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notsponge [240]
2 years ago
11

If EF bisects CD, CG = 5x-1, GD = 7x-13, EF=6x-4, and GF = 13, find EG (MUST SHOW WORK)

Mathematics
1 answer:
iogann1982 [59]2 years ago
3 0

Answer:

EG = 19

Explanation:

Given that a line segment CD. EF bisects the line CD at point G.

Length of CG = 5x-1

Length of GD = 7x - 13

We know that CG = GD

5x-1 = 7x-13

Solve for x,

2x = 12

x = 6

Now given that,

EF = 6x-4

GF = 13

EF = EG + GF

6x - 4 = EG + 13

EG = 6x - 4 - 13

EG = 6x - 17

put the value of x = 6, in order to find the EG

EG = 6*6 - 17

EG = 19

That's the final answer.

I hope it will help you.

Guest
1 year ago
how did you get that equation to find X?
Guest
1 year ago
How did you find X?
Guest
1 year ago
To find x:
Guest
1 year ago
5x-1=7x-13 —> subtract 7x from both sides and add 1 to both sides —> -2x-12 —> divide both sides by -2 —> x=6
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Answer:

b) 0.0608

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As it is mentioned that the next two days i.e 24 hours, the probability of the rain is uniformly distributed

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where,

T = Length of the time interval

Plus, as we know that rain is independent

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So, P1 should be

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2 years ago
The sum of 23.8 and a number is 28.17, as shown below. What number should go in the box to complete the addition problem?
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5 0
2 years ago
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According to the Vivino website, suppose the mean price for a bottle of red wine that scores 4.0 or higher on the Vivino Rating
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Answer:

a) Null and alternative hypothesis

H_0: \mu=32.48\\\\H_a:\mu< 32.48

b) Test statistic t=-1.565

P-value = 0.0612

NOTE: the sample size is n=65.

c) Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

d) Null and alternative hypothesis

H_0: \mu=32.48\\\\H_a:\mu< 32.48

Test statistic t=-1.565

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Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Then, the null and alternative hypothesis are:

H_0: \mu=32.48\\\\H_a:\mu< 32.48

The significance level is 0.05.

The sample has a size n=65.

The sample mean is M=30.15.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{12}{\sqrt{65}}=1.4884

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{30.15-32.48}{1.4884}=\dfrac{-2.33}{1.4884}=-1.565

The degrees of freedom for this sample size are:

df=n-1=65-1=64

This test is a left-tailed test, with 64 degrees of freedom and t=-1.565, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=P(t

As the P-value (0.0612) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

<u>Critical value approach</u>

<u></u>

At a significance level of 0.05, for a left-tailed test, with 64 degrees of freedom, the critical value is t=-1.669.

As the test statistic is greater than the critical value, it falls in the acceptance region.

The null hypothesis failed to be rejected.

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