a. Point estimate for p using the method of moments: P = 1/X, b. MLE estimator for p: P = X, obtained by maximizing the likelihood function. c. The MLE estimator of p is unbiased since E(P) = p, where p is the true population parameter for a geometric distribution.
a. In the method of moments, we equate the sample moments to the corresponding population moments to obtain the point estimate. For a geometric distribution, the population mean is μ = 1/p. Equating this with the sample mean (X), we get the point estimate for p as
P = 1/X.
b. The maximum likelihood estimator (MLE) for p can be obtained by maximizing the likelihood function. For a geometric distribution, the likelihood function is
L(p) =[tex](1-p)^{X1-1} . (1-p)^{X2-1}. ... . (1-p)^{Xn-1} . p^n.[/tex]
Taking the logarithm of the likelihood function, we get
ln(L(p)) = Σ(Xi-1)ln(1-p) + nln(p).
To find the MLE, we differentiate ln(L(p)) with respect to p, set it equal to zero, and solve for p. The MLE estimator for p is P = X.
c. To determine if the MLE estimator of p is unbiased, we need to calculate the expected value of P and check if it equals the true population parameter p. Taking the expectation of P,
E(P) = E(X) = p
(since the sample mean of a geometric distribution is equal to the population mean). Therefore, the MLE estimator of p is unbiased, as
E(P) = p.
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Please answer correctly! I will mark you Brainliest!
Answer:
V=385 cubic units
Step-by-step explanation:
The volume of a rectangular prism is given by the formula [tex]V=lwh[/tex], where l is the length, w is the width, and h is the height. Our dimensions are 5, 7, and 11. So, we have to multiply them to find the volume.
5 × 7 × 11 = 385
Thus, the volume of the rectangular prism is 385 cubic units.
what would be the distance between (-35,20) and (15,20).
Answer:
Well you just have to follow the formula to find the distance between 2 linear plots.
d= √(x2−x1)^2+(y2−y1)^2
Step-by-step explanation:
If you follow it the answer would be
D=50
I need help . Pretend they are labeled a-e
Answer:
a definitely, and maybe e, im not sure
Step-by-step explanation:
let me know if it was right
Based on the following, should a one-tailed or two- tailed test be used? Họ: H = 17,500 Ha: # 17,500 V = 18,000 s= 3000 n = 10
The required correct answer is a two-tailed test should be used based on the following data
Explanation:
To determine whether a one-tailed or two-tailed test should be used based on the following data:
H0: H = 17,500Ha: # 17,500V = 18,000s = 3000n = 10 We must first examine the alternative hypothesis (Ha) to determine whether it is directional (one-tailed) or non-directional (two-tailed).
A directional alternative hypothesis, or a one-tailed test, is a hypothesis that predicts the direction of the difference between the sample mean and the population mean. Ha: < 17,500 or Ha: > 17,500 are examples of a directional hypothesis.
A non-directional alternative hypothesis, or a two-tailed test, is a hypothesis that does not predict the direction of the difference between the sample mean and the population mean.
Ha: ≠ 17,500 is an example of a non-directional hypothesis.Since Ha: # 17,500 is not directional and does not predict the direction of the difference between the sample mean and the population mean, a two-tailed test is required.
Therefore, a two-tailed test should be used based on the following data.
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The change in water level of a lake is modeled by a polynomial function, W(x). Describe how to find the x-intercepts of W(x) and how to construct a rough graph of W(x) so that the Parks Department can predict when there will be no change in the water level. You may create a sample polynomial of degree 3 or higher to use in your explanations.
(Dividing and Solving Polynomials)
First. Finding the x-intercepts of
Let be the change in water level. So to find the x-intercepts of this function we can use The Rational Zero Test that states:
To find the zeros of the polynomial:
We use the Trial-and-Error Method which states that a factor of the constant term:
can be a zero of a polynomial (the x-intercepts).
So let's use an example: Suppose you have the following polynomial:
where the constant term is . The possible zeros are the factors of this term, that is:
.
Thus:
From the foregoing, we can affirm that are zeros of the polynomial.
Second. Construction a rough graph of
Given that this is a polynomial, then the function is continuous. To graph it we set the roots on the coordinate system. We take the interval:
and compute where is a real number between -2 and -1. If , the curve start rising, if not, the curve start falling. For instance:
Therefore the curve start falling and it goes up and down until and from this point it rises without a bound as shown in the figure below
He got a job selling cell phone services. He will be compensated $20 each hour he
works. He also gets a job at a competing cell phone store. His deal is that he will earn
$10 each hour, BUT he will start off with a bonus of $50 at the start of each day.
1) create equations for each man's compensation
2) After how many hours will they earn the same amount?
3) in an 8 hour day, who will earn more? How much more
Answer:
I think you should go with 3
Eat your spinach: Six measurements were made of the mineral content (in percent) of spinach, with the following results. It is reasonable to assume that the population is approximately normal. 19.1 20.8 20.8 21.4 20.5 19.7 a. Construct a 95% confidence interval for the mean mineral content. b. Based on the confidence interval, is it reasonable to believe that the mean mineral content of spinach may be greater than 21%? Explain.
a. The 95% confidence interval for the mean mineral content of spinach can be calculated using the sample data. The formula for the confidence interval is:
Confidence interval = (sample mean) ± (critical value) * (standard deviation / sqrt(sample size))
Using the given data, the sample mean is 20.63 and the standard deviation is 0.676. The critical value for a 95% confidence level can be obtained from the t-distribution table for a sample size of 6 (n-1 degrees of freedom). Calculating the confidence interval using these values gives a range of approximately 19.97% to 21.29%.
b. Based on the 95% confidence interval, the range of the mean mineral content of spinach is between 19.97% and 21.29%. Since this range includes the value of 21%, it is reasonable to believe that the mean mineral content of spinach may be greater than 21%. However, we cannot be certain as the range also includes values below 21%. A larger sample size or narrower confidence interval would provide more precise information about the true mean.
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Consider a population consisting of the following five values, which represent the number of video downloads during the academic year for each of five housemates. 9 15 18 11 12 (a) Compute the mean of this population. H = 13 (b) Select a random sample of size 2 by writing the five numbers in this population on slips of paper, mixing them, and then selecting two. Calculate the mean for your sample. (c) Repeatedly select random samples of size 2, and calculate the x value for each sample until you have the values for 25 samples. Describe your results. This answer has not been graded yet. (d) Construct a density histogram using the 25 x values. Are most of the values near the population mean? Do the values differ a lot from sample to sample, or do they tend to be similar?
(a) Mean = (9 + 15 + 18 + 11 + 12) / 5 = 65 / 5 = 13
(b) Mean of the sample = (9 + 15) / 2 = 24 / 2 = 12
(c) The means of the samples vary, but they tend to be close to the population mean of 13.
(d) Since the problem does not provide the values for the 25 x values, we cannot construct a density histogram or determine if most of the values are near the population mean.
(a) The mean of the population is calculated by summing all the values and dividing by the total number of values:
Mean = (9 + 15 + 18 + 11 + 12) / 5 = 65 / 5 = 13
(b) To calculate the mean for a random sample of size 2, we randomly select two values from the population and calculate their mean:
Random sample: 9, 15
Mean of the sample = (9 + 15) / 2 = 24 / 2 = 12
(c) Repeatedly selecting random samples of size 2 and calculating the mean for each sample:
Here are the means for 25 random samples of size 2 (selected without replacement):
Sample 1: 9, 18 -> Mean = (9 + 18) / 2 = 27 / 2 = 13.5
Sample 2: 11, 9 -> Mean = (11 + 9) / 2 = 20 / 2 = 10
Sample 3: 15, 12 -> Mean = (15 + 12) / 2 = 27 / 2 = 13.5
...
Sample 25: 12, 15 -> Mean = (12 + 15) / 2 = 27 / 2 = 13.5
The means of the samples vary, but they tend to be close to the population mean of 13.
(d) Constructing a density histogram using the 25 x values:
Since the problem does not provide the values for the 25 x values, we cannot construct a density histogram or determine if most of the values are near the population mean.
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What is the area of the triangle with vertices at (1,1), (3,4) and (5,2)?
7 square units
10 square units
14 square units
5 square units
can someone please help me answer these?
Answer: w+x=90
z=y+90
Step-by-step explanation: because the sum of the inner angles of triangle is 180. given that one angle is 90 degrees the sun of x and w has to be 180-90=90
for the second part, you have to use the triangle exterior angle theorem: an exterior angle of a triangle is equal to the sum of the opposite interior angles. The opposite interior angles in this case is 90 degrees and y
(f+g)(2)=
(f-g)(2)=
(fg)(2)=
Answer:
(f + g)(2) = 16
(f - g)(2) = 0
(fg)(2) = 64
Step-by-step explanation:
f(2) = 8 g(2) = 8
(f + g)(2) = f(2) + g(2) = 8 + 8 = 16
(f - g)(2) = f(2) - g(2) = 8 - 8 = 0
(fg)(2) = f(2)g(2) = 8(8) = 64
Use the Laplace transform to solve the initial-value problem x" + 4 = f(t), x(0)=0, x'(0) = 0, if t < 5 f(t) = t25. 3 sin(t-5) if t > 5.
By applying the initial conditions and inverse Laplace transforming, we can obtain the solution x(t) = (1 - cos(2t))u(t-5) + (3 sin(t-5))u(t-5), where u(t) is the unit step function. Therefore, the solution to the initial-value problem is x(t) = (1 - cos(2t))u(t-5) + (3 sin(t-5))u(t-5)
Taking the Laplace transform of the given differential equation x" + 4 = f(t), we obtain the algebraic equation in the Laplace domain: s^2X(s) + 4sX(s) + 4 = F(s), where X(s) is the Laplace transform of x(t) and F(s) is the Laplace transform of f(t).
Next, applying the initial conditions x(0) = 0 and x'(0) = 0, we get X(0) = 0 and sX(0) = 0. Substituting these initial conditions into the Laplace domain equation, we have s^2X(s) + 4sX(s) + 4 = F(s), with X(0) = 0 and sX(0) = 0.
Now, let's consider the Laplace transform of f(t) using the given piecewise function. For [tex]t < 5, f(t) = t^2/5, and for t > 5, f(t) = 3sin(t-5).[/tex]Taking the Laplace transform of f(t) in each interval, we have [tex]F(s) = (1/s^3) + (3/s^2) for t < 5 and F(s) = (3/s^2) * (1/(s^2+1)) for t > 5.[/tex]
Substituting these Laplace transforms into the equation[tex]s^2X(s) + 4sX(s) +[/tex]4 = F(s), we can solve for X(s). Simplifying, we obtain [tex]X(s) = (1/s^3) + (3/s^2) / (s^2 + 4s + 4) + (3/s^2) * (1/(s^2+1)).[/tex]
To find the inverse Laplace transform of X(s), we can split it into partial fractions and apply the inverse Laplace transform formula. The solution is x(t) = (1 - cos(2t))u(t-5) + (3 sin(t-5))u(t-5), where u(t) is the unit step function.
Therefore, the solution to the initial-value problem is x(t) = (1 - cos(2t))u(t-5) + (3 sin(t-5))u(t-5), where u(t) is the unit step function that ensures the piecewise function is activated at t = 5.
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A survey was conducted that asked 967 people how many books they had read in the past year Results indicated that x = 14.8 books and s-16.5 books. Construct a 98confidence Interval for the mean number of books people read. Interpret the interval Construct a 98% confidence interval for the mean number of books people road and interpret the result Select the correct choice below and fill in the answer boxes to complete your choice. (Use ascending order. Round to two decimal places as needed) O A repeated samples are taken, 98% of them will have a sample mean between and OB. There is 98% confidence that the population mean number of books road is between and OC. There is a 98% probability that the true mean number of books road in between and
A 98% confidence interval for the mean number of books people read is (13.76, 15.84).
The given data is: x = 14.8 books, s = 16.5 books, n = 967. Here, we need to find the 98% confidence interval for the mean number of books people read. Now, we know that the confidence interval can be calculated by the formula:
CI = x ± Z_(α/2) (σ/√n), where Z_(α/2) is the z-score value at α/2 level of significance.
Let's calculate the Z_(α/2):
α = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01
`The area in the right tail will be: 0.01 + 0.98 = 0.99 from the z-table.
Looking at the z-table, the corresponding z-score for 0.99 will be "2.33". So, Z_(α/2) = 2.33
Putting all the values in the formula: CI = 14.8 ± 2.33 (16.5/√967)
Calculating this: CI = 14.8 ± 1.96 (0.53)
So, CI = (13.76, 15.84)
Hence, a 98% confidence interval for the mean number of books people read is (13.76, 15.84). The correct choice is: There is 98% confidence that the population mean number of books road is between 13.76 and 15.84.
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Arletta built a cardboard ramp
for her little brother’s toy cars.
Find the volume of this ramp.
Answer:
Volume = 525 in³
Step-by-step explanation:
Volume = 25 x 7 x 6 x 0.5 = 525 in³
With which information can you construct more than one triangle?
A the measurements of two angles
B the measurements of two angles and the length of the included side
C the measurements of all the angles
D the lengths of two sides and the measurement of the included angle
Answer:
B and D
Explanation:
use the limit comparison test to determine whether the series converges. k^2 2/k^3-7
The series ∑[(k² + 2)/(k³ - 7)] diverges using the limit comparison test and the series ∑(1/k) is a well-known harmonic series.
To determine the convergence of the series ∑[(k² + 2)/(k³ - 7)], we can use the limit comparison test. Let's compare it with the series ∑(1/k).
First, we need to find the limit of the ratio of the terms of the two series as k approaches infinity:
lim(k→∞) [(k² + 2)/(k³ - 7)] / (1/k)
Simplifying the expression, we get:
lim(k→∞) [(k² + 2)/(k³ - 7)] × (k/1)
Taking the limit as k approaches infinity, we have:
lim(k→∞) [(k² + 2)/(k³ - 7)] × (k/1) = 1
Since the limit is a finite positive value (1), we can conclude that the given series ∑[(k² + 2)/(k³ - 7)] and the series ∑(1/k) have the same convergence behavior.
The series ∑(1/k) is a well-known harmonic series, which diverges. Therefore, by the limit comparison test, the given series ∑[(k² + 2)/(k³ - 7)] also diverges.
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The question is -
Use the limit comparison test to determine whether the series converges. k² + 2/ k³ - 7
On the first day of the fiscal year, a company issues a $3,000,000, 11%, five-year bond that pays semiannual interest of $165,000 ($3,000,000 × 11% × ½), receiving cash of $2,889,599.
Journalize the first interest payment and the amortization of the related bond discount. Round to the nearest dollar. If an amount box does not require an entry, leave it blank.
On the first day of the fiscal year, a company issued a $3,000,000, 11%, five-year bond that pays semiannual interest of $165,000 ($3,000,000 × 11% × ½), receiving cash of $2,889,599.The journal entries are as follows: July 1Cash Dr2895499Discount on Bonds Payable Dr 10501Bond Payable Cr 3,000,000To record issuance of bond July 31
Interest Expense Dr 165001 Discount on Bonds Payable Dr8751Cash Cr173250To record interest payment ($3,000,000 * 11% * 6/12) - $10501 = $165001 - $8751 = $173,250 December 31 Interest Expense Dr 181501 Discount on Bonds Payable Dr 8581Cash Cr 173250To record interest payment ($3,000,000 * 11% * 6/12) - $7670 = $181501 - $8581 = $173,250Amortization of discount for the first interest period is $10,501 ($173,250 - $165,000). The total discount of $10501 is amortized over the life of the bond, and it will be amortized over 10 interest periods ($10501/10) = $1,050 per period. The journal entry for the bond discount amortization would be:July 31Interest Expense Dr165001Discount on Bonds Payable Dr8751Bond Discount Amortization Dr1050Cash Cr173250The following journal entry for bond discount amortization is on December 31:Interest Expense Dr181501Discount on Bonds Payable Dr8581Bond Discount Amortization Dr1198Cash Cr173250This process will continue until the end of the bond life.
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The complete journal entry for the first interest payment and bond discount amortization would be:
Interest expense Dr. $165,000
Bond discount amortization Dr. $5,520
Cash Cr. $168,520 (rounded to the nearest dollar)
Firstly, let's determine the amount of bond discount.
Bond discount is the difference between the face value of a bond and the amount at which it is sold. Here, the company received cash of $2,889,599, whereas the face value of the bond is $3,000,000.
So, the bond discount is: Face value of bond - Cash received
= $3,000,000 - $2,889,599
= $110,401
Now, let's journalize the first interest payment and the amortization of the related bond discount. The bond has a semi-annual interest rate of 11%, so the first interest payment is: $3,000,000 × 11% × 1/2
= $165,000
The journal entries would be: Interest expense Dr. $165,000
Bond discount amortization Dr. $2,920 Cash Cr. $168,920 (rounded to the nearest dollar)
The bond discount amortization is calculated using the straight-line method. The total bond discount is $110,401, and it is amortized over the term of the bond, which is 5 years or 10 semi-annual periods.
So, the bond discount amortization per period is:
Total bond discount / Total number of periods= $110,401 / 10
= $11,040.10 (rounded to the nearest cent)
This amount is amortized each period, along with the interest payment. So, the bond discount amortization for the first period is: $11,040.10 / 2
= $5,520.05 (rounded to the nearest cent)
Therefore, the complete journal entry for the first interest payment and bond discount amortization would be:
Interest expense Dr. $165,000
Bond discount amortization Dr. $5,520 Cash Cr. $168,520 (rounded to the nearest dollar)
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Can someone help me?!?!
Answer:
b, d, and f are 70°. a, c, g, and e are 110°
Step-by-step explanation:
1. Find the equation of the parabola satisfying the given conditions.
Focus: (3,6); Directrix: x=−1
A. (x−1)2=8(y−6)
B. (y−6)2=8(x−1)
C. (x−1)2=−8(y−6)
D. (y−6)2=−8(x−1)
2. Find the equation of the parabola satisfying the given conditions.
Focus: (−6,3); Directrix: y=1
A. (y−2)2=4(x+6)
B. (x+6)2=4(y−2)
C. (x+6)2=−4(y−2)
D. (y−2)2=−4(x+6)
3. Find the equation of an ellipse that has foci at (−1,0) and (4,0), where the sum of the distances between each point on the ellipse and the two foci is 9.
A. (x+1)2+y2−−−−−−−−−−−√+(x−4)2+y2−−−−−−−−−−−√=9
B. (x−1)2+y2−−−−−−−−−−−√+(x+4)2+y2−−−−−−−−−−−√=9
C. (x+1)2+y2−−−−−−−−−−−√+(x−4)2+y2−−−−−−−−−−−√=81
D. (x−1)2+y2−−−−−−−−−−−√+(x+4)2+y2−−−−−−−−−−−√=81
4. Find the equation of a hyperbola that has foci at (−1,0) and (4,0), where the difference of the distances between each point on the ellipse and the two foci is 5.
A. (x+1)2+y2−−−−−−−−−−−√−(x−4)2+y2−−−−−−−−−−−√=25
B. (x−1)2+y2−−−−−−−−−−−√−(x+4)2+y2−−−−−−−−−−−√=5
C. (x−1)2+y2−−−−−−−−−−−√−(x+4)2+y2−−−−−−−−−−−√=25
D. (x+1)2+y2−−−−−−−−−−−√−(x−4)2+y2−−−−−−−−−−−√=5
Answer:
CABD
Step-by-step explanation:
Can someone plz help me with the answer
Step-by-step explanation:
[tex] \frac{12 \times 8}{2} + {( \frac{8}{2}) }^{2} \pi = \\ = 48 + 16\pi[/tex]
98.24
A sphere has a diameter of 4(x+3) centimeters and a surface area of
784 square centimeters. Find the value of x.
Answer:
78 square centimeters
Step-by-step explanation:
A chocolate muffin recipe serves 12 people. An oatmeal raisin cookie recipe serves 36 people. A lemon cake recipe serves 16 people. Each recipe needs 2 eggs. The muffins need 3 cups of flour, while the cookies need 2 cups. The cake uses 2 lemons and the cookies use 1 cup of raisins and oatmeal. Each recipe needs 1 cup of sugar and milk. If we need to make cookies for 48 people, how much flour is needed?
To make cookies for 48 people, the recipe requires 6 cups of flour.
The oatmeal raisin cookie recipe serves 36 people and requires 2 eggs and 2 cups of flour. Since we need to make cookies for 48 people, we can calculate the amount of flour required as follows:
36 people → 2 cups of flour
1 person → (2 cups of flour) / (36 people) = (1/18) cups of flour
To make cookies for 48 people:
48 people × (1/18) cups of flour = 2.67 cups of flour
Therefore, to make cookies for 48 people, we need approximately 2.67 cups of flour.
Note: Since the result is a fraction of a cup, it is advisable to round up to the nearest whole number, so in this case, 3 cups of flour would be needed.
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Assume that the playbook contains 16 passing plays and 12 running plays. The coach randomly selects 8 plays from the playbook. What is the probability that the coach selects at least 3 passing plays and at least 2 running plays?
The probability that the coach selects at least 3 passing plays and at least 2 running plays out of 8 plays from the playbook is approximately 0.4914 or 49.14%. This means there is a 49.14% chance of the coach choosing a combination that meets the given criteria.
To calculate the probability of the coach selecting at least 3 passing plays and at least 2 running plays out of 8 plays, we need to consider different combinations that satisfy these conditions.
1: Determine the total number of possible combinations of 8 plays from a playbook of 28 plays (16 passing plays + 12 running plays).
Total Combinations = C(28, 8) = 28! / (8! * (28-8)!) = 3,395,685
2: Calculate the number of combinations that have at least 3 passing plays and at least 2 running plays.
First, we calculate the number of combinations with exactly 3 passing plays and 2 running plays:
Number of Combinations with 3 passing and 2 running = C(16, 3) * C(12, 2) = (16! / (3! * (16-3)!) * (12! / (2! * (12-2)!) = 560 * 66 = 36,960
Next, we calculate the number of combinations with exactly 4 passing plays and 2 running plays:
Number of Combinations with 4 passing and 2 running = C(16, 4) * C(12, 2) = (16! / (4! * (16-4)!) * (12! / (2! * (12-2)!) = 1,820 * 66 = 120,120
Finally, we calculate the number of combinations with 5 passing plays and at least 2 running plays:
Number of Combinations with 5 passing and 2 or more running = C(16, 5) * (C(12, 2) + C(12, 3) + C(12, 4) + C(12, 5) + C(12, 6) + C(12, 7) + C(12, 8)) = (16! / (5! * (16-5)!) * (C(12, 2) + C(12, 3) + C(12, 4) + C(12, 5) + C(12, 6) + C(12, 7) + C(12, 8)) = 4368 * (66 + 220 + 495 + 792 + 924 + 792 + 495) = 4368 * 3786 = 16,530,048
Total Number of Combinations with at least 3 passing and 2 running plays = Number of Combinations with 3 passing and 2 running + Number of Combinations with 4 passing and 2 running + Number of Combinations with 5 passing and 2 or more running = 36,960 + 120,120 + 16,530,048 = 16,687,128
3: Calculate the probability.
Probability = (Number of Combinations with at least 3 passing and 2 running plays) / (Total Combinations) = 16,687,128 / 3,395,685 ≈ 0.4914
Therefore, the probability that the coach selects at least 3 passing plays and at least 2 running plays out of 8 plays is approximately 0.4914 or 49.14%.
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What is the Quartile 1 for the Box & Whisker Plot below?
PLSS HELP
Answer:
17
Step-by-step explanation:
The lower quartile Q₁ is positioned at the left side of the box.
The value at the left is Q₁ is 17
please give me the ans
Answer:
how to use a c a l c u l a t o r
Step-by-step explanation:
1. g o o g l e
2. add parenthesis if needed
hurry !!!! I need help
Answer:
x = 66
Step-by-step explanation:
m<5 + m<6 = 180
2x + 48 = 180
2x = 132
x = 66
1. Study Activity 1 on p.301. Then complete the following using the Sampling Distribution
of a Sample Proportion web app.
i. Simulate taking a random sample of 100 voters from a large population of voters of
whom 54% voted for Brown, and record the number out of 100 that voted for Brown.
ii. Report the proportion of your sample that voted for Brown
iii. Insert below the Data Distribution generated by the web app.
According to the question the following using the Sampling Distribution are as follows :
1. Study Activity 1 on p.301. Then complete the following using the Sampling Distribution of a Sample Proportion web app.
i. Simulate taking a random sample of 100 voters from a large population of voters of whom 54\% voted for Brown, and record the number out of 100 that voted for Brown.
ii. Report the proportion of your sample that voted for Brown
iii. Insert below the Data Distribution generated by the web app.
i. Simulate taking a random sample of 100 voters from a large population of voters of whom 54\% voted for Brown, and record the number out of 100 that voted for Brown.
ii. Report the proportion of your sample that voted for Brown: [Insert the proportion value here]
iii. Insert below the Data Distribution generated by the web app:
[Insert the data distribution plot here]
Note: The actual values for the proportion and data distribution will vary based on the simulation results obtained from the Sampling Distribution of a Sample Proportion web app.
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Gerald and Wheatly, Applied Numerical Analysis ▶6. If e¹.3 is approximated by Lagrangian interpolation from the values for eº = 1, el = 2.7183, and e² = 7.3891, what are the minimum and maximum estimates for the error? Compare to the actual error.
Lagrangian interpolation is used to approximate the value of e¹.3 using three known values: eº = 1, el = 2.7183, and e² = 7.3891. We can find the minimum and maximum estimates for the error.
To approximate e¹.3 using Lagrangian interpolation, we construct a polynomial that passes through the three given points: (0, 1), (1, 2.7183), and (2, 7.3891). We can then evaluate this polynomial at x = 1.3 to estimate the value of e¹.3.
Using Lagrangian interpolation, the polynomial P(x) is given by:
P(x) = 1 * L₀(x) + 2.7183 * L₁(x) + 7.3891 * L₂(x),
where L₀(x), L₁(x), and L₂(x) are the Lagrange basis polynomials associated with the three data points.
To find the minimum and maximum estimates for the error, we need to determine the upper bound for the error term in the Lagrangian interpolation formula. The error term is given by:
E(x) = f(x) - P(x),
where f(x) is the actual function we are approximating (in this case, e^x).
To find the upper bound for the error, we can use the maximum value of the absolute value of the n+1st derivative of f(x) in the interval containing the data points.
By calculating the upper bound for the error, we can compare it to the actual error by evaluating the actual function e¹.3 and subtracting the approximation P(1.3) obtained from Lagrangian interpolation.
By analyzing the error estimates and comparing them to the actual error, we can assess the accuracy of the Lagrangian interpolation approximation in this particular case.
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HELPPPPPPPPPPPPPPP!!!!!!!
Answer:
I feel like the answers are 10 and -2
Step-by-step explanation:
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suppose 40% of adults in the u.s. say they get their financial advice from family members. a random sample of 8 adults is selected. what is the probability at least 5 of the 8 say they get their financial advice from family members?
The probability that at least 5 of the 8 adults say they get their financial advice from family members can be calculated using binomial probability distribution.
Formula for binomial probability distribution is:P(X=k) = nCk * pk * (1-p)n-kwhere, P(X=k) is the probability of k successes in n independent trials, p is the probability of success in one trial, q=1-p is the probability of failure in one trial, nCk is the combination of k successes in n independent trials.
In this case, the probability of success is p=0.4 as 40% of adults in the US say they get their financial advice from family members.
Therefore, the probability of failure is q=1-0.4=0.6.So, P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)P(X=k) = nCk * pk * (1-p)n-kwhere, n=8, p=0.4, q=0.6For k = 5, nCk = 8C5P(X = 5) = 8C5 * (0.4)5 * (0.6)3= 0.2787
For k = 6, nCk = 8C6P(X = 6) = 8C6 * (0.4)6 * (0.6)2= 0.1960For k = 7, nCk = 8C7P(X = 7) = 8C7 * (0.4)7 * (0.6)1= 0.0575For k = 8, nCk = 8C8P(X = 8) = 8C8 * (0.4)8 * (0.6)0= 0.0030Therefore,P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.2787 + 0.1960 + 0.0575 + 0.0030 = 0.5352
Hence, the probability that at least 5 of the 8 adults say they get their financial advice from family members is 0.5352.
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Given information that 40% of adults in the US say they get their financial advice from family members. We have to find the probability of at least 5 of the 8 say they get their financial advice from family members.
Hence, the required probability is 0.7530.
The probability of getting financial advice from family members is 40%. Let X be the number of people out of 8, who get their financial advice from family members. Here, X follows a binomial distribution with parameters n = 8 and p = 0.4. The probability of getting atleast 5 people getting their financial advice from family members is
P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
Using binomial distribution formula, we get
P(X = x) = ${n\choose x}p^xq^{n-x}$
Where, n = 8, p = 0.4, q = 0.6
The probability of getting exactly 5 people out of 8 getting their financial advice from family members
P(X = 5) = ${8\choose 5} 0.4^5 (0.6)^{8-5}$
= 0.27869
The probability of getting exactly 6 people out of 8 getting their financial advice from family members
P(X = 6) = ${8\choose 6} 0.4^6 (0.6)^{8-6}$
= 0.29360
The probability of getting exactly 7 people out of 8 getting their financial advice from family members
P(X = 7) = ${8\choose 7} 0.4^7 (0.6)^{8-7}$
= 0.16493
The probability of getting exactly 8 people out of 8 getting their financial advice from family members
P(X = 8) = ${8\choose 8} 0.4^8 (0.6)^{8-8}$
= 0.01678
Therefore, the probability of getting at least 5 people getting their financial advice from family members is
P(X ≥ 5) = 0.27869 + 0.29360 + 0.16493 + 0.01678
= 0.7530
Hence, the required probability is 0.7530.
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