Assume a normal distribution and that the average phone call in a certain town lasted 6 min, with a standard deviation of 1 min. What percentage of the calls lasted less than 5min?What percentage of calls last less than 5min?

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116

1 and 0

step-by-step explanation: