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4 years ago

What is the solution of (8x-8)3/2=64

Mathematics

2 answers:

jasenka [17]4 years ago

8 0

Well x=19/3
Decimal form is 0.6333333....

Vera_Pavlovna [14]4 years ago

4 0

Answer: Is A : X = 3

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Set A of six numbers has a standard deviation of 3 and set B of four numbers has a standard deviation of 5. Both sets of numbers

Alenkinab [10]

Given:

$\sigma_A=3$

$n_A=6$

$\sigma_B=5$

$n_B=4$

$\overline{x}_A=\overline{x}_B$

To find:

The variance. of combined set.

Solution:

Formula for variance is

$\sigma^2=\dfrac{\sum (x_i-\overline{x})^2}{n}$ ...(i)

Using (i), we get

$\sigma_A^2=\dfrac{\sum (x_i-\overline{x}_A)^2}{n_A}$

$(3)^2=\dfrac{\sum (x_i-\overline{x}_A)^2}{6}$

$9=\dfrac{\sum (x_i-\overline{x}_A)^2}{6}$

$54=\sum (x_i-\overline{x}_A)^2$

Similarly,

$\sigma_B^2=\dfrac{\sum (x_i-\overline{x}_B)^2}{n_B}$

$(5)^2=\dfrac{\sum (x_i-\overline{x}_B)^2}{4}$

$25=\dfrac{\sum (x_i-\overline{x}_B)^2}{4}$

$100=\sum (x_i-\overline{x}_B)^2$

Now, after combining both sets, we get

$\sigma^2=\dfrac{\sum (x_i-\overline{x}_A)^2+\sum (x_i-\overline{x}_B)^2}{n_A+n_B}$

$\sigma^2=\dfrac{54+100}{6+4}$

$\sigma^2=\dfrac{154}{10}$

$\sigma^2=15.4$

Therefore, the variance of combined set is 15.4.

7 0

3 years ago

If Fred is x years old now, and his sister is half as old, how old will his sister be in 5 years?

Elanso [62]

Answer: x/2+5 years old

Step-by-step explanation:

His sister would be x/2+5 years old. x/2 is his sisters age right now, and +5 is years later!

6 0

3 years ago

Does the remainder round up or down to get an average whole number

sergij07 [2.7K]

Answer:

It rounds up

Step-by-step explanation:

It rounds up because it just does

8 0

4 years ago

What is the solution to the system of equations?

Dovator [93]

Answer:

Step-by-step explanation:

$\left\{\begin{array}{ccc}x+3y+2z=8&(1)\\3x+y+3z=-10&(2)\\-2x-2y-z=10&(3)\end{array}\right\qquad\text{subtract both sides of the equations (1) from (2)}\\\\\underline{-\left\{\begin{array}{ccc}3x+y+3z=-10\\x+3y+2z=8\end{array}\right }\\.\qquad2x-2y+z=-18\qquad(4)\qquad\text{add both sides of the equations (3) and (4)}\\\\\underline{+\left\{\begin{array}{ccc}-2x-2y-z=10\\2x-2y+z=-18\end{array}\right}\\.\qquad-4y=-8\qquad\text{divide both sides by (-4)}\\.\qquad\qquad y=2\qquad\text{put the value of y to (1) and (3)}$

$\left\{\begin{array}{ccc}x+3(2)+2z=8\\-2x-2(2)-z=10\end{array}\right\\\left\{\begin{array}{ccc}x+6+2z=8&\text{subtract 6 from both sides}\\-2x-4-z=10&\text{add 4 to both sides}\end{array}\right\\\left\{\begin{array}{ccc}x+2z=2&\text{multiply both sides by 2}\\-2x-z=14\end{array}\right\\\underline{+\left\{\begin{array}{ccc}2x+4z=4\\-2x-z=14\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad3z=18\qquad\text{divide both sides by 3}\\.\qquad\qquad z=6\qquad\text{put the value of z to the first equation}$

$x+2(6)=2\\x+12=2\qquad\text{subtract 10 from both sides}\\x=-10$

4 0

3 years ago

I NEED HELP ASAP. PLEASE AND THANK YOU

Andrej [43]

Answer:

1, 2, and 5

Step-by-step explanation:

5 0

3 years ago