How many women must be randomly selected to estimate the mean weight of women in one age group? We want 90% confidence that the sample mean is within 3.7 lbs of the populations mean, and population standard deviation is known to be 28 lbs.

155 women must be randomly selected.

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Z-table as such z has a p-value of .

That is z with a pvalue of , so Z = 1.645.

Now, find the margin of error M as such

In which is the standard deviation of the population and n is the size of the sample.

The population standard deviation is known to be 28 lbs.

This means that

We want 90% confidence that the sample mean is within 3.7 lbs of the populations mean. How many women must be sampled?

This is n for which M = 3.7. So

Rounding up:

155 women must be randomly selected.

"a" determine whether sage found the correct quotient. show your work

step-by-step explanation:

answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation: