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Masja [62]
3 years ago
13

The diagram shows , , and . Which statement can be proven true from the diagram? A. ∠DGB is supplementary to ∠EGA. B. ∠DGB is su

pplementary to ∠CGB. C. ∠CGB is supplementary to ∠AGD. D. ∠EGA is supplementary to ∠EGC. E. ∠EGD is supplementary to ∠CGB.

Mathematics
1 answer:
valkas [14]3 years ago
3 0

Answer:

The correct answer is B: ∠DGB is supplementary to ∠CGB. I got it right on my test.

Step-by-step explanation:

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The endpoints of the diameter of a circle are (-7.3) and (5.1). What is the center of the circle?
Minchanka [31]

Check the picture below.

since we know the endpoints of the diameter of the circle, then we can simply get the midpoint of those endpoints and that'd be the center,

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-7}~,~\stackrel{y_1}{3})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5-7}{2}~~,~~\cfrac{1+3}{2} \right)\implies \left( \cfrac{-2}{2}~~,~~\cfrac{4}{2} \right)\implies (-1~~,~~2)

4 0
3 years ago
A random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 11 and the sample stand
Tpy6a [65]

Answer:

We conclude that the population mean is different from 10.5.

Step-by-step explanation:

We are given that a random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 11 and the sample standard deviation is 2.

<em>We have to test the claim that the population mean is 10.5.</em>

Let, NULL HYPOTHESIS, H_0 : \mu = 10.5  {means that the population mean is 10.5}

ALTERNATE HYPOTHESIS, H_a : \mu \neq 10.5  {means that the population mean is different from 10.5}

The test statistics that will be used here is One-sample t-test;

           T.S. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, \bar X = sample mean = 11

            s = sample standard deviation = 2

            \mu = population mean

            n = sample of values = 16

So, <u>test statistics</u> =  \frac{11-10.5}{\frac{2}{\sqrt{16} } } ~ t_1_5

                            = 1

<em>Now, at 0.05 significance level, t table gives a critical value of 2.131 at 15 degree of freedom. Since our test statistics is way less than the critical value of t so we have insufficient evidence to reject null hypothesis as it will not fall in the rejection region.</em>

Therefore, we conclude that the population mean is different from 10.5.

6 0
4 years ago
Nancy needs to board her dog. Doggy care will take the dog for the initial fee of $5.00 and then $0.75 per hour. Dogs ate us wil
DaniilM [7]
Answer: 10 hours

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These equations intercept at (10,12.5)

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The best thing do is to graph these then to make a table for both and then see when there’s a point on both tables that are the same
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Maksim231197 [3]
U wouldn’t have a answer because it would be ( 0=0 )
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A tree surgeon reduces the height of a beech tree by 32%.
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Let original Hieght be X

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now, 4.42 = X + 32/100 X

4.42 = 132/100 X

X = 4.42 * 100 / 132

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6 0
2 years ago
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