Answer:
Probability that the average mileage of the fleet is greater than 30.7 mpg is 0.7454.
Step-by-step explanation:
We are given that a certain car model has a mean gas mileage of 31 miles per gallon (mpg) with a standard deviation 3 mpg.
A pizza delivery company buys 43 of these cars.
<em>Let </em>
<em> = sample average mileage of the fleet </em>
<em />
The z-score probability distribution of sample average is given by;
Z = 
~ N(0,1)
where, 
= mean gas mileage = 31 miles per gallon (mpg)
= standard deviation = 3 mpg
n = sample of cars = 43
So, probability that the average mileage of the fleet is greater than 30.7 mpg is given by = P(<em> </em>> 30.7 mpg)
P(<em> </em>> 30.7 mpg) = P( > 
) = P(Z > -0.66) = P(Z < 0.66)
= 0.7454
<em>Because in z table area of P(Z > -x) is same as area of P(Z < x). Also, the above probability is calculated using z table by looking at value of x = 0.66 in the z table which have an area of 0.7454.
</em>
Therefore, probability that the average mileage of the fleet is greater than 30.7 mpg is 0.7454.
The -3 shifts the graph 3 units to the right
and The + 9 moves the graph upwards 9 units
Its B
Answer:
2,648.448 cubic feet
Step-by-step explanation:
The cargo box of a truck is a rectangular prism. The volume of a rectangular prism is l*w*h.
If the length is 39.6 ft, the width is 7.6 ft and the height is 8.8 ft then the volume is (39.6)(7.6)(8.8) = 2,648.448 cubic feet
<span><span>35−<span>10/5</span></span>+<span><span>(<span>5+3</span>)</span><span>(4)
</span></span></span><span>=<span><span>35−2</span>+<span><span>(<span>5+3</span>)</span><span>(4)
</span></span></span></span><span>=<span>33+<span><span>(<span>5+3</span>)</span><span>(4)
</span></span></span></span><span>=<span>33+<span><span>(8)</span><span>(4)
</span></span></span></span><span>=<span>33+32
</span></span><span>=<span>65
</span></span>
