This one is tricky! Imagine that you meet a new friend who is also a beginner, and she can run the 5k in 23.5 minutes. You wonder what percentage of the beginner running population could run the 5k faster than your new friend (that is, what percentage of the population has a time that is less than your new friend

38.74%

Step-by-step explanation:

Given the data:

21 21 22 22 23 23 23 24 24 24 24 24 25 25 25 26 26 27 27

We obtain the beginner running population and standard deviation

Population mean, μ = Σx/n = 456/19 = 24

Standard deviation, σ = 1.747 (using calculator)

Friend's Runtime, x = 23.5 minutes

Obtaining the friend's Zscore :

Z = (x - μ) / σ

Z = (23.5 - 24) / 1.747

Z = - 0.286

Obtaining the Pvalue :

Using a standard normal distribution table :

P(Z < - 0.286) = 0.38744

Hence. Percentage of population that has lesser time :

0.38744 * 100% = 38.74%

the answer is (4,3)

since the rule for r0, 90º is (-y,x) and it was already applied to form the image, all you have to do is reverse the answer using this formula.

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632