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)
![\bf a)\qquad \cfrac{dP}{dx}=0.06(200)-3+ \cfrac{2.4}{\sqrt[5]{200}} \\\\\\ b)\qquad \cfrac{dP}{dx}=0.06(2000)-3+\cfrac{2.4}{\sqrt[5]{2000}} \\\\\\ c)\qquad \cfrac{dP}{dx}=0.06(5000)-3+\cfrac{2.4}{\sqrt[5]{5000}} \\\\\\ d)\qquad \cfrac{dP}{dx}=0.06(10000)-3+\cfrac{2.4}{\sqrt[5]{10000}}](https://tex.z-dn.net/?f=%5Cbf%20a%29%5Cqquad%20%0A%5Ccfrac%7BdP%7D%7Bdx%7D%3D0.06%28200%29-3%2B%20%5Ccfrac%7B2.4%7D%7B%5Csqrt%5B5%5D%7B200%7D%7D%0A%5C%5C%5C%5C%5C%5C%0Ab%29%5Cqquad%20%5Ccfrac%7BdP%7D%7Bdx%7D%3D0.06%282000%29-3%2B%5Ccfrac%7B2.4%7D%7B%5Csqrt%5B5%5D%7B2000%7D%7D%0A%5C%5C%5C%5C%5C%5C%0Ac%29%5Cqquad%20%5Ccfrac%7BdP%7D%7Bdx%7D%3D0.06%285000%29-3%2B%5Ccfrac%7B2.4%7D%7B%5Csqrt%5B5%5D%7B5000%7D%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%29%5Cqquad%20%5Ccfrac%7BdP%7D%7Bdx%7D%3D0.06%2810000%29-3%2B%5Ccfrac%7B2.4%7D%7B%5Csqrt%5B5%5D%7B10000%7D%7D)
Descibe the mothod of solving problems with numbers in standard form ?? I have 5 MIN
1 answer:
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Megan:
x to the one third power =
<span>x to the one twelfth power = </span>
<span>The quantity of x to the one third power, over x to the one twelfth power is:
</span>
<span>
Since </span>
then
Now, just subtract exponents:
1/3 - 1/12 = 4/12 - 1/12 = 3/12 = 1/4
Julie:
x times x to the second times x to the fifth = x * x² * x⁵
<span>The thirty second root of the quantity of x times x to the second times x to the fifth is
</span>![\sqrt[32]{x* x^{2} * x^{5} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B32%5D%7Bx%2A%20x%5E%7B2%7D%20%2A%20x%5E%7B5%7D%20%7D%20)
<span>
Since </span>
Then![\sqrt[32]{x* x^{2} * x^{5} }= \sqrt[32]{ x^{1+2+5} } =\sqrt[32]{ x^{8} }](https://tex.z-dn.net/?f=%5Csqrt%5B32%5D%7Bx%2A%20x%5E%7B2%7D%20%2A%20x%5E%7B5%7D%20%7D%3D%20%5Csqrt%5B32%5D%7B%20x%5E%7B1%2B2%2B5%7D%20%7D%20%3D%5Csqrt%5B32%5D%7B%20x%5E%7B8%7D%20%7D)
Since![\sqrt[n]{x^{m}} = x^{m/n} }](https://tex.z-dn.net/?f=%20%5Csqrt%5Bn%5D%7Bx%5E%7Bm%7D%7D%20%3D%20x%5E%7Bm%2Fn%7D%20%7D%20)
Then![\sqrt[32]{ x^{8} }= x^{8/32} = x^{1/4}](https://tex.z-dn.net/?f=%5Csqrt%5B32%5D%7B%20x%5E%7B8%7D%20%7D%3D%20x%5E%7B8%2F32%7D%20%3D%20x%5E%7B1%2F4%7D%20)
Since both Megan and Julie got the same result, it can be concluded that their expressions are equivalent.
x to the one third power =
<span>x to the one twelfth power = </span>
<span>The quantity of x to the one third power, over x to the one twelfth power is:
</span>
<span>
Since </span>
then
Now, just subtract exponents:
1/3 - 1/12 = 4/12 - 1/12 = 3/12 = 1/4
Julie:
x times x to the second times x to the fifth = x * x² * x⁵
<span>The thirty second root of the quantity of x times x to the second times x to the fifth is
</span>
<span>
Since </span>
Then
Since
Then
Since both Megan and Julie got the same result, it can be concluded that their expressions are equivalent.
ANSWER
The length of the altitude is units
<u>EXPLANATION</u>
The altitude of a triangle is the vertical height of the triangle.
From the diagram the altitude is
Method 1: We can use Pythagoras theorem to find
We take the square root of both sides,
Method 2 Using Trigonometry
Answer:
Thomas had the better year relative to their peers.
Step-by-step explanation:
<u>The complete question is</u>: One year Thomas had the lowest ERA (earned-run average, mean number of runs yielded per nine innings pitched) of any male pitcher at his school, with an ERA of 3.31. Also, Karla had the lowest ERA of any female pitcher at the school with an ERA of 3.02. For the males, the mean ERA was 4.837 and the standard deviation was 0.541. For the females, the mean ERA was 4.533 and the standard deviation was 0.539. Find their respective z-scores. Which player had the better year relative to their peers, or ? (Note: In general, the lower the ERA, the better the pitcher.)
We are given that for the males, the mean ERA was 4.837 and the standard deviation was 0.541. For the females, the mean ERA was 4.533 and the standard deviation was 0.539.
As, we know that the z-score is calculated by the following formula;
Z = ~ N(0,1)
where, = population mean
= standard deviation
Now, firstly we will calculate the z score for Thomas;
z-score =
= = -2.823
{Here, the mean ERA for the males was 4.837 and the standard deviation was 0.541}
Similarly, we will calculate the z score for Karla;
z-score =
= = -2.807
{Here, the mean ERA for the females was 4.533 and the standard deviation was 0.539}
Now, it is stated in the question that the lower the ERA, the better the pitcher.
So, we can clearly see that Thomas had a lower ERA of z-score as -2.823 < -2.807. This means that Thomas had the better year relative to their peers.