Lunch break: In a recent survey of 600 working Americans ages 25-34, the average weekly amount spent on lunch as S45.32 with standard deviation S2.87. The weekly amounts are approximately bell-shaped. Part 1 out of 3 Estimate the percentage of amounts that were less than S42.45. Round the answer to one decimal place. Approximately % of the amounts were less than $42.45.

Step-by-step explanation:

given that in a recent survey of 600 working Americans ages 25-34, the average weekly amount spent on lunch as S45.32 with standard deviation S2.87.

The weekly amounts are approximately bell-shaped

so for practical purposes we assume X =  ages of working Americans is N(45.32, 2.87)

Estimate the percentage of amounts that were less than S42.45

First let us find probability

P(X<42.45) = F(42.45)

=

= 0.5-0.3413

=0.1687

So 16.87% of the amounts were less than 42.45 dollars

42 days

Step-by-step explanation:

Given

Leakage = 5 gallons in 3 days

Required

Determine the number of days it finish leaking out

The scenario can be represented as:

5 gallons = 3 days

and

70 gallons = x

Where x represents the number of days to leak out 70 gallons

Cross Multiply:

Divide both sides by 5

Hence, it'll take 42 days