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
Basically we just use this inequality:
70 + 0.02x = 45 + 0.04x
25 + 0.02x = 0.04x
25 = 0.02x
x = 1250
It will be the better choice if you make 1250 copies or less
Answer:
(
+
2
)
(
+
4
)
Step-by-step explanation:
10 in; if the original 100% was 5 in, then you multiply that by 2.
61/5
multiply the 12 and the one on the bottom and then add the one on the top..
or do u mean
__1
12---
__5