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Paul [167]
3 years ago
14

Find some means. Suppose that X is a random variable with mean 30 and standard deviation 4. Also suppose that Y is a random vari

able with mean 50 and standard deviation 8. Find the mean of the random variable Z for each of the following cases. Be sure to show your work.

Mathematics
1 answer:
mr_godi [17]3 years ago
6 0

COMPLETE QUESTION:

Find some means. Suppose that X is a random variable with mean 30 and standard deviation 4. Also suppose that Y is a random variable with mean 50 and standard deviation 8. Find the mean of the random variable Z for each of the following cases. Be sure to show your work.

a) Z = 35-10X

b) Z = 12X-5

c) Z = X+Y

d) Z = X-Y

e) Z = 2X+2Y

Answer:

a) -265

b) 355

c) 80

d) -20

e) 40

Step-by-step explanation:

The attachment below contains the whole calculations that provides for the answers above

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MAVERICK [17]

Answer:

1.22222

Step-by-step explanation:

Use the equation y2-y1/x2-x1=m to find the slope.

Now, just insert your known values.

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1.22222 is the slope.

Hope this helps!

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3 years ago
How to find x using trigonometry
I am Lyosha [343]
Soh Cah Toa

Sine of the angle = opposite side over the hypotenuse

sin(42°) = 6.5/h
rearrange, solve for h
h = 6.5/sin(42°)
h = 9.7 cm

For the other triangle, the angle is unknown. I'd split it into two right triangles; down angle x since it is an isosceles it will be bisected into two equivalent triangles.

The hypotenuse we just found is now split in half. 9.7/2 = 4.9 base of the new smaller right triangle.

The new hypotenuse of the smaller triangle is 7.4 cm

Then you have..

sin(x/2) = 4.9/7.4
sin(x/2) = 0.67
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There are many ways to solve this problem. This is just what I thought of first using trig.
3 0
3 years ago
Which set of numbers does not contain the value below?
Lerok [7]

Answer:

b. Pure Imaginary Numbers.

Step-by-step explanation:

(i \sqrt{3})^2 = i^2 \times (\sqrt{3})^2 = -1 \times 3 = -3

-3 is an integer, a complex number, and a rational number.

Answer: b. Pure Imaginary Numbers.

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3 years ago
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