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.
Let's construct an equation for Canton's population increase as 7220+80n and Holly Spring's population increase with 3200+120n, where n is the number of years. We have to set these equations to each other in order to find when the population of these two places are equal. 3200+120n=7220+80n and n=100.5. After 100 and half years
Answer:
Hiya there!
Step-by-step explanation:
I think it would be one of the first two.
But mostly I think Commutative Property.
Not entirely sure.
Sorry if I am wrong.
Mary has 5 2/5 yards of string. Written as an improper fraction, 5 2/5 = 27/5 yards of string.
The friend was given 2/3 of the string. The friend received 2/3 * 27/5 = 18/5 yards which is the same as 3 3/5 yards and also 3.6 yards.
After Mary gave the friend the string, Mary had 27/5 - 18/5 = 9/5 yards left.
9/5 yards is the same as 1 4/5 yard and also 1.8 yards.