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Novosadov [1.4K]
3 years ago
15

Please help confused

Mathematics
1 answer:
bija089 [108]3 years ago
7 0
The mean is the average, 
add the numbers in the set and divide by the quantity of numbers:
9 + 15 + 22 + 45 + 50 +60 + 65 = 266
266 / 7 = 38
 the mean = 38

the absolute deviation is subtracting the number from the mean: ( all numbers are positive)

38 - 9 = 29

38 - 15 = 23

38-22 = 16

45-38 = 7

50 -38 = 12

60-38 = 22

65-38 = 27


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BRAINLIEST Look at the pattern and decide what number comes next.<br> 2 7 3 8 4 9 5 ?
baherus [9]

Answer:

10

Step-by-step explanation:

2 + 5 = 7

3 + 5 = 8

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5 + 5 = 10

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Read 2 more answers
find the equation for the line that passes through (10, 10) that has slope -1/3. give your answer in point-slope form. you do no
chubhunter [2.5K]

Y - 10 =(-1/3) ( X - 10 )

Y = ( -1/3) X + ( 40/3)

Step-by-step explanation:

Step 1:

The basic form of Slope Intercept equation is

Y - Y_{1}  = m ( X - X_{1})

Step 2:

Points are (10,10) and m=-1/3

Y-10 = (-1/3)(X-10)

Y = ( (-X + 10 )/3) + 10

Y    =( - X / 3 )  +( 40 / 3)

3 0
3 years ago
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
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The answer is B. The shape is staying the exact same, it’s just moving.
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