limit + 0.5The following table shows the classes and their corresponding boundaries:
ClassBoundsFrequency4.1 - 4.3[4.05 - 4.15)1 4.3 - 4.5[4.15 - 4.25)5 4.5 - 4.7[4.25 - 4.35)6 4.7 - 4.9[4.35 - 4.45)2
(b) To determine the frequency distribution, the frequency of each class can be calculated by counting how many data points fall into each class. This can be seen in the table above. There are 1 data point in the class 4.1 - 4.3, 5 data points in the class 4.3 - 4.5, 6 data points in the class 4.5 - 4.7, and 2 data points in the class 4.7 - 4.9.
(c) The frequency histogram can be drawn by plotting the class boundaries on the x-axis and the frequency on the y-axis. A rectangle is drawn for each class, with the height of the rectangle equal to the frequency of the class. The following histogram can be drawn from the data:
Frequency Histogram
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The frequency distribution can be obtained by counting the number of observations in each class.
The results are as follows:
Class interval Frequency
4.0 - 4.4 1
4.5 - 4.9 9
a) Arranging the data into equal classes
The ungrouped data can be arranged into equal classes.
The following class interval can be used:
Class interval Frequency
4.0 - 4.4 1
4.5 - 4.9 9
The range of the data is 4.8 - 4.2 = 0.6 (always round up).
Therefore, we can have the following classes:
Class interval Frequency
4.0 - 4.4 1
4.5 - 4.9 9
b) Determining the frequency distribution
The frequency distribution can be obtained by counting the number of observations in each class.
The results are as follows:
Class interval Frequency
4.0 - 4.4 1
4.5 - 4.9 9
c) Drawing the frequency histogram
A histogram is a graphical representation of a frequency distribution.
The histogram for the frequency distribution of the ungrouped data is given below:
Histogram for the frequency distribution
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PLS HELP! NEED TO RAISE GRADE! WILL GIVE BRAINLIEST AND A LOT OF POINTS!
2. A sequence can be generated by using , where and n is a whole number greater than 1.
(a) What are the first five terms in the sequence?
(b) Write an iterative rule for the sequence. Show your work.
{{{ THE BOLDED CHARACTERS SHOULD BE SMALL. }}}
A sequence can be generated by using an = a(n-1) - 5, where a1 = 100 and n is a whole number greater than 1.
a1 = 100 (given)
a2 = a1 - 5 = 100 - 5 = 95
a3 = a2 - 5 = 95 - 5 = 90
a4 = a3 - 5 = 90 - 5 = 85
a5 = a4 - 5 = 85 - 5 = 80
ANSWER for PART (a): 100, 95, 90, 85, 80
-----------------------------------------------------------------------------
an = a1 + d(n - 1)
a1 = 100d is -5 (common difference, and we know it is -5)an = 100 + -5(n - 1)
an = 100 + -5n + 5
an = 105 - 5n
ANSWER for PART (b): an = 105 - 5n
Answer:
I agree with the person above me and here give him brainliest
Step-by-step explanation:
Help! A password with 6 characters is randomly selected from the 26 letters of the alphabet. What is the probability that the password does not have repeated letters, expressed to the nearest tenth of a percent?
Enter your answer as a number, like this: 42.3
The probability that the password does not have repeated letters, expressed to the nearest tenth of a percent is 0.0018%.
Given that a password with 6 characters is randomly selected from the 26 letters of the alphabet.
The number of ways to choose the first letter is 26 since all 26 letters are available.
The number of ways to choose the second letter is 25 since one letter has already been chosen and there are only 25 letters remaining.
Similarly, the number of ways to choose the third, fourth, fifth, and sixth letters are 24, 23, 22, and 21, respectively.
So, the total number of possible passwords is given by: 26 × 25 × 24 × 23 × 22 × 21 = 26P6
We want to find the probability that the password does not have repeated letters.
Let's calculate this probability now.
The first letter can be any of the 26 letters.
The second letter, however, can be one of the remaining 25 letters.
The third letter can be one of the remaining 24 letters, and so on.
So, the number of possible passwords that do not have repeated letters is given by: 26 × 25 × 24 × 23 × 22 × 21 / (6 × 5 × 4 × 3 × 2 × 1) = 26P6/6P6
So, the probability that the password does not have repeated letters is given by: P(A) = 26P6/6P6≈ 0.000018449 or 0.0018% (to the nearest tenth of a percent)
Therefore, the probability that the password does not have repeated letters, expressed to the nearest tenth of a percent is 0.0018%.
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Let v = - [3 1] and u=[2 1]. Write v as the sum of a vector in Span{u} and a vector orthogonal to u. (2) Find the distance from v to the line through u and origin.
The vector v can be written as the sum of a vector in Span{u} and a vector orthogonal to u as follows: v = (1/5)u + (-4/5)[1 -3].
The main answer can be obtained by decomposing the vector v into two components: one component lies in the span of vector u, and the other component is orthogonal to u. To find the vector in the span of u, we scale the vector u by the scalar (1/5) since v = - [3 1] can be written as (-1/5)[2 1]. This scaled vector lies in the span of u and can be denoted as (1/5)u.
To find the vector orthogonal to u, we subtract the vector in the span of u from v. This can be calculated by multiplying the vector u by the scalar (-4/5) and subtracting the result from v. The orthogonal component is obtained as (-4/5)[1 -3].
Thus, we have successfully decomposed vector v as v = (1/5)u + (-4/5)[1 -3], where (1/5)u lies in the span of u and (-4/5)[1 -3] is orthogonal to u.
In linear algebra, vector decomposition is a fundamental concept that allows us to express a given vector as a sum of vectors that have specific properties. The decomposition involves finding a vector in the span of a given vector and another vector that is orthogonal to it. This process enables us to analyze the behavior and properties of vectors more effectively.
In the context of this problem, the vector v is decomposed into two components. The first component, (1/5)u, lies in the span of the vector u. The span of a vector u is the set of all vectors that can be obtained by scaling u by any scalar value. Therefore, (1/5)u represents the part of v that can be expressed as a linear combination of u.
The second component, (-4/5)[1 -3], is orthogonal to u. Two vectors are orthogonal if their dot product is zero. In this case, we subtract the vector in the span of u from v to obtain the orthogonal component. By choosing the scalar (-4/5), we ensure that the resulting vector is orthogonal to u.
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find the value of a and b.
Answer:
2a°+2a°=180°[opposite angle of cyclic quadrilateralis 180°]
4a°=180
a=180°/4
a=45°
4b+2b=180°[similarly]
6b=180°
b=180°/6
b=30
COMPUTE THE AREA PARALLELOGRAM DETERMINED BY U= (4,-1) AND V = (-6,-2).
To compute the area of a parallelogram determined by two vectors U = (4, -1) and V = (-6, -2), we can use the formula that states the area of a parallelogram is equal to the magnitude of the cross product of the two vectors. Therefore, the area of the parallelogram determined by U = (4, -1) and V = (-6, -2) is 28 square units.
The formula to compute the area of a parallelogram determined by two vectors U and V is given by:
Area = |U x V|
To calculate the cross product U x V, we can use the following determinant:
| i j k |
| 4 -1 0 |
|-6 -2 0 |
Expanding the determinant, we get:
i * (0 * -2 - 0 * -2) - j * (4 * -2 - 0 * -6) + k * (4 * -2 - (-1) * -6)
= -12i + 24j + 8k
Taking the magnitude of the cross product, we have:
|U x V| = √((-12)^2 + 24^2 + 8^2) = √(144 + 576 + 64) = √784 = 28
Therefore, the area of the parallelogram determined by U = (4, -1) and V = (-6, -2) is 28 square units.
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Consider a regular surface S in R given by x2 + y2 = 2022. Is S orientable ? Justify your answer.
S is not orientable.
Given a regular surface S in R given by x² + y² = 2022, we need to find out whether S is orientable or not.
The surface is given by, x² + y² = 2022.
Rearranging the terms, we get, y² = 2022 - x²
Let the differentiable function g(x, y) = y, and the set U be the upper hemisphere (upper half) of the surface S.
Then, U = {(x, y, z) : x² + y² = 2022, z ≥ 0}
We know that the partial derivatives of the above function are continuous in U and it follows that U is a regular surface.
We now compute the partial derivatives of g(x, y) :∂g/∂x = 0, and ∂g/∂y = 1
Taking the cross-product of the two partial derivatives, we get : (0i - j + 0k) which is -j.
Now, if we define the positive normal to U to be the upward-pointing unit normal, then we see that (-j) points downward at all points on U.
Thus, U is not orientable.
Therefore, we conclude that the given surface S in R is not orientable.
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a door delivery florist wishes to estimate the proportion of people in his city that will purchase his flowers. suppose the true proportion is 0.07 . if 259 are sampled, what is the probability that the sample proportion will be less than 0.05 ? round your answer to four decimal places.
The probability that the proportion will be less than 0.05 is approximately 0.1056, rounded to four decimal places.
We have,
To calculate the probability that the sample proportion will be less than 0.05, we can use the sampling distribution of the sample proportion.
Given that the true proportion is 0.07 and a sample of size 259 is taken, we can assume that the distribution of the sample proportion follows a normal distribution with a mean equal to the true proportion (0.07) and a standard deviation equal to the square root of (p(1-p)/n), where p is the true proportion and n is the sample size.
In this case, the mean is 0.07 and the standard deviation is:
= √((0.07 x (1 - 0.07)) / 259).
To find the probability that the sample proportion will be less than 0.05, we can standardize the value using the z-score formula:
z = (x - mean) / standard deviation
In this case, we want to find P(X < 0.05), which is equivalent to finding P(z < (0.05 - mean) / standard deviation).
Calculating the z-score and using a standard normal distribution table or a calculator, we can find the corresponding probability.
Substituting the values into the formula:
z = (0.05 - 0.07) / √((0.07 x (1 - 0.07)) / 259)
Now, we can find the probability by looking up the corresponding
z-value in the standard normal distribution table or using a calculator.
The probability that the sample proportion will be less than 0.05 is the probability corresponding to the calculated z-value.
Round the answer to four decimal places to get the final result.
Therefore,
The probability that the proportion will be less than 0.05 is approximately 0.1056, rounded to four decimal places.
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Please Answer This, the question is on the picture. it needs to be a fraction
will mark brainllest if its right, no links!
(for hamster dude)
Answer:
x = 27.2 or 27 1/5
Step-by-step explanation:
cos 54° = 16/x
x = 27.2 or 27 1/5
8. Find the standard deviation, to one decimal place, of the test marks tabulated below. 41-50 51-60 61-70 71-80 81-90 Mark Frequency 5 0 10 8 2
The standard deviation of the test marks is 6.5
To calculate the standard deviation of the test marks, we need to follow a few steps. Let's go through them:
Step 1: Calculate the midpoint for each interval.
The midpoint is calculated by adding the lower and upper limits of each interval and dividing by 2.
Midpoint for 41-50: (41 + 50) / 2 = 45.5
Midpoint for 51-60: (51 + 60) / 2 = 55.5
Midpoint for 61-70: (61 + 70) / 2 = 65.5
Midpoint for 71-80: (71 + 80) / 2 = 75.5
Midpoint for 81-90: (81 + 90) / 2 = 85.5
Step 2: Calculate the deviation for each midpoint.
The deviation is calculated by subtracting the mean (average) from each midpoint.
Mean = ((45.5 * 5) + (55.5 * 0) + (65.5 * 10) + (75.5 * 8) + (85.5 * 2)) / (5 + 0 + 10 + 8 + 2)
= (227.5 + 0 + 655 + 604 + 171) / 25
= 1657.5 / 25
= 66.3
Deviation for 45.5: 45.5 - 66.3 = -20.8
Deviation for 55.5: 55.5 - 66.3 = -10.8
Deviation for 65.5: 65.5 - 66.3 = -0.8
Deviation for 75.5: 75.5 - 66.3 = 9.2
Deviation for 85.5: 85.5 - 66.3 = 19.2
Step 3: Square each deviation.
(-20.8)^2 = 432.64
(-10.8)^2 = 116.64
(-0.8)^2 = 0.64
(9.2)^2 = 84.64
(19.2)^2 = 368.64
Step 4: Calculate the squared deviation sum.
Sum of squared deviations = 432.64 + 116.64 + 0.64 + 84.64 + 368.64 = 1003.2
Step 5: Calculate the variance.
Variance = (Sum of squared deviations) / (Number of data points - 1) = 1003.2 / (25 - 1) = 1003.2 / 24 = 41.8
Step 6: Calculate the standard deviation.
Standard deviation = √(Variance) ≈ √(41.8) ≈ 6.5 (rounded to one decimal place)
Therefore, the standard deviation of the test marks is approximately 6.5 (to one decimal place).
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Consider random variables (X, Y ) with joint p.d.f.
fX,Y (x, y) = 1/3 x ≥ 0, y ≥ 0, 2x + 3y ≤ 6
0 otherwise.
(a) Let W = X + Y . Compute FW (w) and fw(w).
(b) Compute E[W] and V ar[W].
(c) Let Z = Y − X. What are the minimum and maximum of Z?
(d) Write FZ(z) in terms of double integral on x and y. You want to consider two separate cases for w ≥ 0 and w < 0.
(e) Find fZ(z).
(f) Compute E[Z] and V ar[Z].
(a) To compute the cumulative distribution function (CDF) of W, denoted as FW(w), we integrate the joint probability density function (PDF) over the appropriate region. The region is defined by the inequalities x ≥ 0, y ≥ 0, and 2x + 3y ≤ 6. The CDF is given by: FW(w) = P(W ≤ w) = ∫∫[fX,Y(x, y)] dy dx
To find the PDF fw(w), we differentiate FW(w) with respect to w.
(b) To compute E[W], we integrate the product of w and the PDF fw(w) over the range of W. The variance V ar[W] is calculated by finding E[W^2] and subtracting (E[W])^2.
(c) To find the minimum and maximum values of Z, we need to determine the range of Y - X. We consider the range of x and y that satisfy the given conditions. By substituting the limits of x and y, we can calculate the minimum and maximum values of Z.
(d) The cumulative distribution function FZ(z) can be written as a double integral over the joint PDF fX,Y(x, y). We consider two cases: w ≥ 0 and w < 0. For each case, we determine the appropriate region and integrate the PDF accordingly.
(e) To find the PDF fZ(z), we differentiate FZ(z) with respect to z.
(f) To calculate E[Z], we integrate the product of z and the PDF fZ(z) over the range of Z. The variance V ar[Z] is computed by finding E[Z^2] and subtracting (E[Z])^2.
Please note that without the specific range or shape of the region defined by the inequalities, it is not possible to provide detailed numerical calculations for each part.
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the heights of mature pecan trees are approximately normally distributes with a mean of 42 feet and a standard deviation of 7.5 feet. what proportion are between 43 and 46 feet tall.
The proportion of mature pecan trees between 43 and 46 feet tall can be calculated using the normal distribution with a mean of 42 feet and a standard deviation of 7.5 feet.
To find the proportion, we need to calculate the z-scores corresponding to the given heights and then find the area under the normal curve between those z-scores.
First, we calculate the z-score for 43 feet:
z1 = (43 - 42) / 7.5 = 0.1333
Next, we calculate the z-score for 46 feet:
z2 = (46 - 42) / 7.5 = 0.5333
Using a standard normal distribution table or a calculator, we can find the area between these two z-scores. The area corresponds to the proportion of trees between 43 and 46 feet tall.
The explanation would involve using a standard normal distribution table or a calculator to find the area under the normal curve between the z-scores of 0.1333 and 0.5333. This area represents the proportion of mature pecan trees between 43 and 46 feet tall.
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Find the distance between the points (3, -8) and (8,4).
Answer:
13
Step-by-step explanation:
Answer:
13
Step-by-step explanation:
Hello There!
We can calculate the distance between two points using the distance formula
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
where the x and y values are derived from the coordinates in which you are trying to find the distance between
The points we need to find the distance between is (3,-8) and (8,4)
So we plug the x values and y values into the formula
[tex]d=\sqrt{(8-3)^2+(4-(-8)^2} \\8-3=5\\4-(-8)=12\\d=\sqrt{5^2+12^2} \\5^2=25\\12^2=144\\144+25=169\\\sqrt{169} =13[/tex]
so we can conclude that the distance between the two points is 13 units
Please help me with this I am stuck
Answer:
450 cm ^3
Step-by-step explanation:
In R3, for the vectors , V1 := (1, 2, -3), V2 := (2,0,–2), V3 := (1,1,–2), and w := (2,3, -5), = find all possible representations of w as a linear combination of V1, V2, V3?
Two possible representations of vector w as a linear combination of V1, V2, and V3 are: w = 2V1 + V2 + 3V3 and w = -3V1 + 2V2 - 4V3.
The vector w can be represented as a linear combination of V1, V2, and V3 in the following ways:
w = 2V1 + V2 + 3V3
w = -3V1 + 2V2 - 4V3
Explanation:
To find the possible representations of w as a linear combination of V1, V2, and V3, we need to determine the coefficients that satisfy the equation w = aV1 + bV2 + cV3, where a, b, and c are scalars.
We can set up a system of equations to solve for a, b, and c. Using the given vectors and coefficients, we get:
2a + 2b + c = 2
a + c = 3
-3a + 2b - 4c = -5
By solving this system of equations, we find the values of a, b, and c that satisfy the equation. In this case, we have two possible solutions: a = 2, b = 1, and c = 3 for the first representation, and a = -3, b = 2, and c = -4 for the second representation.
Therefore, the possible representations of w as a linear combination of V1, V2, and V3 are w = 2V1 + V2 + 3V3 and w = -3V1 + 2V2 - 4V3.
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Find the area of the triangle below.
Answer:
20.25
Step-by-step explanation:
Just multiply the base and height then divide the results in half.
Hope this helps!
We formally define the length function f(w) of a string w = ww2...W), (where ne N, and Vi= 1,..., n W; € 2) as 1. if w = €, then f(w) = 0. 2. if w = au for some a € and some string u over 2, then f(x) = 1 + f(u). 1, ..., Prove using proof by induction: For any strings w = wW2...Wy. (where n € N, and Vi W: € 9), f(w) = n.
The length function f(w) of a string w = w₁w₂...W), where n ∈ N and Vi ∈ W: € 9, is equal to n.
The length function f(w) is defined recursively based on the structure of the string w. In the base case, if w is an empty string (ε), the length is defined as 0. In the recursive case, if w can be written as au, where a is a character from the alphabet and u is a string over the alphabet Σ, then the length is defined as 1 plus the length of u.
To prove that for any string w =w₁w₂...wy, where n ∈ N and Vi ∈ W: € 9, the length function f(w) is equal to n, we will use a proof by induction.
Base case:For w = ε (an empty string), we have f(ε) = 0, which satisfies the condition when n = 0.
Inductive step:Assume that for any string w = w₁w₂...wn, where n ∈ N and Vi ∈ W: € 9, the length function f(w) = n.
Now, consider a string w' = w₁w₂...wn+1. By the recursive definition, we can write w' as au, where a is the last character wn+1 and u is the string w₁w₂...wn. From our assumption, we know that f(u) = n.
Therefore, f(w') = 1 + f(u) = 1 + n = n + 1.
Since we have established that for any string w = w₁w₂...wy, where n ∈ N and Vi ∈ W: € 9, the length function f(w) = n, we can conclude that f(w) = n.
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I work in quality control for a company and I need to compare two processes our company is using. I sample the results of 100 runs for each process and find that for Process A the average is 277 (standard deviation is 9.2), while for process B the average is 274 (standard deviation is 8).
What is the mean difference (1 decimal place)?
The mean difference between Process A and Process B is 3.0 (rounded to 1 decimal place).
To calculate the mean difference between two processes, we subtract the average of Process B from the average of Process A.
Mean difference = Average of Process A - Average of Process B
Mean difference = 277 - 274 = 3.0
Therefore, the mean difference between Process A and Process B is 3.0.
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An instructor gives four 1-hour exams and one final exam, which counts as three 1-hour exams. Find a student's grade if she received 65, 84, 98, and 91 on the 1-hour exams and 82 on the final exam.
The student's grade is approximately 83.43.
To calculate the student's grade, we need to consider the weight of each exam. The four 1-hour exams are worth 1 hour each, and the final exam is equivalent to three 1-hour exams.
Let's break down the calculation step by step:
Calculate the sum of the 1-hour exams:
65 + 84 + 98 + 91 = 338
Calculate the weighted sum of the exams by multiplying the sum of the 1-hour exams by 1 (since each 1-hour exam has a weight of 1):
Weighted sum of 1-hour exams = 338×1 = 338
Calculate the weighted score for the final exam by multiplying the final exam score (82) by 3 (since it counts as three 1-hour exams):
Weighted score for the final exam = 82× 3 = 246
Add the weighted sum of the 1-hour exams and the weighted score for the final exam to obtain the total weighted sum:
Total weighted sum = Weighted sum of 1-hour exams + Weighted score for the final exam
= 338 + 246 = 584
Calculate the total weight of all the exams by summing the individual weights:
Total weight = Weight of 1-hour exams + Weight of the final exam
= 4 + 3 = 7
Finally, calculate the student's grade by dividing the total weighted sum by the total weight:
Student's grade = Total weighted sum / Total weight
= 584 / 7 ≈ 83.43
Therefore, the student's grade is approximately 83.43.
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Your game publishing company is looking to find out how much lower-income users spend each month on average on a game's in-app purchases so that they can make marketing decisions. You managed to collect 404 valid survey responses and calculate the mean monthly purchase amount as $5.93. From past studies of this nature, you know that the standard deviation is $1.05. Construct a 99% confidence interval to estimate the average monthly purchases of lower-income players. The lower limit is A The upper limit is B Enter an answer.
The 99% confidence interval for estimating the average monthly purchases of lower-income players is A = $5.74 and B = $6.12.
To construct a confidence interval, we can use the formula:
Confidence interval = sample mean ± (critical value * standard error)
Given that the sample size is 404, the mean monthly purchase amount is $5.93, and the standard deviation is $1.05, we can calculate the standard error using the formula:
Standard error = standard deviation / √(sample size)
Substituting the values, we find that the standard error is approximately $0.052.
To determine the critical value, we need to consider the desired confidence level. For a 99% confidence interval, we have 1 - (0.99) = 0.01, and dividing this by 2 gives us 0.005 for each tail. Consulting a t-distribution table or using a statistical software, we find that the critical value is approximately 2.626.
Substituting the values into the confidence interval formula, we get:
Confidence interval = $5.93 ± (2.626 * $0.052)
Calculating the lower and upper limits, we find that A = $5.74 and B = $6.12. Therefore, we can estimate with 99% confidence that the average monthly purchases of lower-income players fall between $5.74 and $6.12.
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(a) Find to.005 when v= 19.
(b) Find to.10 when v= 14.
(c) Find to 975 when v = 20.
Click here to view page 1 of the table of critical values of the t-distribution.
Click here to view page 2 of the table of critical values of the t-distribution.
(a) to.005 = ___ (Round to three decimal places as needed.)
(a) We get the result as:
to.005 = 2.539
(b) The required value is to.
10 = 1.345
(c) From the distribution we get:
to.975 = 2.086.
Given: v=19, α= 0.005
For finding to.005 when v= 19, we need to follow the below steps:
The t-distribution table has two tails and it is symmetric about the mean.
So, the area in one tail is (α/2), and in the second tail is also (α/2).
Step 1: First of all we need to find the row of the t-distribution table and this will be equal to the degree of freedom (v) which is given to be 19.
In this case, we will find the value in row 19 in the table of critical values of the t-distribution which is shown below:
Step 2: Now, look for the value of α at the top of the table (at 0.005).
Step 3: Since the table is showing the area in the right-hand tail, the value of to.005 will be a positive value.
Therefore, we have to use the positive row of the table and for this, we can find the to.005 by looking at the intersection of row 19 and the column corresponding to α=0.005.
Therefore, to.005 = 2.539 (approximately) (Rounded to three decimal places)
Hence, the correct option is to.005 = 2.539
(b) v=14, α= 0.10
For finding to.10 when v= 14, we need to follow the same steps that we followed in part (a).
The table of critical values of the t-distribution is shown below:
Step 1: Find the row corresponding to the v=14 in the t-distribution table.
Step 2: Look for the α=0.10 at the top of the table.
Since the area in one tail is (α/2) which is equal to 0.05, therefore we need to find the critical values that will cut off the top 5% of the curve.
Step 3: Since the table is showing the area in the right-hand tail, the value of to.10 will be a positive value.
Therefore, we have to use the positive row of the table and for this, we can find the to.10 by looking at the intersection of row 14 and the column corresponding to α=0.10 .
Therefore, to.10 = 1.345 (approximately) (Rounded to three decimal places)
Hence, the correct option is to.10 = 1.345
(c) v = 20, α = 0.025
For finding to.025 when v= 20, we need to follow the same steps that we followed in part (a).
The table of critical values of the t-distribution is shown below:
Step 1: Find the row corresponding to the v=20 in the t-distribution table.
Step 2: Look for the α=0.025 at the top of the table.
Since the area in one tail is (α/2) which is equal to 0.0125, therefore we need to find the critical values that will cut off the top 1.25% of the curve.
Step 3: Since the table is showing the area in the right-hand tail, the value of to.975 will be a positive value.
Therefore, we have to use the positive row of the table and for this, we can find the to.975 by looking at the intersection of row 20 and the column corresponding to α=0.025 .
Therefore, to.025 = 2.086 (approximately) (Rounded to three decimal places)
Hence, the correct option is to.975 = 2.086.
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Solve the problem. The function D(h) = 5e^-0.4h can be used to determine the milligrams D of a certain drug in a patient's bloodstream h hours after the drug has been given. How many milligrams (to two decimals) will be present after 9 hours? a. 182.99 mg b. 0.14 mg c. 1.22 mg d. 3.35 mg
B. 0.14 mg will be present after 9 hours.
The given function is D(h) = 5e^(-0.4h), which can be used to determine the milligrams D of a certain drug in a patient's bloodstream h hours after the drug has been given.
To find the milligrams after 9 hours, we need to plug in h = 9 in the function D(h) = 5e^(-0.4h).
D(h) = 5e^(-0.4h)
D(9) = 5e^(-0.4(9))
D(9) = 5e^(-3.6)
D(9) = 5 × 0.024419
D(9) = 0.1220 ≈ 0.12 mg
Hence, the answer is option (c) 1.22 mg.
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ok soooo
when you rrly think abt it we all have kicked a pregnant lady
Answer:
Step-by-step explanation:
lol true
Assume that a randomly selected subject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1. Draw a graph and find Upper P9, the 9th percentile. This is the bone density score separating the bottom 9% from the top 91%.
The bone density score corresponding to the 9th percentile, separating the bottom 9% from the top 91%, is approximately -1.34.
To find this value, we can refer to the standard normal distribution table or use a statistical calculator. The standard normal distribution has a mean of 0 and a standard deviation of 1. The area to the left of any given z-score represents the cumulative probability up to that point.
In this case, since we want to find the 9th percentile, we are interested in the value that separates the bottom 9% (cumulative probability) from the top 91%. This means that we need to find the z-score that corresponds to an area of 0.09 under the curve.
By referencing the standard normal distribution table or using a calculator, we find that the z-score corresponding to a cumulative probability of 0.09 is approximately -1.34. This z-score represents the bone density score at the 9th percentile, separating the bottom 9% from the top 91%.
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An article in Transactions of the Institution of Chemical Engineers (1956, Vol. 34, pp. 280-293) reported data from an experiment investigating the effect of several process variables on the vapor phase oxidation of naphthalene. A sample of the percentage mole conversion of naphthalene to maleic anhydride follows: 4.2, 4.7,4.7, 5.0, 3.8, 3.6, 3.0, 5.1, 3.1, 3.8, 48, 4.0, 5.2, 4.3, 2.8, 2.0, 2.8, 3.3, 4.8, 5.0. a. (5 points) Calculate the sample mean, sample variance, and sample standard deviation.
The "sample-mean" is 4, the sample-variance is 0.876, and the sample standard-deviation is 0.935.
To calculate the sample-mean, sample variance, and sample standard deviation, we use the formulas:
Sample mean (x') = (sum of all values)/(number of values)
Sample variance (s²) = [(sum of (each value - sample mean)²/(number of values - 1)],
Sample standard deviation (s) = √(sample variance),
Given the data: 4.2, 4.7, 4.7, 5.0, 3.8, 3.6, 3.0, 5.1, 3.1, 3.8, 4.8, 4.0, 5.2, 4.3, 2.8, 2.0, 2.8, 3.3, 4.8, 5.0.
⇒ Calculate the sample-mean (x'):
x' = (4.2 + 4.7 + 4.7 + 5.0 + 3.8 + 3.6 + 3.0 + 5.1 + 3.1 + 3.8 + 4.8 + 4.0 + 5.2 + 4.3 + 2.8 + 2.0 + 2.8 + 3.3 + 4.8 + 5.0) / 20
x' ≈ 80/20 = 4
So, mean is 4,
⇒ Calculate the sample-variance (s²):
s² = [(4.2 - 4)² + (4.7 - 4)² + (4.7 - 4)² + (5.0 - 4)² + (3.8 - 4)² + (3.6 - 4)² + (3.0 - 4)² + (5.1 - 4)² + (3.1 - 4)² + (3.8 - 4)² + (4.8 - 4)² + (4.0 - 4)² + (5.2 - 4)² + (4.3 - 4)² + (2.8 - 4)² + (2.0 - 4)² + (2.8 - 4)² + (3.3 - 4)² + (4.8 - 4)² + (5.0 - 4)²]/(20-1),
s² = [0.04 + 0.49 + 0.49 + 1.0 + 0.16 + 0.64 + 1.0 + 1.21 + 0.81 + 0.16 + 0.64 + 0.0 + 1.44 + 0.09 + 1.44 + 4.0 + 1.44 + 0.49 + 0.64 + 1.0]/19,
s² = 16.64/19 ≈ 0.876,
⇒ Calculate the sample standard-deviation (s):
s = √(s²)
s ≈ 0.935
Therefore, the standard-deviation is 0.935.
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The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 7 8 cm. Find the probability that an individual distance is greater than 214.
The probability that an individual distance is greater than 214 is 0.4554.
The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 78 cm. To find the probability that an individual distance is greater than 214, we need to use the standard normal distribution, which is a normal distribution with a mean of 0 and a standard deviation of 1. To do this, we will first calculate the z-score for 214:z = (214 - 205) / 78z = 0.1154Then, we will use a standard normal distribution table or calculator to find the probability of a z-score greater than 0.1154. The area to the right of the z-score is the probability of an individual distance being greater than 214.Using a standard normal distribution table, we find that the probability of a z-score greater than 0.1154 is 0.4554. Therefore, the probability that an individual distance is greater than 214 is 0.4554.
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Given that the overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 78 cm. We need to find the probability that an individual distance is greater than 214. So, we have to standardize the given value 214 as follows:
Z = (X - μ)/σZ = (214 - 205)/78Z = 0.1154
We have to find the probability that an individual distance is greater than 214.
So, P(X > 214) = P(Z > 0.1154)
Using the standard normal distribution table, the area under the curve to the right of Z = 0.1154 is 0.4573.Approximately, the probability that an individual distance is greater than 214 is 0.4573.Hence, the correct option is (a) 0.4573.
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The area of an unknown shape is represented by the expression 1672 - 104x + 169. If X represents an integer, what shape could this figure be? b) What is the smallest possible area of the figure? c) What is the smallest possible perimeter of the figure? 6. The length of the base of a rectangular prism is 2 m more than its width, and the height of the prism is 15 m. a) Write an algebraic expression for the volume of the rectangular prism. b) If the volume of the prism is 2145 m, write an equation to model the situation. c) Solve the equation by factoring. What are the dimensions of the base of the rectangular prism? d) Make up a similar problem complete with a solution. 7. Jacinta pilots a small plane. Last weekend she was flying at an altitude of 1500 m parallel to the ground at a horizontal distance of 4000 m from the beginning of the landing strip. Jacinta flew in to land at a constant angle of depression. a) What was the angle of depression, to the nearest tenth of a degree? b) How far did she fly before touching down, rounded to the nearest metre? 8. An isosceles triangle has a base of 22 cm and the angle opposite the base measuring 36º. Find the perimeter of the triangle to the nearest tenth of a centimetre.
1) The area of an unknown shape is represented by the expression 1672 - 104x + 169. a) The expression represents a quadratic equation of the form ax² + bx + c. Thus, the unknown shape could be a quadratic shape. b) The smallest possible area of the figure is obtained when the value of x is at the vertex of the quadratic equation, and the formula to find the vertex is -b/2a. On substituting the values, we get the smallest possible area to be 1665 sq units. c) The perimeter of the shape can not be found from the given expression.
2) The length of the base of a rectangular prism is 2 m more than its width, and the height of the prism is 15 m. a) The algebraic expression for the volume of the rectangular prism is V = lwh, where l represents the length of the base, w represents the width of the base, and h represents the height of the prism. Thus, V = (w + 2)wh + 15m³. b) On substituting the given value of the volume, we get the quadratic equation 0 = w² + 2w - 429. c) Factoring the quadratic equation, we get the dimensions of the base of the rectangular prism to be 21 m × 11 m. d) Find the dimensions of a rectangular prism with height 8 m, and volume 864 m³.
3) Jacinta pilots a small plane. Last weekend she was flying at an altitude of 1500 m parallel to the ground at a horizontal distance of 4000 m from the beginning of the landing strip. Jacinta flew in to land at a constant angle of depression. a) The angle of depression can be calculated using the formula tanθ = opposite/adjacent. Thus, tanθ = 1500/4000, giving θ ≈ 20.6°. b) Using trigonometry, we can find the distance Jacinta flew before touching down to be 4201 m.
4) An isosceles triangle has a base of 22 cm and the angle opposite the base measuring 36º. Find the perimeter of the triangle to the nearest tenth of a centimetre.The perimeter of the isosceles triangle can be found using the formula P = 2a + b, where a is the length of the two equal sides and b is the length of the base. Using trigonometry, we can find the length of the equal sides to be a ≈ 16.9 cm. Thus, the perimeter of the triangle is P ≈ 56.8 cm.
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Suppose data collected by observers at randomly selected intersections across the country revealed that in a sample of 100 drivers, 30 were using their cell phone. a. Give a point estimate of the true driver cell phone use rate that is, the true proportion-or-population porportion of drivers who are using their cell phone while driving). b. Computea 90% confidence interval for c. Give a practical interpretation of the interval, part b.
a. the point estimate of the true driver cell phone use rate is 0.3 or 30%.
b. the 90% confidence interval for the true driver cell phone use rate is approximately (21.5%, 38.5%).
c. The practical interpretation of the confidence interval is that we are 90% confident that the true driver cell phone use rate falls within the range of 21.5% to 38.5%
a. The point estimate of the true driver cell phone use rate (population proportion) can be calculated by dividing the number of drivers using their cell phone by the total sample size. In this case, the sample size is 100, and 30 drivers were using their cell phone.
Point estimate = Number of drivers using their cell phone / Total sample size
Point estimate = 30/100 = 0.3 (or 30%)
Therefore, the point estimate of the true driver cell phone use rate is 0.3 or 30%.
b. To compute a 90% confidence interval for the true driver cell phone use rate, we can use the formula for a confidence interval for a proportion. The formula is:
Confidence interval = Point estimate ± (Critical value × Standard error)
The critical value depends on the desired level of confidence. For a 90% confidence interval, the critical value is typically obtained from the standard normal distribution and is approximately 1.645.
The standard error can be calculated using the formula:
Standard error = sqrt((point estimate * (1 - point estimate)) / sample size)
In this case, the point estimate is 0.3, and the sample size is 100.
Standard error = sqrt((0.3 * (1 - 0.3)) / 100) ≈ 0.048
Plugging in the values, we can calculate the confidence interval:
Confidence interval = 0.3 ± (1.645 * 0.048)
Confidence interval = (0.215, 0.385)
Therefore, the 90% confidence interval for the true driver cell phone use rate is approximately (21.5%, 38.5%).
c. The practical interpretation of the confidence interval is that we are 90% confident that the true driver cell phone use rate falls within the range of 21.5% to 38.5%. This means that based on the sample data, we can estimate with 90% confidence that the proportion of drivers using their cell phone while driving in the entire population lies between these two percentages. It provides a range of likely values for the true population proportion.
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The sequence (a_n) is defined recursively by a_1 = - 36, a_n+1 = a-n/2 + 72/a-n 1) Find the term a_3 of this sequence. a3 = _________
2) Prove by induction that for all n ∈ N, a_n < 0.
1) The term a_3 of this sequence a3 = -362/37.2
2) By the principle of mathematical induction, for all n ∈ N, aₙ < 0.
1) We are given the recursive formula:
a₁ = -36, aₙ₊₁ = aₙ/2 + 72/aₙ.
We need to find the term a₃ of this sequence. a₂ is given by the recursive formula as:
a₂ = a₁/2 + 72/a₁a₂ = -36/2 + 72/(-36) = -37/2
a₃ is given by the recursive formula as:
a₃ = a₂/2 + 72/a₂= (-37/2)/2 + 72/(-37/2)= -74/37 + (-288/37) = -362/37
Therefore, a₃ = -362/37.2
2) We need to prove by induction that for all n ∈ N, aₙ < 0.
Base case:
For n = 1, we have a₁ = -36 < 0. So, the base case is true.
Inductive step:
Let's assume that for some arbitrary n = k, aₖ < 0.
We need to show that aₖ₊₁ < 0.
Using the recursive formula: aₖ₊₁ = aₖ/2 + 72/aₖ
Since aₖ < 0, -aₖ > 0 and aₖ/2 < 0.Hence, aₖ/2 + 72/aₖ < 0
Therefore, aₖ₊₁ < 0.So, the statement that for all n ∈ N, aₙ < 0 is true for n = 1 and if it's true for n = k, then it's true for n = k + 1.
Therefore, by the principle of mathematical induction, for all n ∈ N, aₙ < 0.
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What is the distance to the nearest tenth A unit, between point M (-8,-1) and point N (3,5)?
Answer: 12.5
Step-by-step explanation:
what is the approximate percentage of a 10c sample left after the time it took a to walk one lap around the gym, where 5 laps takes 200 seconds
The approximate percentage of a 10c sample left after the time it took to walk one lap around the gym is 100 - 25c.
Let x be the time it takes to walk one lap around the gym.
We know that 5 laps take 200 seconds.
Therefore, x can be found by dividing 200 by 5:
x = 200/5 = 40 seconds.
Now, let's find the percentage of the sample left after walking one lap around the gym.
Since x is the time it takes to walk one lap around the gym, we know that the sample decreases at a rate of 10c/x per second.
Therefore, after x seconds, the percentage of the sample remaining is given by: 100(1 - 10c/x)
Substituting x = 40, we get:
100(1 - 10c/40) = 100(1 - 0.25c) = 100 - 25c
So the approximate percentage of a 10c sample left after the time it took to walk one lap around the gym is 100 - 25c.
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