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3 years ago

Factorise each of the following.

Mathematics

1 answer:

Alborosie3 years ago

7 0

I will do (a) only.

6rt - 3st + 6ru - 3su

6rt - 3st = 3t(2r - s)

6ru - 3su = 3u(2r - s)

Answer: (3t + 3u)(2r - s)

Do the rest likewise.

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Sedaia [141]

Complete Question

Answer:

$SE = 0.66}$

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

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

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

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

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

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Answer:

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h(g(-4)) aka h(2)= 2(2)+3= 4+3=7

Answer is 7

Step-by-step explanation:

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