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Norma-Jean [14]

4 years ago

11

During their first round of gulf, Imani was 7 strokes over par and Peter was 8 strokes below par. Name the integers representing

the situation.

1 answer:

Serjik [45]4 years ago

5 0

-1 >2&×8;@*÷92*1]+?×<;2>+,×

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What's 5*(-2/3+3x) ?

sdas [7]

<span>5*(-2/3+3x)
= 15x - 10/3

hope it helps</span>

7 0

3 years ago

Nina makes $12 per hour at her job. She also earns commission. Last week she earned $78 in commission and made a total of $444.

Vesnalui [34]

Answer:

30.5 hours

Step-by-step explanation:

444-78=366

366/12=30.5

30.5 x 12= 366

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3 years ago

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Use the interactive to graph a line with a slope of zero and passing through the 0,4

Gwar [14]

Answer:

You need to draw a horizontal line that goes through the point (0,4).

Step-by-step explanation:

Slope of zero means it does not go up or down. It looks like this ↔, but stretched out to the two sides of the graph.

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3 years ago

Ill give brainlist please i beg ulol

denis23 [38]

GHE i think because it’s the same degrees

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3 years ago

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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