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IgorLugansk [536]
3 years ago
13

What is the MAD of the data set? {10, 9, 16, 5, 10}

Mathematics
1 answer:
Oksanka [162]3 years ago
5 0
First, you'd find the mean of all the numbers by doing 10+9+16+5+10 = 50/5 = 10. Second, you'd subtract all values from the mean by doing 
16 - 10 = 6
10 - 10 = 0
10 - 10 = 0
10 - 9 = 1
<span>10 - 5 = 5
Lastly, finding the mean of all 5 numbers would be, 2.4! Hope this helps!! :D</span>
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Solve the system of equations by<br> y= 5x + 8<br> y = 8x + 9
rodikova [14]
You should’ve put a picture to make it easier for us
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3 years ago
marcel rides his bicycle to school and back on Tuesdays and Thursdays. He lives 3.62 kilometers away from school. Marcels gym te
Nonamiya [84]

It should be 14.48 please correct me if wrong!

5 0
3 years ago
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Solve: x^+4/x-1 = 5/x-1
AlexFokin [52]
The first step for solving this equation is to determine the defined range.
\frac{ x^{4}  }{x-1} = \frac{5}{x-1}, x ≠ 1
Remember that when the denominators of both fractions are the same,, you need to set the numerators equal. This will look like the following:
x^{4} = 5
Take the root of both sides of the equation and remember to use both positive and negative roots.
x +/- \sqrt[4]{5}
Separate the solutions.
x = \sqrt[4]{5}           , x ≠ 1
x = -\sqrt[4]{5} 
Check if the solution is in the defined range.
x = \sqrt[4]{5} 
x = -\sqrt[4]{5} 
This means that the final solution to your question are the following:
x = \sqrt[4]{5}          
x = -\sqrt[4]{5} 
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7 0
3 years ago
Explain answer ! <br><br> Will give brainlst
Ne4ueva [31]

Answer:

x = 4, y = 2

Step-by-step explanation:

Start by multiplying the first equation by 2:

2x + 2y = 12 --> 4x + 4y = 24

Subtract the second from the first:

   4x + 4y = 24

-   5x + 4y = 28

4x - 5x = -x

4y = 4y = 0

24 - 28 = -4

so you end with -x + 0 = -4

Solve for x to get x = 4

Plug x = 4 back into 2x + 2y = 12 to find y.

2(4) + 2y = 12

8 + 2y = 12

2y = 4

y = 2

5 0
3 years ago
I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur
Butoxors [25]
\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx

Setting x=5\sin\theta, you have \mathrm dx=5\cos\theta\,\mathrm d\theta. Then the integral becomes

\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta
\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta
\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta
\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta

Now, \sqrt{x^2}=|x| in general. But since we want our substitution x=5\sin\theta to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means \theta=\sin^{-1}\dfrac x5, which implies that \left|\dfrac x5\right|\le1, or equivalently that |\theta|\le\dfrac\pi2. Over this domain, \cos\theta\ge0, so \sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta.

Long story short, this allows us to go from

\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta

to

\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta
\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta

Then integrate term-by-term to get

\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)
=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C

Now undo the substitution to get the antiderivative back in terms of x.

=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C
4 0
2 years ago
Read 2 more answers
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