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3 years ago

Please i need help asapl

Mathematics

1 answer:

Marizza181 [45]3 years ago

7 0

Answer:

Ella should not have included 5 in absolute value portion of the equation

Step-by-step explanation:

Because -1 is a negative number in order to find the true distance between A and B you have to make it positive.

But, because Ella included 5 in the absolute value portion of the equation it messed up the answer.

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Please help, will give extra points

asambeis [7]

Answer:

59 degrees

Step-by-step explanation:

Since angle BGC is 40 degrees and angle AGC is 19 degrees and they are asking for angle AGB which is the sum of both angles together, you just add the 40 degrees plus 19 degrees.

6 0

3 years ago

Can I get the answer for the questions at the bottom

aniked [119]

27)

Equation of the existing water pipe's line

a) slope, m = rise / run = Δy / Δx = 3/2

b) y-intercept, b = 3

equation: y = mx + b = (3/2)x + 3

Equation of the new water pipe's line

slope, m = - 1 / (slope of the perpendicular line) = - 1 / (3/2) = - 2/3

point (0,2)

=> y - 2 = (-2/3) (x - 0) => y = (-2/3)x + 2 <---- equation of the new pipe

28)

two parallel lines have the same slope =>

slope, m = rise / run = Δy / Δx = [4 - 0] / [11 - 8] = 4 / 3

point (4,5) => y - 5 = (4/3) (x - 4)

=> y = (4/3)x - 16/3 + 5 = (4/3)x - 1/3

y = (4/3)x - 1/3 <--- equation of the new bike path

6 0

3 years ago

There is a photo attached<br><br>tan 270 is incorrect!

wlad13 [49]

$\dfrac{1-\cos135^\circ}{\sin135^\circ}=\dfrac{1-\left(\cos^2 67.5^\circ-\sin^2 67.5^\circ\right)}{2\sin67.5^\circ\cos67.5^\circ}=\dfrac{1-\left(1-\sin^2 67.5^\circ-\sin^2 67.5^\circ\right)}{2\sin67.5^\circ\cos67.5^\circ}=\medskip\\=\dfrac{2\sin^2 67.5^\circ}{2\sin67.5^\circ\cos67.5^\circ}=\dfrac{\sin 67.5^\circ}{\cos 67.5^\circ}=\tan 67.5^\circ$

3 0

3 years ago

If 400 patrons visit the park in March and 550 patrons visit in April, the total number of patrons who

Zielflug [23.3K]

950 is a real number, and it is a rational number since it can be expressed as 950/1. Therefore, it falls into all of the categories expect irrational numbers.

7 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