Answer:
probability that a single car of this model emits more than 86 mg/mi of NOX + NMOG = 0.8668
Step-by-step explanation:
mean, μ = 80 mg/mi
Standard deviation, ![\sigma = 4 mg/mi](https://tex.z-dn.net/?f=%5Csigma%20%3D%204%20mg%2Fmi)
probability that a single car of this model emits more than 86 mg/mi of NOX + NMOG
![P(X > x ) = P(z > \frac{x - \mu}{\sigma})](https://tex.z-dn.net/?f=P%28X%20%3E%20x%20%29%20%3D%20P%28z%20%3E%20%5Cfrac%7Bx%20-%20%5Cmu%7D%7B%5Csigma%7D%29)
![P(X > x ) =1 - P(z < \frac{x - \mu}{\sigma})](https://tex.z-dn.net/?f=P%28X%20%3E%20x%20%29%20%3D1%20-%20%20P%28z%20%3C%20%5Cfrac%7Bx%20-%20%5Cmu%7D%7B%5Csigma%7D%29)
![P(X > 86 ) =1 - P(z < \frac{86 - 80}{4})](https://tex.z-dn.net/?f=P%28X%20%3E%2086%20%29%20%3D1%20-%20%20P%28z%20%3C%20%5Cfrac%7B86%20-%2080%7D%7B4%7D%29)
P(X > 86) = 1 - P(z < 1.5)
From the standard normal table, P(z < 1.5) = 0.9332
P(X > 86) = 1 - 0.9332
P(X > 86) = 0.0668
Answer:
4 and 5 hope this helps
Step-by-step explanation:
Answer:
The rate of the boat in still water is 40 miles per hour
The rate of the current is 10 miles per hour
Step-by-step explanation:
we know that
The speed or rate is equal to divide the distance by the time
Let
x ----> the rate of the current (miles per hour)
y ----> the rate of the boat in still water (miles per hour)
we have that
<em>going upstream </em>
----> equation A
<em>going downstream </em>
----> equation B
Solve the system of equations by elimination
Adds equation A and equation B
<em>Find the value of x </em>
therefore
The rate of the boat in still water is 40 miles per hour
The rate of the current is 10 miles per hour
9/20 as a decimal= 0.45
In short, your answer is 0.45!
Hope this helps!