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3 years ago

13

I’m stuck on this...I got it wrong on the first try

1 answer:

Shtirlitz [24]3 years ago

5 0

What's the question?

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What is the product?

viva [34]

Answer:

its C

Step-by-step explanation:

I hope this helps :)

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2 years ago

Find the complete factored form of the polynomial<br> 28a^4b^6 + 21a^3b^6

n200080 [17]

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2 years ago

Write the equation that represents the relationship between n the number of triangles and p the perimeter of the figures formed.

Klio2033 [76]

1 triangle: 2*6 + 1*5

2 triangles: 2*6 + 2*5

3 triangles: 2*6 + 3*5

So for n triangles: p = 2*6 + n*5 = 12 + 5n

4 0

2 years ago

Read 2 more answers

If fx()=8-10x and g(x)=5x+4, what is the value of (fg)(-2)

lyudmila [28]

So, we are looking for F of G of 2. So, we need to work backwards and find G(x) first.

$g(-2)=5(-2)+4\\g(-2)=-6$

So we know that g(-2) = -6, now we take thatt -6 and plug in into f(x)

$f(g(-2))=f(-6)\\f(-6)=8-10(-6)\\f(-6)=8+60\\f(-6)=68$

68 is the answer

7 0

2 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