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cricket20 [7]
3 years ago
15

Write an integer for this situation 1 . $50 deposit

Mathematics
2 answers:
Ierofanga [76]3 years ago
6 0
50 i think i need points lol
valentina_108 [34]3 years ago
6 0
Answer: 50
explanation: integers are numbers that can be written without a fraction or decimal. so if it was a withdrawal of $50 the integer would be -50 but a deposit would be a positive 50.

hope this helps
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I really don’t get this , it’s graphing linear inequalities
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Next, we need to figure out which side to shade. To do so, simply pick any point (I like to use the point (0,0) because it makes the calculations easy) and see whether it satisfies the inequality. If it does, shade the side with that point, and if not, shade the opposite side of the graph.

Here we see that the point (0,0) does satisfy the inequality, since -(0) + 4(0) is 0, and 0 ≥ -8, so the top half of the graph should be shaded (see the second image).

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Midpoint is the center point on the line segment <br> True or false?
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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)

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2 years ago
Small frustrations and minor irritations are called
Sonbull [250]
Annoying????????????
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3 years ago
EXPERT HELP, I'LL GIVE BRAINLIEST:
DanielleElmas [232]

Answer:

B or A.

Step-by-step explanation:

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3 years ago
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