$19,574 at 7.5% for 3 1/4 years

The weights (to the nearest pound) of some boxes to be shipped are found to be:. Weight 65 68 69 70 71 72 90 95 frequency 1 2 5

Lena [83]

So the mean is 72.97

We need to subtract the mean from each value and square it.
(65-72.97)^2= 63.5209
(68-72.97)^2=24.7009
(69-72.97)^2=15.7609
(70-72.97)^2=8.8209
(71-72.97)^2= 3.8809
(72-72.97)^2=0.9409
(90-72.97)^2=290.0209
(95-72.97)^2=485.3209

Now we add up the new values ( also consider their frequency) and find their mean.
Add the values
63.5209+(2 •24.7009=49.4018)+(5•15.7609=78.8045)+(8•8.8209=70.5672)+(7•3.8809=27.1663)+(3•0.9409=2.8227)+(2•290.0209=580.0418)+(2•485.3209=970.6418)= 1,842.967
Divide by total numburs to find the mean
1,842.967/ 30=61.43223333

The standar deviation is the square root of the mean so is
Square root of 61.43223333=7.837871735
Round to the nearest tenth
Standard Deviation is 7.8

7 0

3 years ago

What is the solution to the system of equations below? y = negative one-third x + 6 and x = –6

PSYCHO15rus [73]

Answer:2

26x + y = 23 → y = 23 – 6x.

7x + y = 25 → 7x + (23 – 6x) = 25 → x + 23 = 25 → x = 2

4 0

3 years ago

What are the angles of rotation for a 20-gon?

Basile [38]

There are 360 degrees so divide 360 by 20 and get 18 degrees

4 0

3 years ago

Read 2 more answers

What is the solution of the proportion? 6/a = 18/27

Marta_Voda [28]

Answer:

a=9

Step-by-step explanation:

To solve this proportion, we have to get the variable, a, by itself.

First, cross multiply.

6/a=18/27

Multiply the denominator of the first fraction by the numerator of the second, and the numerator of the second by the denominator of the first.

a*18=6*27

18a=162

Now, 18 and a are being multiplied. In order to get a by itself, perform the opposite of what is being done. They are being multiplied, so the opposite would be division. Divide both sides by 18.

18a/18=162/18

a=162/18

a=9

So, the proportion, with 9 substituted in for a, will be:

6/9=18/27

6 0

3 years ago

Read 2 more answers

(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

3 years ago