8. A statement of an employee's biweekly earnings is given below.

I need help ASAP. Thank you (:

Anarel [89]

Answer: volume = 1282.2 ft³

Step-by-step explanation:

The composite figure consists of a cone and a cylinder. The volume of the composite figure is

Volume of cylinder + volume of cone.

The formula for determining the volume of a cylinder is expressed as

Volume = πr²h

Where

r represents the radius of the cylinder.

h represents the height of the cylinder.

π is a constant whose value is 3.14

From the information given,

Radius = 7 ft

Height = 6 ft

Volume = 3.14 × 7² × 6 = 923.16ft³

The formula for determining the volume of a cone is expressed as

Volume = 1/3πr²h

From the information given,

Radius = 7 ft

Height = 7ft

Volume = 1/3 × 3.14 × 7² × 7 = 359ft³

Volume of composite solid =

923.16 + 359 = 1282.2 ft³

3 0

3 years ago

Suppose we take a random sample of 41 state college students. Then we measure the length of their right foot in centimeters. We

Debora [2.8K]

Answer:

$ME = \frac{25.09-21.01}{2}= 1.69$

The general formula for the margin of error is given by:

$ME= t_{\alpha/2} \frac{s}{\sqrt{n}}$

And for this case the width is:

$Width= 2*t_{\alpha/2} \frac{s}{\sqrt{n}}$

And if we decrease the confidence level from 95% to 90% then the critical value $t_{\alpha/2}$ would decrease and in effect the width for this new confidence interval decreases.

As confidence level decreases, the interval width decreases

Step-by-step explanation:

For this cae we know that the sample size selected is n =41

And we have a confidence interva for the true mean of foot length for students at a college selected.

The confidence interval is given by this formula:

$\bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}$

And for this case the 95% confidence interval is given by: (21.71,25.09)

A point of etimate for the true mean is given by:

$\bar X = \frac{21.71+25.09}{2}= 23.4$

And the margin of error would be:

$ME = \frac{25.09-21.01}{2}= 1.69$

The general formula for the margin of error is given by:

$ME= t_{\alpha/2} \frac{s}{\sqrt{n}}$

And for this case the width is:

$Width= 2*t_{\alpha/2} \frac{s}{\sqrt{n}}$

And if we decrease the confidence level from 95% to 90% then the critical value $t_{\alpha/2}$ would decrease and in effect the width for this new confidence interval decreases.

As confidence level decreases, the interval width decreases

3 0

3 years ago

What is the 8th term 28, 33 ,38, 48

Orlov [11]

Do you have a typo?
Besides that, is this problem a multiple choice question

5 0

3 years ago

Read 2 more answers

Emma buys packets of beads

Serhud [2]

Answer: