Hey i need some help with this question but i dont like the history people section. i trust math more.

If b is the midpoint of ac, ac=cd, ab=3x+4, ac=11x-17, and ce=49, find de

IrinaK [193]

First, we draw our line.

|------------------------------------------------------------------------------------|

a e

Next, break up this line into segments using the information.

|----------------------|----------------------|--------------------|------------------|

a b c d e

The entire line is 29.

ab + bc + cd + de = ae

ab + bc + cd + de = 29

You also know that

bd = bc + cd

Due to midpoint theorem,

ab = bc

cd = de

Then,

2ab + 2cd = 29

The equations we will use are

bd = bc + cd eq1

2bc + 2cd = 29 eq2

Dividing both sides of the equation in eq2 yields

bc + cd = 14.5

bd = bc + cd

bd = 14.5

5 0

4 years ago

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. Suppose the coordinate of A is 0 and AR = 5 and AT = 7.

ASHA 777 [7]

<h3>Possible coordinates of R = -5 or 5</h3><h3>Possible coordinates of T = -7 or 7</h3>

==================================================

Explanation:

If we're on a number line, then R could be at either R = 5 or R = -5. This is so the distance from A to R is 5 units. Distance is never negative. You count out the spaces to get the distance, or use subtraction and absolute value.

Saying "distance from A to R is 5" can be written as AR = 5. Meaning segment AR is 5 units long.

Now if AT = 7, then T could be at 7 or -7 on the number line. The reasoning as similar as to why R could be at -5 or 5.

8 0

4 years ago

What is the cost of buying 5 kg of cheese worth<br> $9.00 per kg?

topjm [15]

$45

Hope this helps you

6 0

3 years ago

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within a kiambu county,students were randomly assigned to one of two mathematics teachers.Mrs.Elite and Mrs. Bright. After the a

DerKrebs [107]

Using the t-distribution, as we have the standard deviation for the samples, it is found that since the absolute value of the test statistic is greater than the critical value, we reject the claim that Mrs.Elite and Mrs. Bright are equally effective teachers.

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

At the null hypothesis, we test if they are equally effective teachers, that is, the subtraction of their means is 0, hence:

$H_0: \mu_1 - \mu_2 = 0$

At the alternative hypothesis, we test if they are not equally effective teachers, that is, the subtraction of their means is not 0, hence:

$H_1: \mu_1 - \mu_2 \neq 0$

<h3>What is the distribution of the differences?</h3>

For Mrs. Elite, we have that:

$\mu_1 = 78, \sigma_1 = 10, n_1 = 30, s_1 = \frac{10}{\sqrt{30}} = 1.82574$

For Mrs. Bright, we have that:

$\mu_2 = 85, \sigma_2 = 15, n_2 = 25, s_2 = \frac{15}{\sqrt{25}} = 3$

For the distribution of differences, we have that:

$\overline{x} = \mu_1 - \mu_2 = 78 - 85 = -7$

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{1.82574^2 + 3^2} = 3.5119$

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

The test statistic is given by:

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

In which $\mu = 0$ is the value tested at the null hypothesis.

Hence:

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

$t = \frac{-7 - 0}{3.5119}$

$t = -1.99$

Considering a<em> two-tailed test</em>, as we are testing if the mean is different of a value, with 30 + 25 - 2 = <em>53 df and a significance level of 0.1</em>, the critical value is of $|t^{\ast}| = 1.6741$ .

Since the absolute value of the test statistic is greater than the critical value, we reject the claim that Mrs.Elite and Mrs. Bright are equally effective teachers.

To learn more about the t-distribution, you can take a look at brainly.com/question/13873630

3 0

3 years ago

Explain the difference in meaning between I-3I and -3.

dexar [7]

|-3| is an absolute value. This means it will always be positive. -3 is an integer and is negative.

7 0

3 years ago

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