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2 years ago

The first person to answer this question and get it right will be marked as Brainiest!

Mathematics

2 answers:

Advocard [28]2 years ago

8 0

Answer:

There is not enough info to draw valid conclusion

Step-by-step explanation:

Hope this helps

iris [78.8K]2 years ago

8 0

Answer:

Not enough information to draw a valid conclusion.

Step-by-step explanation:

As the mean is 17 the total value if you for example have 5 numbers will be 5*17 = 85.

I can't see that you can conclude 1, 3 or 4:

For example the numbers could be 2, 8, 24, 25 and 26.

- you could have a mode if you include 2 16's instead of the 8 and 24 but you cant say that's not probable.

The median is not 17 nor is it a typical value.

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Find the mean, median, mode, and range of the data set.

kompoz [17]

Answer:

mean 11.14

Median 10

mode 10.

Explanation

Mean is sum of data divide by number of data.

Mode is the most appearing number.

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2 years ago

Simplify the following 2(×+8k)+2k=

omeli [17]

Answer:

2x + 18k = X= -9k

Step-by-step explanation:

you have to solve the rational equation by combining expressions and isolating the variable x.

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2 years ago

A college freshman started the year with $4000 in spendingmoney and after three months, she has $2800 left. Assume thatshe will

Bezzdna [24]

She would have $800 left.
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6 0

2 years ago

Read 2 more answers

How do I find the integral<br> ∫10(x−1)(x2+9)dxint10/((x-1)(x^2+9))dx ?

defon

$\int\frac{10}{(x-1)(x^2+9)}\ dx=(*)\\\\\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}=\frac{A(x^2+9)+(Bx+C)(x-1)}{(x-1)(x^2+9)}\\\\=\frac{Ax^2+9A+Bx^2-Bx+Cx-C}{(x-1)(x^2+9)}=\frac{(A+B)x^2+(-B+C)x+(9A-C)}{(x-1)(x^2+9)}\\\Updownarrow\\10=(A+B)x^2+(-B+C)x+(9A-C)\\\Updownarrow\\A+B=0\ and\ -B+C=0\ and\ 9A-C=10\\A=-B\ and\ C=B\to9(-B)-B=10\to-10B=10\to B=-1\\A=-(-1)=1\ and\ C=-1$

$(*)=\int\left(\frac{1}{x-1}+\frac{-x-1}{x^2+9}\right)\ dx=\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)\ dx\\\\=\int\frac{1}{x-1}\ dx-\int\frac{x+1}{x^2+9}\ dx=\int\frac{1}{x-1}-\int\frac{x}{x^2+9}\ dx-\int\frac{1}{x^2+9}\ dx=(**)\\\\\#1\ \int\frac{1}{x-1}\ dx\Rightarrow\left|\begin{array}{ccc}x-1=t\\dx=dt\end{array}\right|\Rightarrow\int\frac{1}{t}\ dt=lnt+C_1=ln(x-1)+C_1$

$\#2\ \int\frac{x}{x^2+9}\ dx\Rightarrow \left|\begin{array}{ccc}x^2+9=u\\2x\ dx=du\\x\ dx=\frac{1}{2}\ du\end{array}\right|\Rightarrow\int\left(\frac{1}{2}\cdot\frac{1}{u}\right)\ du=\frac{1}{2}\int\frac{1}{u}\ du\\\\\\=\frac{1}{2}ln(u)+C_2=\frac{1}{2}ln(x^2+9)+C_2$

$\#3\ \int\frac{1}{x^2+9}\ dx=\int\frac{1}{x^2+3^2}\ dx=\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3\\\\therefore:\\\\\#1;\ \#2;\ \#3\Rightarrow(**)=ln(x-1)+C_1-\frac{1}{2}ln(x^2+9)+C_2-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3$

$\boxed{=ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C}$

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3 years ago

How to know what to do next in a geometry two column proof?

GREYUIT [131]

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2 years ago