Since it takes 10 minutes to peel 4 potatoes, you can assume it would take 30 minutes to peel 12 potatoes (mulitiplying both numbers by 3). Then you'd add 5, instead of 10 because you are peeling 2 more potatoes than 12, instead of peeling 4, you'd divide 10 by 2 to get half the time it would take to peel half as many potatoes. Finally, add 30 and 5 to get you final answer of 35.

I hope this helps!

8 0

3 years ago

Find the area of the triangle below.

krek1111 [17]

Answer:

26 yd^2

Step-by-step explanation:

Area of Triangle = 1/2(b*h)

base = b = 13 yd

height = h = 4yd

Area of Triangle = 1/2(13*4) = 26 yd^2

3 0

3 years ago

Read 2 more answers

The mathematics section of a standardized college entrance exam had a mean of 19.8 and an SD of 5.1 for a recent year. Assume th

Fudgin [204]

Answer:

a) 0.84% of students scored over 32

b) 29.12% of students scored under 17

c) 70.04% of students scored between 17 and 32.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 19.8, \sigma = 5.1$

a) About what percent of students scored over 32?

This is 1 subtracted by the pvalue of Z when X = 32. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{32 - 19.8}{5.1}$

$Z = 2.39$

$Z = 2.39$ has a pvalue of 0.9916

1 - 0.9916 = 0.0084

So 0.84% of students scored over 32

b) About what percent of students scored under 17?

This is the pvalue of Z when X = 17. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{17 - 19.8}{5.1}$

$Z = -0.55$

$Z = -0.55$ has a pvalue of 0.2912

So 29.12% of students scored under 17

c) About what percent of students scored between 17 and 32?

This is the pvalue of Z when X = 32 subtracted by the pvalue of Z when X = 12. So

X = 32

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{32 - 19.8}{5.1}$

$Z = 2.39$

$Z = 2.39$ has a pvalue of 0.9916

X = 17

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{17 - 19.8}{5.1}$

$Z = -0.55$

$Z = -0.55$ has a pvalue of 0.2912

0.9916 - 0.2912 = 0.7004

70.04% of students scored between 17 and 32.

5 0

3 years ago