Square A has an area of 18a2 + 862

A circular merry-go-round has a radius of 5 feet. Find the distance of a spin of one full rotation. Use 3.14 for pi and round th

Sedbober [7]

Option C

The distance of a spin of one full rotation is 31.4 feet

Solution:

Given that,

A circular merry-go-round has a radius of 5 feet

Radius = 5 feet

We have to find the distance of spin for one full rotation

Distance covered in 1 full of rotation is equal to circumference of circle

The circumference of circle is given as:

$C = 2 \pi r$

Where "r" is the radius

$C = 2 \times 3.14 \times 5\\\\C = 6.28 \times 5\\\\C = 31.4$

Thus the distance of a spin of one full rotation is 31.4 feet

6 0

3 years ago

I really need help with this. I'd really appreciate anyone's help.

vekshin1

Answer:

I think it's G

Step-by-step explanation:

Since a=8 root 3=8×3=24cm

Yh so I multiplied the 24cm by 12cm

12cm×24cm=288cm squared

and the square root of 288cm is 96 root 3 cm squared

4 0

3 years ago

5b-b-1 > or equal too-11

vova2212 [387]

Answer:

-2.5

Step-by-step explanation:

5b - b - 1 ≥ -11

4b - 1 ≥ -11

+1 +1

4b ≥ -10

4/4 -10/4

b ≥ -2.5

4 0

3 years ago

Read 2 more answers

Can someone help me with this please?

Gnom [1K]

Answer:3:7

Step-by-step explanation:

8 0

2 years ago

Small Sample:

agasfer [191]

Answer:

The 80% confidence interval for the proportion of companies that are planning to increase their workforce is (0.033, 0.331).

A sample size of 8852 is needed.

Step-by-step explanation:

First question:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Only 2 of the 11 companies were planning to increase their workforce.

This means that $n = 11, \pi = \frac{2}{11} = 0.1818$

80% confidence level

So $\alpha = 0.2$ , z is the value of Z that has a pvalue of $1 - \frac{0.2}{2} = 0.9$ , so $Z = 1.28$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1818 - 1.28\sqrt{\frac{0.1818*0.8182}{11}} = 0.033$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1818 + 1.28\sqrt{\frac{0.1818*0.8182}{11}} = 0.331$

The 80% confidence interval for the proportion of companies that are planning to increase their workforce is (0.033, 0.331).

Second question:

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

A poll taken in July 2010 estimates this proportion to be 0.36.

This means that $\pi = 0.36$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.01?

This is n for which M = 0.01. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.01 = 1.96\sqrt{\frac{0.36*0.64}{n}}$

$0.01\sqrt{n} = 1.96\sqrt{0.36*0.64}$

$\sqrt{n} = \frac{1.96\sqrt{0.36*0.64}}{0.01}$

$(\sqrt{n})^2 = (\frac{1.96\sqrt{0.36*0.64}}{0.01})^2$

$n = 8851.04$

Rounding up(as a sample of 8851 will have a margin of error slightly over 0.01):

A sample size of 8852 is needed.

4 0

2 years ago

Square A has an area of 18a2 + 862 – 24ab.