What is the area of the following polygon? Explain your process for how you arrived at the final area. Show your work

Segment AB is rotated to form A'B'¯¯¯¯¯¯¯ . The coordinates of point A are (1, 5) and the coordinates of point B are (−6, 4)

dem82 [27]

It is rotated 180 degrees, and it doesn't matter wether clockwise or counterclockwise. 180 degrees results in the same thing both ways

6 0

3 years ago

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What is another way to name angle 3?

fiasKO [112]

Answer:

intimately there are three types of angles a cute angle and angle between zero and 90° right angle and angle 90 degree angle of two angle and angle between 90 and 180 degree

7 0

2 years ago

The shorter leg of a right triangle is 14 feet less than the other leg. Find the length of the two legs of the hypotenuse is 25

irakobra [83]

Answer:

9.233 ft, 23.233 ft

Step-by-step explanation:

If the shorter leg is x, then the longer leg is x+14 and the Pythagorean theorem tells you ...

x^2 + (x +14)^2 = 25^2

2x^2 +28x +196 = 625

x^2 +14x = 214.5

x^2 +14x +49 = 263.5

(x +7)^2 = 263.5

x = -7 +√263.5 ≈ 9.23268

The two leg lengths are √263.5 ± 7 feet, {9.23 ft, 23.23 ft}.

6 0

4 years ago

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A sociologist was investigating the ages of grandparents of high school students. From a random sample of 10 high school student

umka2103 [35]

Answer:

$(69.8-71.6) -2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= -8.908$

$(69.8-71.6) +2.101 \sqrt{\frac{8.38^2}{10} +\frac{6.65^2}{10}}= 5.308$

So then the confidence interval for the difference of means is given by:

$-8.908 \leq \mu_M -\mu_F \leq 5.308$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_{M}= 69.8$ represent the mean for the age of grandmother

$\bar X_{F}= 71.6$ represent the mean for the age of grandfather

$s_{M}= 8.38$ represent the sample deviation for the age of grandmother

$s_{F}= 6.65$ represent the sample deviation for the age of grandfather

$n_M = n_F= 10$

Solution to the problem

For this case the confidence interval is given by:

$(\bar X_{M} -\bar X_F) \pm t_{\alpha/2} \sqrt{\frac{s^2_M}{n_M} +\frac{s^2_F}{n_F}}$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n_M +n_F-2=10+10-2=18$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,18)".And we see that $t_{\alpha/2}=2.101$