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3 years ago

Constance is constructing a perpendicular bisector using a compass and a straightedge.

Mathematics

2 answers:

Leviafan [203]3 years ago

4 0

hope this helps !

:):):):):)

zloy xaker [14]3 years ago

3 0

<span>B. 2
</span>B. The chords must bisect one another.

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What is the midpoint between (3, -1) and (7, -5)?<br> Group of answer choices

Hatshy [7]

Answer:

The answer is

Step-by-step explanation:

The midpoint M of two endpoints of a line segment can be found by using the formula

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(3, -1) and (7, -5)

The midpoint is

$M = ( \frac{3 + 7}{2} , \: \frac{ - 1 - 5}{2} ) \\ = ( \frac{10}{2} \: , \: - \frac{6}{2} )$

We have the final answer as

Hope this helps you

6 0

3 years ago

How does the graph of y = 3^–x compare to the graph of y = (1/3)^x?

oee [108]

Answer:

A. The graphs are the same.

Step-by-step explanation:

4 0

3 years ago

According to the Knot, 22% of couples meet online. Assume the sampling distribution of p follows a normal distribution and answe

Ann [662]

Using the <em>normal distribution and the central limit theorem</em>, we have that:

a) The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

b) There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

c) There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1 - p)}{n}}$ , as long as $np \geq 10$ and $n(1 - p) \geq 10$ .

In this problem:

22% of couples meet online, hence p = 0.22.

A sample of 150 couples is taken, hence n = 150.

Item a:

The mean and the standard error are given by:

$\mu = p = 0.22$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.22(0.78)}{150}} = 0.0338$

The sampling distribution is approximately normal, with mean 0.22 and standard error 0.0338.

Item b:

The probability is <u>one subtracted by the p-value of Z when X = 0.25</u>, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem:

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.25 - 0.22}{0.0338}$

Z = 0.89

Z = 0.89 has a p-value of 0.8133.

1 - 0.8133 = 0.1867.

There is a 0.1867 = 18.67% probability that in a random sample of 150 couples more than 25% met online.

Item c:

The probability is the <u>p-value of Z when X = 0.2 subtracted by the p-value of Z when X = 0.15</u>, hence:

X = 0.2:

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.2 - 0.22}{0.0338}$

Z = -0.59

Z = -0.59 has a p-value of 0.2776.

X = 0.15:

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.15 - 0.22}{0.0338}$

Z = -2.07

Z = -2.07 has a p-value of 0.0192.

0.2776 - 0.0192 = 0.2584.

There is a 0.2584 = 25.84% probability that in a random sample of 150 couples between 15% and 20% met online.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can check brainly.com/question/24663213

4 0

2 years ago

You are graphing Square ABCDABCDA, B, C, D in the coordinate plane. The following are three of the vertices of the square: A(4,

Vsevolod [243]

Answer:

D(4,-3)

Step-by-step explanation:

Given three of the vertices of the square: A(4, -7), B(8, -7),C(8, -3)

Let the coordinate of the fourth vertex be D(x,y).

We know that diagonals of a square are perpendicular bisector. So, the midpoint of both diagonals is the same.

The diagonals are BD and AC

Midpoint of BD = Midpoint of AC

$\left(\dfrac{8+x}{2},\dfrac{-7+y}{2}\right) =\left(\dfrac{4+8}{2},\dfrac{-7+(-3)}{2}\right)\\ \left(\dfrac{8+x}{2},\dfrac{y-7}{2}\right) =\left(\dfrac{12}{2},\dfrac{-10}{2}\right)\\ \left(\dfrac{8+x}{2},\dfrac{y-7}{2}\right) =\left(6,-5\right)\\$Therefore$:\\\dfrac{8+x}{2}=6\\8+x=12\\x=12-8\\x=4\\$Similarly$\\\dfrac{y-7}{2}=-5\\y-7=-5*2\\y-7=-10\\y=-10+7=-3$