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 perimeter is increased by a factor of 2

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation: