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2 years ago

Find the mean absolute deviation of the data below: 7,4,7,7,2,10,12,8,7,5

Mathematics

1 answer:

Virty [35]2 years ago

5 0

Answer:

1.9

Step by step answer

step1 finding the mean

(7+4+7+7+2+10+12+8+7+5)/10 = 6.9

step2 Difference between each data

Difference between 7 and 6.9 is 0.1

Difference between 4 and 6.9 is 2.9

Difference between 7 and 6.9 is 0.1

Difference between 2 and 6.9 is 4.9

Difference between 10 and 6.9 is 3.1

Difference between 12 and 6.9 is 5.1

Difference between 8 and 6.9 is 1.1

Difference between 7 and 6.9 is 0.1

Difference between 5 and 6.9 is 1.9

step3 add all the Difference

0.1 + 2.9+0.1+0.1+4.9+3.1+5.1+1.1+1.9 = 19.4

step4 divde it by n

19.4/10 = 1.9

so the answer is 1.9

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3 years ago

Read 2 more answers

Help me please!! Also explain you answer since this is a study guide for a important test!!

il63 [147K]

Answer:

<h3>the answer is A</h3>

Step-by-step explanation:

8x+6is equivalent to-7(1-x)+x+13 because

-7(1-x)+x+13

we use-7to multiply the numbers in the bracket which is equal to -7+7x(minus ❌times minus =+)

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3 years ago

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Let X have the uniform distribution U(0, 2) and let the conditional distribution of Y , given that X = x, be U(0, x). Find the j

Elina [12.6K]

Answer:

$f(x,y) = \frac{1}{x} \frac{1}{2}= \frac{1}{2x} , 0\leq x \leq 2 , 0\leq y \leq x$

$E(Y|x) = \int_{x=y}^2 y \frac{1}{x} dx= y ln x \Big|_{x=y}^2 =y ln 2 -y ln y = y(1-lny) \$

Step-by-step explanation:

We have two random variables X and Y. $X \sim Unif(0,2)$ and given that X=x, Y has uniform distribution (0,x)

From the definition of the uniform distribution we have the densities for each random variable given by:

$f_X (x) =\frac{1}{2} , 0\leq x\leq 2$

$f_{Y|X} (y|x) = \frac{1}{x}, 0\leq y \leq x$

And on this case we can find the joint density with the following formula:

$f(x,y) = f_{Y|X}(y|x) f_X (x)$

And multiplying the densities we got this:

$f(x,y) = \frac{1}{x} \frac{1}{2}= \frac{1}{2x} , 0\leq x \leq 2 , 0\leq y \leq x$

Now with the joint density we can find the expected value E(Y|x) with the following formula:

$E(Y|x) = \int y f_{Y|X}(y|x)dx$

And replacing we got:

$E(Y|x) = \int_{x=y}^2 y \frac{1}{x} dx= y ln x \Big|_{x=y}^2 =y ln 2 -y ln y = y(1-lny) \$

5 0

4 years ago

A cylinder with a 2 centimeter diameter is drilled through the triangular prism shown on the left find the volume of the prism

saw5 [17]

Answer:

The volume of a prism on its own = 248.5cm^3

The volume of a cylinder = 87.92cm^3

Step-by-step explanation:

The volume does not change for the prism as the hole was drilled.

We can show the drilled space with v(l) and (p) for prism V

v(l) 248.5p - 87.92p + 87.92p

= 248.5p

3 0

3 years ago

Evaluate the expression for the given value of the variable. d=75 2d+−45d−−−−−√3=

Ahat [919]

Answer:

The expression for the given value of the variable d=75 is, 135

Step-by-step explanation:

Given the expression:

$2d+\sqrt[3]{-45d}$ ....[1]

We have to find the value for the given expression when d = 75

Substitute the value of d in [1] we have

$2(75)+\sqrt[3]{-45(75)}$

⇒ $150+\sqrt[3]{-3375}$

⇒ $150-15$

Simplify:

⇒ $135$

Therefore, the expression for the given value of the variable d=75 is, 135

3 0

4 years ago