A recent study found that the average life expectancy of a person living in Africa is 53 years with a standard deviation of 7.5 years. If a person in Africa is selected at random, what is the probability that the person will die before the age of 65?

0.945200708

Step-by-step explanation:

In this question, we are asked to calculate the probability that a person selected at random would live less than 65.

To solve this problem, what we need to do is to first calculate the z-score.

Mathematically, z = (x-mean)/standard deviation

According to the question, the mean is 53 years and the standard deviation is 7.5

Substituting these values in the z-score equation:

Z-score = (65-53)/7.5 = 12/7.5 = 1.6

We proceed to use the z-score table to obtain the probability. Using the table, z score of 1.6 corresponds to a probability of

0.945200708

0.85 or 85%

step-by-step explanation:

1) divide 23 by 27

2) round the decimal to the nearest 10

3) so your answer would be 0.85 or 85%

c: y > - 1/6x + 1

step-by-step explanation:

dashed line - meaning this < or > also not equal

solid line - meaning this < or > also equal

your answer is c because y > means up & it's shaded up, but y < means down

hope this ! : )