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3 years ago

HELP please guys ................

Mathematics

2 answers:

alisha [4.7K]3 years ago

4 0

Answer:

2nd choice

Step-by-step explanation:

Montano1993 [528]3 years ago

4 0

Im pretty sure its 1!

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Brad says that if a second number is 125% of the first number than the second number must be 75% of the second number is he corr

elena-s [515]

Answer:

Incorrect.

Step-by-step explanation:

Let, the first number is x, and the second number is y. Given, the second number is 125% of the first number. So, the equation will be

y = 1.25 x

Or, y= 5/4x

Now, multiplying it with 4/5, we get, x= 4/5y

Or, x= 0.80 y

But, Brad is saying the first number must be 75% of the second number. But here we see that the first number is 80% of the second number. That's why Brad is incorrect.

5 0

3 years ago

Evaluate 8^-2 pls only real answers

Burka [1]

Step-by-step explanation:

By Law of Indices,

8^(-2) = 8^0 / 8^2

= 1 / 64.

8 0

3 years ago

You mow your neighbor’a lawn in 5 hours and earn $45. what is your hourly wage

Talja [164]

$9.00 per hour. take 45 and divide by 5 you get 9.

6 0

3 years ago

Read 2 more answers

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

<img src="https://tex.z-dn.net/?f=7x%20%2B%202%20%3D%202" id="TexFormula1" title="7x + 2 = 2" alt="7x + 2 = 2" align="absmiddle"

Anastaziya [24]

Easy.

7x + 2 = 2

7x = 2 - 2

7x = 0

x = 0 / 7

x = 0

Verify this by putting x = 0 ...

7 ( 0 ) + 2 = 2

0 + 2 = 2

2 = 2

Hence, "0" is the final answer.

Hope helps ........

4 0

3 years ago