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3 years ago

8

During the first two hours, how far does the car travel?

2 answers:

Yanka [14]3 years ago

6 0

In the two hours, he drives 15 (km)

Hope I helped : )

BartSMP [9]3 years ago

6 0

It most definitly looks like fifteen km to me.

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Let $G$ denote the centroid of triangle $ABC$. If triangle $ABG$ is equilateral with side length 2, then determine the perimeter

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Hey there!

You are trying to find the perimeter of triangle ABC. Each side is 2. Since there are 3 sides on a triangle, multiply 2 by 3 to get 6. You do not need the centroid for this problem.

I hope this helps!

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3 years ago

12x-y when x= 1/3, y=-2

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Answer:

Step-by-step explanation:

12 x 1/3-2

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3 years ago

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What is the value of the expression ? (243^2)^1/10<br><br> A.9<br><br> B.81<br><br> C.3<br><br> D.27

podryga [215]

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2 years ago

Sec (x) = -2.3. If x and y are complementary, what is csc (y)?

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3 years ago

Two shooters, Rodney and Philip, practice at a shooting range. They fire rounds each at separate targets. The targets are mark

Sedaia [141]

Complete Question

Answer:

a

$SE = 0.66}$

b

$-3.29 < \mu_1 - \mu_2 < -0.70$

Step-by-step explanation:

From the question we are told that

The sample size is n = 60

The first sample mean is $\= x _1 = 8$

The second sample mean is $\= x _2 = 10$

The first variance is $v_1 = 0.25$

The first variance is $v_2 = 0.55$

Given that the confidence level is 95% then the level of significance is 5% = 0.05

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the first standard deviation is

$\sigma_1 = \sqrt{v_1}$

=> $\sigma_1 = \sqrt{0.25}$

=> $\sigma_1 = 0.5$

Generally the second standard deviation is

$\sigma_2 = \sqrt{v_2}$

=> $\sigma_2 = \sqrt{0.55}$

=> $\sigma_2 = 0.742$

Generally the first standard error is

$SE_1 = \frac{\sigma_1}{\sqrt{n} }$

$SE_1 = \frac{0.5}{\sqrt{60} }$

$SE_1 = 0.06$

Generally the second standard error is

$SE_2 = \frac{\sigma_2}{\sqrt{n} }$

$SE_2 = \frac{0.742}{\sqrt{60} }$

$SE_2 = 0.09$

Generally the standard error of the difference between their mean scores is mathematically represented as

$SE = \sqrt{SE_1^2 + SE_2^2 }$

=> $SE = \sqrt{0.06^2 +0.09^2 }$

=> $SE = 0.66}$

Generally 95% confidence interval is mathematically represented as

$(\= x_1 -\= x_2) -(Z_{\frac{\alpha }{2} } * SE) < \mu_1 - \mu_2 < (\= x_1 -\= x_2) +(Z_{\frac{\alpha }{2} } * SE)$

=> $(8 -10) -(1.96 * 0.66) < \mu_1 - \mu_2 < (8-10) +(Z_{\frac{\alpha }{2} } * 0.66)$

=> $-3.29 < \mu_1 - \mu_2 < -0.70$

5 0

3 years ago