54 - 4a^2 3b^3 when a = -2 and b = 4?

Mathematics

1 answer:

Bumek [7]3 years ago

7 0

54 - 4a^2 + 3b^3 when a = -2 and b = 4 is 54 - 4(-2)^2 + 3(4)^3 = 54 - 4(4) + 3(64) = 54 - 16 + 192 = 230

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A certain change in a process for manufacture of component parts is being considered. Samples are taken using both the existing

Katyanochek1 [597]

Answer:

a) The confidence interval is $0.00\leq\pi_1-\pi_2\leq0.02$ .

Step-by-step explanation:

We have to calculate a confidence interval (CI) of a difference of proportions.

For the existing procedure, the proportion is:

$p_1=75/1500=0.05$

For the new procedure, the proportion is:

$p_2=80/2000=0.04$

To calculate the CI, we need to estimate the standard deviation

$s=\sqrt{\frac{p_1(1-p_1)}{n_1} +\frac{p_2(1-p_2)}{n_2} }\\\\s=\sqrt{\frac{0.05(1-0.05)}{1500} +\frac{0.04(1-0.04)}{2000} }=\sqrt{ 0.000032 + 0.000019 }=\sqrt{ 0.000051 }\\\\s=0.007$

For a 90% CI, the z-value is 1.64.

Then, the CI is:

$(p_1-p_2)-z*\sigma \leq\pi_1-\pi_2\leq(p_1-p_2)+z*\sigma \\\\(0.05-0.04)-1.64*0.007\leq\pi_1-\pi_2\leq(0.05-0.04)+1.64*0.007\\\\0.01-0.01\leq\pi_1-\pi_2\leq0.01+0.01\\\\0.00\leq\pi_1-\pi_2\leq0.02$