Answer: b. 134
Step-by-step explanation:
Given : A minimum usual value of 135.8 and a maximum usual value of 155.9.
Let x denotes a usual value.
i.e. 135.8< x < 155.9
Therefore , the interval for the usual values is [135.8, 155.9] .
If interval for any usual value is [135.8, 155.9] , then any value should lie in this otherwise we call it unusual.
Let's check all options
a. 137 ,
since 135.8< 137 < 155.9
So , it is usual.
b. 134
since 134<135.8 (Minimum value)
So , it is unusual.
c. 146
since 135.8< 146 < 155.9
So , it is usual.
d. 155
since 135.8< 1155 < 155.9
So , it is usual.
Hence, the correct answer is b. 134 .
Keeping in mind that for a cost C(x) and profit P(x) and revenue R(x), the marginal cost, marginal profit and marginal revenue are respectively dC/dx, dP/dx and dR/dx, then
![\bf P(x)=0.03x^2-3x+3x^{0.8}-4400 \\\\\\ \stackrel{marginal~profit}{\cfrac{dP}{dx}}=0.06x-3+2.4x^{-0.2} \\\\\\ \cfrac{dP}{dx}=0.06x-3+2.4\cdot \cfrac{1}{x^{0.2}}\implies \cfrac{dP}{dx}=0.06x-3+2.4\cdot \cfrac{1}{x^{\frac{1}{5}}} \\\\\\ \cfrac{dP}{dx}=0.06x-3+\cfrac{2.4}{\sqrt[5]{x}}](https://tex.z-dn.net/?f=%5Cbf%20P%28x%29%3D0.03x%5E2-3x%2B3x%5E%7B0.8%7D-4400%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bmarginal~profit%7D%7B%5Ccfrac%7BdP%7D%7Bdx%7D%7D%3D0.06x-3%2B2.4x%5E%7B-0.2%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7BdP%7D%7Bdx%7D%3D0.06x-3%2B2.4%5Ccdot%20%5Ccfrac%7B1%7D%7Bx%5E%7B0.2%7D%7D%5Cimplies%20%5Ccfrac%7BdP%7D%7Bdx%7D%3D0.06x-3%2B2.4%5Ccdot%20%5Ccfrac%7B1%7D%7Bx%5E%7B%5Cfrac%7B1%7D%7B5%7D%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7BdP%7D%7Bdx%7D%3D0.06x-3%2B%5Ccfrac%7B2.4%7D%7B%5Csqrt%5B5%5D%7Bx%7D%7D)
The answer here is 275 miles, because it is going at 55 miles per hour. you can find this out by dividing the miles by the hours. Hope this helps.
Answer:The answer is D
Step-by-step explanation: