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anygoal [31]
3 years ago
9

A company has determined that its weekly profit is a function of the number of items that it sells. Which equation could represe

nt the weekly profit in thousands of dollars, y, when the company sells x items? y squared = 4 x squared minus 100 y = negative x squared + 50 x minus 300 x = negative y squared minus 400 x squared = negative 6 y squared + 200
Mathematics
2 answers:
stira [4]3 years ago
8 0

Answer:

Its B

Step-by-step explanation:

Give the credit to the guy above me :)

Cerrena [4.2K]3 years ago
7 0

Answer:

B. y= -x^2 + 50x - 300

Step-by-step explanation:

Options given

A. y^2=4x^2 - 100

B. y= -x^2 + 50x - 300

C. x= -y^2 - 400

D. x^2= -6y^2 + 200

We are to find profits (y) function in thousands of dollars if the company sells x items.

We will use elimination method to eliminate the wrong options above in order to find the correct answer

Option A. y^2= 4x^2 - 100 is used to find the square of the profit function

Option B. y= -x^2 + 50x - 300 is used to calculate profit (y) function

Option C. x= -y^2 - 400 is for the calculation of the items x sold

Option D. x^2= -6y^2 + 200 is to find items x sold squared.

Since, we are required to find the profit (y) function, option A, C and D will be eliminated.

We are left with option B which is the correct answer to the question.

The weekly profits (y) function in thousands of dollars when the company sells x items is

B. y= -x^2 + 50x - 300

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X and y are normal random variables with e(x) = 2, v(x) = 5, e(y) = 6, v(y) = 8 and cov(x,y)=2. determine the following: e(3x 2y
andriy [413]

The result for the given normal random variables are as follows;

a. E(3X + 2Y) = 18

b. V(3X + 2Y) = 77

c. P(3X + 2Y < 18) = 0.5

d. P(3X + 2Y < 28) = 0.8729

<h3>What is normal random variables?</h3>

Any normally distributed random variable having mean = 0 and standard deviation = 1 is referred to as a standard normal random variable. The letter Z will always be used to represent it.

Now, according to the question;

The given normal random variables are;

E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8.

Part a.

Consider E(3X + 2Y)

\begin{aligned}E(3 X+2 Y) &=3 E(X)+2 E(Y) \\&=(3) (2)+(2)(6 )\\&=18\end{aligned}

Part b.

Consider V(3X + 2Y)

\begin{aligned}V(3 X+2 Y) &=3^{2} V(X)+2^{2} V(Y) \\&=(9)(5)+(4)(8) \\&=77\end{aligned}

Part c.

Consider P(3X + 2Y < 18)

A normal random variable is also linear combination of two independent normal random variables.

3 X+2 Y \sim N(18,77)

Thus,

P(3 X+2 Y < 18)=0.5

Part d.

Consider P(3X + 2Y < 28)

Z=\frac{(3 X+2 Y-18)}{\sqrt{77}}

\begin{aligned} P(3X + 2Y < 28)&=P\left(\frac{3 X+2 Y-18}{\sqrt{77}} < \frac{28-18}{\sqrt{77}}\right) \\&=P(Z < 1.14) \\&=0.8729\end{aligned}

Therefore, the values for the given normal random variables are found.

To know more about the normal random variables, here

brainly.com/question/23836881

#SPJ4

The correct question is-

X and Y are independent, normal random variables with E(X) = 2, V(X) = 5, E(Y) = 6, and V(Y) = 8. Determine the following:

a. E(3X + 2Y)

b. V(3X + 2Y)

c. P(3X + 2Y < 18)

d. P(3X + 2Y < 28)

8 0
2 years ago
Which is the correct answer? <br> A) 3.41 m<br> B) 2.29 m<br> C) 1.59 m<br> D) 2.71 m
PtichkaEL [24]

Answer:

its b or c ( im not for sure tho)

Step-by-step explanation:

5 0
3 years ago
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