The average yearly snowfall in Chillyville is approximately Normally distributed with a mean of 55 inches. nIf the snowfall in Chillyville exceeds 60 inches in 15% of the years, what is the standard deviation? a. 8.93 inches

b. 4.83 inches

c. 6.04 inches

d. The standard deviation cannot be computed from the given information

e. 5.18 inches

45 is the only number in both sets

Okay so n is going to represent our first number. since the second number of five times greater than n, the second number will be represented as 5n. for us to figure out the two  numbers we need to have  the value of n first. this is going to be done by adding the two numbers together, n and 5n. it can be simplified as 6n, which the problem states is equal to 72. divide 72 by 6n and you can solve for n.  n+5n=726n=72n=(72/6)n=12  now that we know the value of n, we have to multiply 5 times 12 which equals 60. so our two numbers/integers are 12 and 60