Find the mean of the following probability distribution? Round your answer to one decimal. P(2) 0 0.0017 1 0.3421 2 0.065 3 0.4106 4 0.1806 mean = ___

Answer:

The removed numbers are 13 and 19, and the product is:

13*19 = 247

Step-by-step explanation:

We have the set:

{9, 13, 15, 17, 19, 23, 31, 49}

The original median is the number that is just in the middle of the set (in a set of 8 numbers, we take the average between the fourth and fifth numbers)

then the median is:

(17 + 19)/2 = 18

and the mean is:

(9 + 13 + 15 + 17 + 19 + 23 + 31 + 49)/8 = 22

We want to remove two numbers such that the mean and the median increase by two.

Is immediate to notice that if we want the median to increase by two, we need to remove the number 19 and one number smaller than 17.

Then the median will be equal to:

(17 + 23)/2 = 20

which is 2 more than the previous median.

because 19 assume that we remove the 19 and number N.

To find the value of N, we can solve for the new mean:

((9 + 13 + 15 + 17  + 23 + 31 + 49 - N)/6 = 22 + 2

(this means that if we remove the number 19 and the number N, the mean increases by 2.

(9 + 13 + 15 + 17  + 23 + 31 + 49 - N)/6 = 22 + 2

(9 + 13 + 15 + 17  + 23 + 31 + 49 - N) = 24*6 = 144

157 - N = 144

157 - 144 = N = 13

This means that the other number we need to remove is 13

Then we remove the numbers 13 and 19

The product of the two removed numbers is:

13*19  =247

Answer:

247

Step-by-step explanation:

The average of the original numbers is 176/8 = 22. The median of the original numbers is the average of the middle two numbers: 17 + 19/2 = 18.

Thus, after removing two numbers, we should obtain a list of six numbers whose average is 24 and whose median is 20.

For the median of six numbers to be 20, the middle two numbers in that list must add up to 40. Searching our original list for pairs of numbers that add up to 40, we find two such pairs: 9, 31 and 17, 23. But 9 and 32 can't be the middle numbers after we remove two numbers, so 17 and 23 must be the middle numbers. This tells us that one of the removed numbers must be 19.

For the average of six numbers to be 24, the sum of the six numbers must be 6 ∙ 24 = 144. This is 32 less than the original sum of 176, so if one of the removed numbers is 19, the other must be 32 - 19 = 13.

Therefore, the product of the two removed numbers is 13 ∙ 19 = 247.

Answer:

Hi Bunni  ,  :D

Step-by-step explanation:

a =2

b= -8

c = 9

axis of symmetry  is

-(-8) / 2(2)

8 / 4

2

vertex = ( 2, 1)  

:)

Given function is `f(x) = √√√x³ + x² + x + 1`. Now, we are going to find out the first derivative of f(x).f(x) = √√√x³ + x² + x + 1

Therefore, `(f¹)' (2) = 693 / 16√√√21`This is how the value of `(f¹)' (2)` for the function `f(x) = √√√x³ + x² + x + 1` is evaluated.

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Answer:

a) x= -5

b) x= 3

c) x= -5

Answer:

24,27,30 and 33 and so on

Step-by-step explanation:

The first 5 terms of this sequence represented by f(n) = f(n − 1) + f(n − 2). is 1, 3, 2, -1 and -3

A function is an expression that shows the relationship between two or more variables and numbers.

Given the function:

f(n) = f(n − 1) + f(n − 2)

f(1) = 1, f(2) = 3

f(3) = f(2) - f(1) = 3 - 1 = 2

f(4) = f(3) - f(2) = 2 - 3 = -1

f(5) = f(4) - f(3) = -1 - 2 = -3

The first 5 terms of this sequence represented by f(n) = f(n − 1) + f(n − 2). is 1, 3, 2, -1 and -3

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Answer:

48in^2

Step-by-step explanation:

The area of a triangle is calculated by the formula A= b*h1/2. Substitute b for 24, h for 4, and solve.

Answer:

H and J

Step-by-step explanation:

Answer:

You need to find the radius first,

Half of 23 is = 11.5

Formula you must remember when finding area of circle is:

πr^2

(pie (3.14 or 22/7) x radius squared)

Our pie is 3.14 because they said to use it in this particular question.

3.14 x 11.5^2 (remember we are making our radius squared) <--- Before

Let's evaluate first:

3.14 x 132.25 <--- After evaluating

So, 3.14 x 132.25 = 415.265 <----- your answer

                               _______

Answer:

U scammer scaming peoiple ans wasting their points!

Step-by-step explanation:

The coordinate matrix of x in the basis B' is [4, -1, 3, 2].

To find the coordinate matrix of x in the basis B', we need to express x as a linear combination of the basis vectors in B'.

Let's denote the coordinate matrix of x in the basis B' as Xb'. It will have the form:

Xb' = [a1]

     [a2]

     [a3]

     [a4]

To find the values of a1, a2, a3, and a4, we solve the equation:

x = a1 * (1, -1, 2, 1) + a2 * (1, 1, -4, 3) + a3 * (1, 2, 0, 3) + a4 * (1, 2, -2, 0)

Expanding the equation, we get:

(8, 9, -12, 2) = (a1 + a2 + a3 + a4, -a1 + a2 + 2a3 + 2a4, 2a1 - 4a2, a1 + 3a2 + 3a3)

Equating the corresponding components, we have the following system of equations:

a1 + a2 + a3 + a4 = 8 ...(1)

-a1 + a2 + 2a3 + 2a4 = 9 ...(2)

2a1 - 4a2 = -12 ...(3)

a1 + 3a2 + 3a3 = 2 ...(4)

To solve this system of equations, we can represent it in matrix form:

| 1 1 1 1 | | a1 | | 8 |

| -1 1 2 2 | * | a2 | = | 9 |

| 2 -4 0 0 | | a3 | | -12 |

| 1 3 3 0 | | a4 | | 2 |

We can solve this matrix equation to find the values of a1, a2, a3, and a4.

Solving the matrix equation, we find:

a1 = 4

a2 = -1

a3 = 3

a4 = 2

Therefore, the coordinate matrix of x in the basis B' is:

Xb' = [4]

[-1]

[3]

[2]

Hence, Xb' = [[4], [-1], [3], [2]].

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a) The probability that both cards are Kings, if the drawing is done with replacement is 1/169. b) The probability that both cards are hearts, if the drawing is done without replacement is 3/52.

a) If the drawing is done with replacement, then the probability of drawing a King is 4/52 = 1/13. Since there are 4 Kings in the deck, the probability of drawing two Kings is:

P(King and then King) = P(King) × P(King) = (1/13) × (1/13) = 1/169

b) If the drawing is done without replacement, then the probability of drawing a heart is 13/52 = 1/4. Since there are 13 hearts in the deck, the probability of drawing a second heart after drawing the first heart is 12/51 because there are only 12 hearts left in the deck out of 51 cards remaining. So, the probability of drawing two hearts is:

P(Heart and then Heart) = P(Heart) × P(Heart|Heart was drawn first) = (1/4) × (12/51) = 3/52

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Answer:

73% = parameter

68% = statistic

Step-by-step explanation:

Parameter and statistic differs from one another in statistical parlance in that, parameter refers to nemwrical characteristic derived from the population data. In the scenario described above, 73% describes the percentage of all registered Voters (population of interest). On the other hand, statistic refers to numerical characteristic derived from the sample data. 68% represents the percentage of democrats from the sample surveyed from the the larger population.

Hence,

68% is a sample statistic and 73% is a population parameter.

If A is proper "nonempty-subset" of "connected-space" X, then boundary of A, Bd(A), is nonempty because every point in A is either an interior or exterior point.

To prove that if A is a proper nonempty subset of "connected-space" X, then boundary of A, denoted Bd(A), is nonempty, we can use a proof by contradiction.

We assume that A is proper "nonempty-subset" of "connected-space" X, and suppose, for sake of contradiction, that Bd(A) is empty,

Since Bd(A) is set of all "boundary-points" of A, the assumption that Bd(A) is empty implies that there are no boundary points in A.

If there are no boundary points in A, it means that every point in A is either an interior-point or an exterior-point of A.

Consider the sets U = A ∪ X' and V = X\A, where X' represents the set of exterior points of A. Both U and V are open sets since A is a proper nonempty subset of X.

U and V are disjoint sets that cover X, i.e., X = U ∪ V,

Since X is a connected space, the only way for X to be written as a union of two nonempty disjoint open sets is if one of them is empty. Both U and V are nonempty since A is proper and nonempty.

So, the assumption that Bd(A) is empty leads to a contradiction with the connectedness of X.

Thus, Bd(A) must be nonempty when A is a proper nonempty subset of a connected space X.

By contradiction, we have shown that if A is a proper nonempty subset of a connected space X, then the boundary of A, Bd(A), is nonempty.

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The approximate probability of obtaining at least 37 ones is -0.000005.

Given that, a fair die is tossed 180 times. A one refers to the face with one dot.

Thus, P(one) = 1/6.The probability of obtaining at least 37 ones can be approximated using the normal distribution.

Using the central limit theorem, we have,μ = n * P = 180 * (1/6) = 30 andσ² = n * P * (1 - P) = 180 * (1/6) * (1 - 1/6) = 25/6

Since we cannot use a normal distribution directly since we need to find the probability of at least 37 ones, we use a continuity correction by considering the range from 36.5 to 37.5 ones.

Now, using the standard normal distribution, we have, z = (37.5 - 30) / √(25/6) = 5.40 and z = (36.5 - 30) / √(25/6) = 4.87

Using a table of standard normal distribution, the probability corresponding to z = 5.40 and z = 4.87 are 0.0000034 and 0.0000084, respectively.Using continuity correction, the approximate probability of obtaining at least 37 ones is given by,P(X ≥ 37) ≈ P(36.5 ≤ X ≤ 37.5) = P( z ≤ 5.40) - P(z ≤ 4.87)= 0.0000034 - 0.0000084= -0.000005

Therefore, the approximate probability of obtaining at least 37 ones is -0.000005.

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The approximate probability of obtaining at least 37 ones is 0.8051.

Given that a fair die is tossed 180 times where the probability of getting a one is 1/6.

We need to find the approximate probability of obtaining at least 37 ones.

Using the binomial probability distribution, the probability of getting r successes in n independent trials can be given by:

[tex]P(r) = nCr x (p^r) x (1-p)^(n-r)[/tex]

Where n = 180, p = 1/6, q = 1- p = 5/6

We want to find the probability of getting at least 37 ones.

Therefore, we can subtract the probability of getting less than 37 ones from 1.

[tex]P(X≥37) = 1 - P(X<37) = 1 - P(X≤36)So, P(X≤36) = P(X=0) + P(X=1) + P(X=2) + .....+ P(X=36)P(X= r) = nCr x (p^r) x (q)^(n-r)[/tex]

On substituting the given values, we get:

P(X≤36) = ∑P(X = r)

n = 180

r = 0 to 36

p = 1/6, q = 5/6

[tex]P(X≤36) = ∑(nCr) × (p^r) × (q)^(n-r)[/tex]

n = 180, p = 1/6, q = 5/6= 0.1949

On subtracting it from 1, we get:

[tex]P(X≥37) = 1 - P(X<37) = 1 - P(X≤36)≈ 1 - 0.1949≈ 0.8051[/tex]

Therefore, the approximate probability of obtaining at least 37 ones is 0.8051.

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i) Writing the process {x} in backshift operator notation:

{x_t} = 3{x_{t-1}} - 1 + 57 - 2^2 + 3{-2} + 2{x_{t-2}} + 52{-1} - {-2}^2

Using the backshift operator (B), we can rewrite the process as:

{x_t} = 3B{x_t} - 1 + 57 - 2^2 + 3(-2) + 2B^2{x_t} + 52B{x_t} - (-2)^2

ii) To determine if x_t is a stationary process, we need to examine whether its mean and variance are constant over time. Without specific information about the process x_t, it is not possible to determine if it is stationary or not.

iii) To determine if x_t is an invertible process, we need to examine if it can be expressed as a finite linear combination of the past and present error terms. Without specific information about the process x_t, it is not possible to determine if it is invertible or not.

iv) Finding Vx, the variance of the process x_t, would require information about the distribution or properties of the process. Without specific information, it is not possible to calculate Vx.

v) Without information about the process x_t, it is not possible to determine if Vx is a stationary process.

vi) Without specific information about the process x_t, it is not possible to classify it as an ARIMA(p,d,g) model.

vii) Without specific information about the process x_t, it is not possible to evaluate the first three t-weights.

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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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Answer: Greater than or equal to 2.

Step-by-step explanation:

Given

Function is  [tex]f(x)=2x^2+5\sqrt{x-2}[/tex]

The entity inside a square root is always positive i.e. greater than equal to zero

[tex]\therefore \ x-2\geq 0\\\Rightarrow x\geq2[/tex]

So, the domain of the function is all real numbers greater than or equal to 2.

Answer:

I suggest method 1 because it is unbiased and systematic.  

The second method requires a lot on the initiatives of customers, and likely to have extreme cases only (liked very much or disliked very much).  

The third is biased towards teenagers, which may not be the only category of customers who ordered pizzas.

Again, the fourth requires initiative from the customer, so biased towards customers who had something to say.

Step-by-step explanation:

Answer:

What do you mean lucky?

Answer:

Yes

Step-by-step explanation:

The coefficient of determination, denoted by R², is the ratio of the explained variation to the total variation in the dependent variable, Y. R² is calculated by dividing the sum of squares of the regression by the total sum of squares.

Here, the R² from a regression of consumption on income is 0.75, which means that 75% of the variation in consumption is explained by the variation in income. The Type 1 error is an error that occurs when we reject a null hypothesis that is actually true. The level of significance in a hypothesis test is the probability of making a Type 1 error. It is the probability of rejecting the null hypothesis when it is true.

The level of significance is usually set at 0.05 or 0.01, which means that the probability of making a Type 1 error is 5% or 1%. If we set a higher level of significance, the probability of making a Type 1 error increases. If we set a lower level of significance, the probability of making a Type 1 error decreases.

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The  f(x) = [tex]e^x[/tex] and g(x) = x form a linearly independent set of functions.

To show that the functions f(x) = [tex]e^x[/tex] and g(x) = x are linearly independent, we can calculate their Wronskian and verify that it is nonzero for all values of x.

The Wronskian of two functions f(x) and g(x) is defined as the determinant of the matrix:

| f(x) g(x) |

| f'(x) g'(x) |

Let's calculate the Wronskian of f(x) = [tex]e^x[/tex] and g(x) = x:

f(x) = [tex]e^x[/tex]

f'(x) = [tex]e^x[/tex]

g(x) = x

g'(x) = 1

Now we can form the Wronskian matrix:

| [tex]e^x[/tex] x |

| [tex]e^x[/tex] 1 |

The determinant of this matrix is:

Det = ([tex]e^x[/tex] * 1) - ([tex]e^x[/tex] * x)

      = [tex]e^x[/tex] - x[tex]e^x[/tex]

      = [tex]e^x[/tex](1 - x)

We can see that the determinant of the Wronskian matrix is not zero for all values of x. Since the Wronskian is nonzero for all x, it implies that the functions f(x) = [tex]e^x[/tex]and g(x) = x are linearly independent.

Therefore, f(x) = [tex]e^x[/tex] and g(x) = x form a linearly independent set of functions.

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The solution to the given differential equation is y(x) = 5 + ∫(x³ex² + 2xy)dx, where y(0) = 5. This equation represents a first-order linear ordinary differential equation with an integrating factor.

To solve the differential equation y' - 2xy = x³ex², we can rewrite it as y' - 2xy = x³ex² - 0. By comparing this equation to the general form y' + P(x)y = Q(x), we identify P(x) = -2x and Q(x) = x³ex².

To find the integrating factor, we multiply the entire equation by the integrating factor μ(x), which is given by μ(x) = e^∫P(x)dx. In this case, μ(x) = e^∫(-2x)dx = e^(-x²).

Multiplying the given equation by μ(x), we have e^(-x²)y' - 2xey^2 = x³ex²e^(-x²). We can simplify this equation to d(e^(-x²)y)/dx = x³.

Now, we integrate both sides with respect to x: ∫d(e^(-x²)y)/dx dx = ∫x³ dx. This gives us e^(-x²)y = x⁴/4 + C, where C is the constant of integration.

Solving for y, we have y(x) = (x⁴/4 + C)e^(x²). Applying the initial condition y(0) = 5, we find that C = 5. Therefore, the solution to the differential equation is y(x) = 5 + (x⁴/4 + 5)e^(x²).

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Answer:

Coordinates of R = (1,2)

Step-by-step explanation:

Let the coordinate of R be (x, y)

Since coordinates of S and T are S(-9,-4) and T(6,5). Then we can use the Formula for length of line from coordinates to find coordinates of R since SR/RT = 2/1

Thus; 1(S) = 2(T)

Coordinates of R = [(1(-9) + 2(6))/(1 + 2)], [(1(-4) + 2(5))/(1 + 2)

Coordinates of R = (3/3), (6/3)

>> (1, 2)

Answer:

5

Step-by-step explanation:

An absolute value is NEVER negative.

|9 - 14| = |-5| = 5

Answer: 5

The probability that a randomly selected student obtained a score between 150 and 162 on the verbal part of the GRE exam is approximately 0.383, or 38.3%.

To find the probability that a student obtained a score between 150 and 162, we need to calculate the area under the normal distribution curve between these two scores. Since the distribution is normal with a mean of 160 and a standard deviation of 8, we can use the Z-score formula to standardize the scores.

First, we calculate the Z-score for a score of 150:

Z1 = (150 - 160) / 8 = -1.25

Next, we calculate the Z-score for a score of 162:

Z2 = (162 - 160) / 8 = 0.25

Using a standard normal distribution table or a statistical calculator, we can find the area under the curve between these two Z-scores. The probability of obtaining a score between 150 and 162 is equal to the area to the right of Z1 minus the area to the right of Z2.

P(150 ≤ X ≤ 162) = P(Z1 ≤ Z ≤ Z2) ≈ P(Z ≤ 0.25) - P(Z ≤ -1.25)

Looking up these values in the standard normal distribution table, we find that P(Z ≤ 0.25) is approximately 0.5987 and P(Z ≤ -1.25) is approximately 0.1056. Subtracting the second probability from the first, we get:

P(150 ≤ X ≤ 162) ≈ 0.5987 - 0.1056 ≈ 0.4931

Therefore, the probability that a randomly selected student obtained a score between 150 and 162 is approximately 0.4931, or 49.3%.

To find the score of a student who scored in the 82nd percentile, we need to find the Z-score that corresponds to the 82nd percentile. The Z-score can be found using the inverse standard normal distribution (also known as the Z-score table or a statistical calculator).

The Z-score corresponding to the 82nd percentile is approximately 0.905. We can use this Z-score to find the corresponding score on the distribution using the formula:

X = Z * σ + μ

Substituting the values, we get:

X = 0.905 * 8 + 160 ≈ 167

Therefore, the score of a student who scored in the 82nd percentile on the verbal part of the GRE is approximately 167.

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9514 1404 393

Answer:

  down: $436.50

  payment total: $2,908.08

  total paid: $3,344.58

  finance charge: $434.58

Step-by-step explanation:

A) The down payment is 15% of the cost, so is ...

  0.15 × $2910 = $436.50

__

B) The total of the 18 monthly payments of $161.56 each is ...

  18 × $161.56 = $2908.08

__

C) The total amount paid for the car is the sum of the down payment and the  monthly payments:

  $436.50 +2,908.08 = $3,344.58

__

D) The finance charge is the difference between the amount paid and the original cost of the car:

  $3,344.58 -2,910 = $434.58

Answer:

A. 6.25x + 7.50 = 26.25

Step-by-step explanation:

Clara paid 7.50 for an admission ticket, each round was 6.25 and she did it for x rounds. 6.25x is the same thing as 6.25 times x rounds.

Answer:

whered all the time go? :(

Step-by-step explanation:

Answer:

2,000(1 + 1)⁵ = 64000, or in alternate form, 40³

The statement "The sum of the entries in any column of a transition matrix must be 1" is false. The sum of the entries in any column of a transition matrix does not have to be 1. Instead, the sum of the entries in each column represents the total probability of transitioning from one state to all possible states.

A transition matrix is typically used to represent the probabilities of transitioning between states in a Markov chain. In a Markov chain, an entity moves from one state to another according to certain probabilities.

Let's consider a transition matrix T. Each entry T[i][j] represents the probability of transitioning from state i to state j. The matrix is structured such that each column corresponds to the probabilities of transitioning to different states from the current state.

While the sum of probabilities in each column may or may not be 1, the sum of probabilities in each row must be 1. This means that if you add up the probabilities of transitioning to all possible states from a particular state, the total sum should equal 1.

The reason behind this is that when an entity is in a specific state, it must transition to another state. Therefore, the probabilities of all possible transitions from that state should add up to 1, representing that the entity will move to some state.

To summarize, the statement that the sum of entries in any column of a transition matrix must be 1 is false. Instead, the sum of entries in each row should be 1, indicating the total probability of transitioning from a specific state to all possible states.

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