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2 years ago

PLEASE HELP ME!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

jekas [21]2 years ago

4 0

Answer:

smal and blurry

Step-by-step explanation:

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(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

7 0

2 years ago

The price of an item yesterday was 130. Today, the price fell to 91 . Find the percentage decrease.

Shalnov [3]

Answer:

30%

Step-by-step explanation:

130-91=39

39 / 130 =0.3

0.3x100=30

and you got the answer!

7 0

2 years ago

Read 2 more answers

Solve for a. 1/5(25−5a)=4−a

aalyn [17]

1/5(25-5a)=4-a
⇔5-a=4-a
<span>NO SOLUTION</span>

3 0

3 years ago

Mona solved an equation incorrectly, as shown below:

Vesnalui [34]

Answer:

She did not add 5 to 30. ... Mona subtracted 5 where she should have divided 5.

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Watch help video Given mn, find the value of x. (6x-3)⁰ (2x+13)°

Ilia_Sergeevich [38]

Answer:

x = 4

Step-by-step explanation:

(6x - 3)° = (2x + 13)° are vertical angles that equal each other get x by itself and solve for x.

(6x - 3)° = (2x + 13)°

6x - 2x = 13 + 3

4x = 16

x = 4

Substitute x for the angles.

(6x - 3)° = ?

(6*4 - 3)° = ?

(24 - 3)° = 21°

(2x + 13)° = ?

(2*4 + 13)° = ?

(8 + 13)° = 21°

8 0

1 year ago