In 2008, the per capita consumption of soft drinks in country a was reported to be 18.11 gallons. assume that the per capita consumption of soft drinks in country a is approximately normally distributed, with a mean of 18.11 gallons and a standard deviation of 4 gallons. complete parts (a) through (d) below. a. what is the probability that someone in country a consumed more than 13 gallons of soft drinks in 2008? the probability is nothing. (round to four decimal places as needed.)

There is a 89.97% probability that someone in Country A consumed more than 13 gallons of soft drinks in 2008.

Step-by-step explanation:

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

In a set with mean and standard deviation , the zscore of a measure X is given by

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

The per capita consumption of soft drinks in Country A is approximately normally distributed, with a mean of 18.11 gallons and a standard deviation of 4 gallons., so .

a. What is the probability that someone in Country A consumed more than 13 gallons of soft drinks in 2008?

The first step is finding the z score of .

has a pvalue of 0.1003.

This means that there is a 1-0.1003 = 0.8997 = 89.97% probability that someone in Country A consumed more than 13 gallons of soft drinks in 2008.

b.an increase in a causes b to decrease

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2