We can set up a proportion, and let x be the amount 16kg would cost: 14/154=16/x. We simplify 14/154 = 1/11. Next, we cross-multiply to get (1)(x)=(16)(11), so x=176. Therefore, 16kg of apples cost $176.
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Brewed decaffeinated coffee contains some caffeine. We want to estimate the amount of caffeine in 8 ounce cups of decaf coffee a

rusak2 [61]

Answer:

We would have to take a sample of 62 to achieve this result.

Step-by-step explanation:

Confidence level of 95%.

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

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.025 = 0.975$ , so Z = 1.96.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

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

Assume that the standard deviation in the amount of caffeine in 8 ounces of decaf coffee is known to be 2 mg.

This means that $\sigma = 2$

If we wanted to estimate the true mean amount of caffeine in 8 ounce cups of decaf coffee to within /- 0.5 mg, how large a sample would we have to take to achieve this result?

We would need a sample of n.

n is found when $M = 0.5$ . So