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

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18

a orange at the grocery store idk what other product that could be