Find the sum of the distinct prime factors of $5^5 - 5^3$.

1 answer:

5 0

The prime factorization is $2016=2^5\times3^2\times7$. Since the problem is only asking us for the distinct prime factors, we have $2,3,7$. Their desired sum is then $\boxed{\textbf{(B) }12}$.

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12. If a manufacturer conducted a survey among randomly selected target market households and wanted to be 95% confident that th

atroni [7]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11$

And rounded up we have that n=1068

Step-by-step explanation:

We have the following info given:

$Confidence= 0.95$ the confidence level desired

$ME =0.03$ represent the margin of error desired

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

The confidence level is 95% or 0.95, the significance is $\alpha=0.05$ and the critical value for this case using the normal standard distribution would be $z_{\alpha/2}=1.96$