Please help picture included

I didnt put the answer choices because ppl be putting random stuff

lesya692 [45]

I’m pretty sure the answer is 127.48 ,but I could be wrong.

5 0

3 years ago

How many units long is a segment whose endpoints are (2,3) and (7,15)?

Ahat [919]

Answer:

13 units

Step-by-step explanation:

Distance formula: $d = \sqrt{(x2-x1)^2 + (y2-y1)^2}$

Simply plug in your variables

$d = \sqrt{(7-2)^2 + (15-3)^2}$

$d = \sqrt{5^2 + 12^2}$

$d = \sqrt{25 + 144}$

$d = \sqrt{169}$

d = 13

5 0

3 years ago

A track has the shapes shown. what is the perimeter ?

USPshnik [31]

Break this problem down first. You have two lines each with length of 5. You also have two halves of a circle, combine them to get a full circle with a radius of 2. The formula to find the perimeter of a circle is
$2\pi \times r$
and this gives a perimeter of 12.566 for the circle. Add this to your sides and you get a total of 22.566 which is answer B

4 0

3 years ago

I want to know how to get this factored and simplified the easiest way. I know that I can multiply out and then factor but that

Reptile [31]

Answer:

Enter the expression you want to factor in the editor. The Factoring Calculator transforms complex expressions into a product of simpler factors. It can factor expressions with polynomials involving any number of vaiables as well as more complex functions.

Step-by-step explanation:

8 0

3 years ago

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0

3 years ago