Answer:
The percent of the parts are expected to fail before the 2100 hours is 0.15.
Step-by-step explanation:
Given :Life tests on a helicopter rotor bearing give a population mean value of 2500 hours and a population standard deviation of 135 hours.
To Find : If the specification requires that the bearing lasts at least 2100 hours, what percent of the parts are expected to fail before the 2100 hours?.
Solution:
We will use z score formula

Mean value = 
Standard deviation = 
We are supposed to find If the specification requires that the bearing lasts at least 2100 hours, what percent of the parts are expected to fail before the 2100 hours?
So we are supposed to find P(z<2100)
so, x = 2100
Substitute the values in the formula


Now to find P(z<2100) we will use z table
At z = −2.96 the value is 0.0015
So, In percent = 
Hence The percent of the parts are expected to fail before the 2100 hours is 0.15.
Answer:
4 2/3
Step-by-step explanation:
Answer:
41.67hrs
Step-by-step explanation:
if the snail moves 0.12km in 1hr
the snail will move 5km in Xhrs
0.12 × X = 5 × 1
0.12X= 5
X= 5/0.12 = 41.67hrs
Answer:
D
Step-by-step explanation:
QG=GS
x+5=2x-2
x=7
QG=7+5=12
3(8x - 5) = -4(7 - 6x)
Distribute 3 and -4 inside their respective parentheses.
24x - 15 = -28 + 24x
Add 28 to both sides.
24x + 13 = 24x
Subtract 24x from both sides.
13 = 0
This is an untrue statement, so that means there are no solutions to this equation.
Your answer is no solution.