Acredit card company receives numerous phone calls throughout the day from customers reporting fraud and billing disputes. most of these callers are put "on hold" until a company operator is free to them. the company has determined that the length of time a caller is on hold is normally distributed with a mean of 2.5 minutes and a standard deviation 0.5 minutes. if 1.5% of the callers are put on hold for longer than x minutes, what is the value of x?

The value of X is 3.6 minutes.

Step-by-step explanation:

We assume that the time on hold follows a normal distribution with

To calculate how many minutes the 1.5% of the customers are put on hold, we have to determine the z-value for this probability (P=0.015).

Looking in a z-table, the z value is 2.17:

This can be translated to our distribution as:

The value of X is 3.6 minutes.

the answer is 3/5

step-by-step explanation:

first divide 24 divided by 8 because that is the least common multiple

then divide 40 by 8

then you will get 3/5.

example 2 – simplify: step 1: completely factor both the numerators and denominators of all fractions. step 2: change the division sign to a multiplication sign and flip (or reciprocate) the fraction after the division sign; essential you need to multiply by the reciprocal. step 3 : cancel or reduce the fractions.