Answer:
1811.6
Step-by-step explanation:
647x28
18116
move once to the left bc of the point in 64.7
1811.6
Complete Question
The quality control manager at a computer manufacturing company believes that the mean life of a computer is 91 months with a standard deviation of 10 months if he is correct. what is the probability that the mean of a sample of 68 computers would differ from the population mean by less than 2.08 months? Round your answer to four decimal places. Answer How to enter your answer Tables Keypad
Answer:

Step-by-step explanation:
From the question we are told that:
Population mean \mu=91
Sample Mean \=x =2.08
Standard Deviation \sigma=10
Sample size n=68
Generally the Probability that The sample mean would differ from the population mean
P(|\=x-\mu|<2.08)
From Table

T Test




Therefore From Table

C
6x^2-13X-5=0
(3X+1)(2X-5)
3x-1=0
X=1/3. 2x-5=0. X= 5/2
The goal is to put the equation in the simplest form to solve:
2x²+17=179
Subtract both sides by 17.
179-17= 162
2x²=162
Square root 162.
√162 = 12.7 (rounded)
2x=12.7
Divide both sides by 2.
12.7/2= 6.4 (rounded)
x=6.4
The variable x is 6.4, rounded to the nearest tenth for simple purposes.
Hope this was helpful.
The answer is c. Because there inside and opposite