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1 year ago

I don't have a question but I was wondering if you could check my work

Mathematics

1 answer:

cestrela7 [59]1 year ago

8 0

The mean of a data sample can be found by adding all the terms in the data sample and dividing the result by the number of terms. In the first case we have the following data sample:

$12,8,15,11,9$

The mean of this sample is:

$\operatorname{mean}=\frac{12+8+15+11+9}{5}=\frac{55}{5}=11$

The mean absolute deviation is the sum of the differences between each data value and the mean, divided by the number of data values, the formula is:

$MAD=\frac{\sum ^{}_{}\mleft|x_i-\bar{x}\mright|}{n}$

Therefore, we must calculate the positive difference between each value and the mean, add them up, and divide the result by the number of data points in the sample. In the case of the first sample we get:

$\text{MAD}=\frac{(12-11)+(11-8)+(15-11)+(11-9)}{5}$

Solving the operation:

$\text{MAD}=\frac{10}{5}=2$

If done this way, all the results for each data set must be correct.

You might be interested in

Type the correct answer for each.

xxTIMURxx [149]

If point $\left(x,\ \frac{\sqrt{7}}{3}\right)$ is on the unit ircle, then:

$x^2+\left( \frac{ \sqrt{7} }{3} \right)^2=1 \\ \\ \Rightarrow x^2+ \frac{7}{9} =1 \\ \\ \Rightarrow x^2=1- \frac{7}{9} = \frac{2}{9} \\ \\ \Rightarrow x= \sqrt{ \frac{2}{9} } = \frac{ \sqrt{2} }{3}$

Since, the point is in the second quadrant, x is negative.

Thus, $(x,\ y)=\left(x,\ \frac{\sqrt{7}}{3}\right)=\left(-\frac{ \sqrt{2} }{3},\ \frac{\sqrt{7}}{3}\right)$

Part A:

$3 \sqrt{7} =6\left( \frac{ \sqrt{7} }{2} \right) \\ \\ =6\left( \frac{ \sqrt{7} }{3} \cdot \frac{3}{ \sqrt{2} } \right) \\ \\ =6\left( \frac{ \frac{ \sqrt{7} }{3} }{ \frac{ \sqrt{2} }{3} } \right)=-6\left( \frac{ \frac{ \sqrt{7} }{3} }{ -\frac{ \sqrt{2} }{3} } \right) \\ \\ =-6\left( \frac{y}{x} \right)=-6\tan\theta$

Therefore, $-6\tan\theta=3\sqrt{7}$ .

Part B:

$- \frac{ \sqrt{7} }{2} =- \frac{ \sqrt{7} }{3} \cdot \frac{3}{ \sqrt{2} } \\ \\ =- \frac{ \frac{ \sqrt{7} }{3} }{ \frac{ \sqrt{2} }{3} } = \frac{ \frac{ \sqrt{7} }{3} }{ -\frac{ \sqrt{2} }{3} } = \frac{y}{x} \\ \\ =\tan\theta$

Therefore, $\tan\theta=- \frac{ \sqrt{7} }{2}$

8 0

3 years ago

I don’t know how to solve this

g100num [7]

The two equations graphs intersect and the points where they are touching are belonging to both graphs therefore solutions for both equations.

(2) points (x,y) are
(-1,0) (-1)^2. +0^2=1; 1=1 ✔️
0=-1+1; 0=0✔️
(0, 1). (0)^2. +1^2=1; 1=1 ✔️
1=0+1; 1=1 ✔️

6 0

2 years ago

Gemma recently rode her bicycle to visit her friend who lives 6 miles away. On her way there, her average speed was 16 miles per

DochEvi [55]

first speed --- x mph

return speed -- x+16 mph

6/x + 6/(x+16) = 1

times each term by x(x+16)

6(x+16) + 6x = x(x+16)

x^2 + 4x - 96 = 0

(x-8)(x+12) = 0

x = 8 or x is a negative

her first speed was 8 mph

her return speed was 24 mph

check:

6/8 + 6/24 = 1 , that's good!

8 0

2 years ago

What is x. pls help me

ehidna [41]

The answer might be 40

8 0

2 years ago

Read 2 more answers

This is for algebra 2 and I got stuck on this anyone know this?

AnnZ [28]

You just simplify the fraction (so, 8/6 which is 4/3)

So your comes 4(square root of 35)/ 3 (square root of 15)

5 0

2 years ago