All categories

3 years ago

Which best describes the relationship between the line that passes through the points (-6,5) and (-

Mathematics

1 answer:

natita [175]3 years ago

3 0

Answer:

Slope passes through (-6, 5) and (-2, 7)

$m_1=\frac{y_2-y_1}{x_2-x_1} =m_1=\frac{7-5}{-2+6}$

$m_1=\frac{2}{4} =\frac{1}{2}$

Slope passes through (4, 2) and (6, 6)

$m_2=\frac{6-2}{6-4}$

$\frac{4}{2} =2$

$m_1\times m_2 \neq -1$ $m_1\neq m_2$

<u>Answer:- A) neither perpendicular nor parallel.</u>

<u>OAmalOHopeO</u>

You might be interested in

Someones answer q3 ill mark as brainliest pleaseee

il63 [147K]

The length of the rectangle is x and the height is x-2

To find the area of a rectangle you multiply length times width

l(w) = 146

x(x-2) = 146

x^2 - 2x = 146

5 0

3 years ago

Write one hundred eleven in expanded form

nydimaria [60]

Answer: 100+10+1

You basically place zeroes instead of the actual digit to find the value of each number.

Example:

855

800+50+5

6 0

3 years ago

2 equivalent forms of 0.7875

nadezda [96]

Idk what that answer is soooo

4 0

3 years ago

10. A manufacturer wanted to know if more coupons would be redeemed if they were mailed to the

Mandarinka [93]

Using the t-distribution, it is found that the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if there is no difference, that is, the mean is of 0, hence:

$H_0: \mu = 0$

At the alternative hypothesis, it is tested if the mean number is greater for females, that is, the mean is greater than 0, hence:

$H_1: \mu > 0$

<h3>What is the test statistic?</h3>

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.

$\mu$ is the value tested at the null hypothesis.

s is the standard deviation of the sample.

n is the sample size.

In this problem, the parameters are given as follows:

$\overline{x} = 1.5, \mu = 0, s = 4.75, n = 50$ .

Hence, the test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{1.5 - 0}{\frac{4.75}{\sqrt{50}}}$

t = 2.23

<h3>What is the conclusion?</h3>

Considering a <em>right-tailed test</em>, as we are testing if the mean is greater than a value, with a <em>significance level of 0.01 and 50 - 1 = 49 df</em>, the critical value is given by $t^{\ast} = 2.4$ .

Since the test statistic is less than the critical value for the right-tailed test, the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

More can be learned about the t-distribution at brainly.com/question/26454209

5 0

2 years ago

This holiday season, Ms. Bastarache organized to give each 9th grader a $10 Dunkin Donuts gift card. Because she bought a lot of

Mila [183]

Answer:

1. $950

2. 80

Step-by-step explanation:

8 0

3 years ago