Answer:
0.6915
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
![\mu = 88, \sigma = 4[/ex]What is the probability that a single car of this model emits more than 86 mg/mi of NOX + NMOG?This is 1 subtracted by the pvalue of Z when X = 86. So[tex]Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=%5Cmu%20%3D%2088%2C%20%5Csigma%20%3D%204%5B%2Fex%5D%3C%2Fp%3E%3Cp%3E%3Cstrong%3EWhat%20is%20the%20probability%20that%20a%20single%20car%20of%20this%20model%20emits%20more%20than%2086%20mg%2Fmi%20of%20NOX%20%2B%20NMOG%3F%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3EThis%20is%201%20subtracted%20by%20the%20pvalue%20of%20Z%20when%20X%20%3D%2086.%20So%3C%2Fp%3E%3Cp%3E%5Btex%5DZ%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)


has a pvalue of 0.3085
1 - 0.3085 = 0.6915
The answer is 0.6915
Answer:
-23
Step-by-step explanation:
Let 70% of X = 98
70%.x = 98
(70/100).x = 98
0,7x = 98
x = 98/0,7
x = 140
Then, 90%x = 90%.140
= 0,9.(140)
= 126
The answer to this question is Letter D)
I hope this has helped.
we know that
A relationship between two variables, x, and y, represent an inverse variation if it can be expressed in the form
or 
Let
x-------> the pressure in PSI
y------> the volume of the gas in cubic inches
In this problem we have the point 
so

Find the constant k

substitute the values of x and y


the equation is

therefore
<u>the answer is</u>

Answer:
C and D
Step-by-step explanation:
using the rules of exponents
A
=
≠ ![\sqrt[3]{125^7}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B125%5E7%7D)
B
(
)^7 =
≠ 
C
(
)^5 =
← correct
D
(
)^9 =
← correct