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Zina [86]
3 years ago
15

Given the function LaTeX: f\left(x\right)=-5x^2-x+20f ( x ) = − 5 x 2 − x + 20, find LaTeX: f\left(2\right)f ( 2 ). Group of ans

wer choices -2 -28 -12 38
Mathematics
1 answer:
BartSMP [9]3 years ago
6 0

The value of f(2) is -2 ⇒ 1st answer

Step-by-step explanation:

The form of the quadratic function is;

f(x) = ax² + bx + c, where

  • a is the coefficient of x²
  • b is the coefficient of x
  • c is the numerical term
  • x is the domain of the function and f(x) is the range of the function

∵ f(x) = -5x² - x + 20

- f(2) means value f(x) at x = 2

∵ x = 2

- Substitute x in the function by 2 to find f(2)

∵ f(2) = -5(2)² - (2) + 20

∴ f(2) = -5(4) - 2 + 20

∴ f(2) = -20 - 2 + 20

∴ f(2) = -2

The value of f(2) is -2

Learn more:

You can learn more about the function in brainly.com/question/12363217

#LearnwithBrainly

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I have 100 items of product in stock. The probability mass function for the product's demand D is P(D=90)=P(D=100)=P(D=110)=1/3.
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Answer:

The probability mass function for the items sold is

P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.

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b) The probability mass function for the unfilled demand due to lack of stock is

P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.

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Step-by-step explanation:

If the demand is higher than 100, then you will sell 100 items only. Thus, there is a probability of 1/3+1/3 = 2/3 that you will sell 100 items, while there is a probability of 1/3 that you will sell 90.

The probability mass function for the items sold is

P_X(k) = \left \{ {\frac{1}{3} \, \, \, {k=90} \atop \, \frac{2}{3} \, \, \, {k=100}} \right.

The mean is 1/3 * 90 + 2/3 * 100 = 290/3 = 96.667

The variance is V(X) = E(X²)-E(X)² = (1/3*90² + 2/3*100²) - (290/3)² = 200/9 = 22.222

b) If order to be unfilled demand, you need to have a demand of 110, which happens with probability 1/3. In that case, the value of the variable, lets call it Y, that counts the amount of unfilled demand due to lack of stock is 110-100 = 10. In any other case, the value of Y is 0, which would happen with probability 1-1/3 = 2/3. Thus

P_Y(k) = \left \{ {\frac{2}{3} \, \, \, {k=0} \atop \, \frac{1}{3} \, \, \, {k=10}} \right.

The mean is 2/3 * 0 + 1/3 * 10 = 10/3 = 3.333

The variance is 2/3*0² + 1/3*10² = 100/3 = 33.333

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