MATHEMATICS COLLEGE

A lamina occupies the disk x squared space plus space y squared space less or equal than 4. Find the mass, if the density at any point is proportional to its distance from the origin, where the proportionality constant is 3.

Answers

Answer 1

Answer:

Answer: Mass of Lamina is (K/3)

Centre of mass is (3/8, 3pi/16)

Step-by-step explanation:

Find explanation in the attachments

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Suppose that is in standard position and the given point is on the terminal side of 0. Give the exact value of the indicated trig function for 0. 15) (-5, 12); Find sin

Answers

Answer:

12/13

Step-by-step explanation:

The length of the segment from the origin to the terminal point is ...

r = √((-5)² +12²) = √169 = 13

The sine of the angle is the ratio of the y-coordinate to this distance

sin(θ) = y/r = 12/13

_____

Additional comment

The other trig functions are ...

cos(θ) = x/r = -5/13

tan(θ) = y/x = -12/5

This is a 2nd-quadrant angle, where the sine is positive, but the cosine and tangent are negative.

Suppose you wanted to estimate the difference between two population means correct to within 4.8 at the 92% confidence level. If prior information suggests that both population variances are approximately equal to 12 and you want to select independent random samples of equal size from the populations, how large should the sample sizes be?Critical Value: 1.75
The sample sizes should be: n1=___n2=_____?

Answers

Answer: $n_1=n_2=4$

Step-by-step explanation:

Given : Margin of error : E= 4.8

Confidence level : 92%

Significance level : $1-0.92=0.08$

$\sigma_1^2=\sigma_1^2\approx12$

Two-tailed critical value :-

$z_(\alpha/2)=z_(0.08/2)=z_(0.04)=1.75$

If we want to select independent random samples of equal size from the populations,

Formula for the sample size :

$n_1=n_2=((z_(\alpha/2))/(E))^2(\sigma_1^2+\sigma_2^2)$

Then buy using given values , we have

$n_1=n_2=((1.75)/(4.8))^2(12+12)$

Simplify ,

$n_1=n_2=((1.75)/(4.8))^2(12+12)=3.190\approx4$ [Round to the next integer.]

Hence, the The sample sizes should be: $n_1=n_2=4$

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Answers

Answer:

I’ll neve switch on u dxxdy

Step-by-step explanation:

In a survey of 1016 ?adults, a polling agency? asked, "When you? retire, do you think you will have enough money to live comfortably or not. Of the 1016 ?surveyed, 535 stated that they were worried about having enough money to live comfortably in retirement. Construct a 99?% confidence interval for the proportion of adults who are worried about having enough money to live comfortably in retirement.A. There is a 99?% probability that the true proportion of worried adults is between ___ and ___.

B. 99?% of the population lies in the interval between ___ and ___.

C. There is 99?% confidence that the proportion of worried adults is between ___ and ___.

Answers

Answer:

C. There is 99% confidence that the proportion of worried adults is between 0.487 and 0.567

Step-by-step explanation:

1) Data given and notation

n=1016 represent the random sample taken

X=535 represent the people stated that they were worried about having enough money to live comfortably in retirement

$\hat p=(535)/(1016)=0.527$ estimated proportion of people stated that they were worried about having enough money to live comfortably in retirement

$\alpha=0.01$ represent the significance level

Confidence =0.99 or 99%

z would represent the statistic

p= population proportion of people stated that they were worried about having enough money to live comfortably in retirement

2) Confidence interval

The confidence interval would be given by this formula

$\hat p \pm z_(\alpha/2) \sqrt{(\hat p(1-\hat p))/(n)}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_(\alpha/2)=2.58$

And replacing into the confidence interval formula we got:

$0.527 - 2.58 \sqrt{(0.527(1-0.527))/(1016)}=0.487$

$0.527 + 2.58 \sqrt{(0.527(1-0.527))/(1016)}=0.567$

And the 99% confidence interval would be given (0.487;0.567).

There is 99% confidence that the proportion of worried adults is between 0.487 and 0.567

Final answer:

To build a 99% confidence interval, we first calculate our sample proportion by dividing the number of such instances by the total sample size. Next, we determine the standard error of the proportion, then our margin of error by multiplying the standard error by the Z value of the selected confidence level. Lastly, we determine the confidence interval by adding and subtracting the margin of error from the sample proportion.

Explanation:

To construct a 99% confidence interval for the proportion of adults worried about having enough money to live comfortably in retirement, we will utilize statistical methods and proportions. First, we must calculate the sample proportion. The sample proportion (p) is equal to 535 (the number who are worried) divided by 1016 (the total number of adults surveyed).

Then, we find the standard error of the proportion which we get by multiplying the square root of ((p*(1-p))/n) where n is the number of adults sampled. The margin of error is found using the Z value corresponding to the desired confidence level, in this case, 99%. Multiply the standard error by this Z value. Lastly, we construct the confidence interval by taking the sample proportion (p) ± the margin of error.

The result will give you the 99% confidence interval - meaning we are 99% confident that the true proportion of adults who are worried about having enough money to live comfortably in retirement lies within this interval.

Learn more about Confidence Interval here:

brainly.com/question/34700241

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3. Kirk bought a bag of candy and took 10pieces. He split the rest evenly among 12
friends. Each friend received 5 pieces. Letc
represent the number of pieces in a bag.
Equation:
Solve it to find how many pieces of candy were in the bag.
Type here
Show your work
Write and solve the equation

Answers

70, Multiply 5 pieces by 12 friends and you get 60, then add the 10 pieces he originally took out, and you get 70

A cylindrical can without a top is made to contain 25 3 cm of liquid. What are the dimensions of the can that will minimize the cost to make the can if the metal for the sides will cost $1.25 per 2 cm and the metal for the bottom will cost $2.00 per 2 cm ?