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Akimi4 [234]
3 years ago
5

Find the mean, median, mode, and range of the data 10, 13, 7, 6, 9, 4, 6, 3, 5​

Mathematics
1 answer:
elena-s [515]3 years ago
7 0
Median is 6 range is 9 mode is 6 and mean is 7
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First make a substitution and then use integration by parts to evaluate the integral. (Use C for the constant of integration.) x
e-lub [12.9K]

Answer:

(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C

Step-by-step explanation:

Ok, so we start by setting the integral up. The integral we need to solve is:

\int x ln(5+x)dx

so according to the instructions of the problem, we need to start by using some substitution. The substitution will be done as follows:

U=5+x

du=dx

x=U-5

so when substituting the integral will look like this:

\int (U-5) ln(U)dU

now we can go ahead and integrate by parts, remember the integration by parts formula looks like this:

\int (pq')=pq-\int qp'

so we must define p, q, p' and q':

p=ln U

p'=\frac{1}{U}dU

q=\frac{U^{2}}{2}-5U

q'=U-5

and now we plug these into the formula:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int \frac{\frac{U^{2}}{2}-5U}{U}dU

Which simplifies to:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int (\frac{U}{2}-5)dU

Which solves to:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\frac{U^{2}}{4}+5U+C

so we can substitute U back, so we get:

\int xln(x+5)dU=(\frac{(x+5)^{2}}{2}-5(x+5))ln(x+5)-\frac{(x+5)^{2}}{4}+5(x+5)+C

and now we can simplify:

\int xln(x+5)dU=(\frac{x^{2}}{2}+5x+\frac{25}{2}-25-5x)ln(5+x)-\frac{x^{2}+10x+25}{4}+25+5x+C

\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}-\frac{5x}{2}-\frac{25}{4}+25+5x+C

\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C

notice how all the constants were combined into one big constant C.

7 0
3 years ago
Harry buys 9 dozen eggs how many eggs does he have in all
bixtya [17]
A dozen= 12
12x9= 108 eggs
4 0
4 years ago
Read 2 more answers
I need help with this one
Reika [66]

Answer:

look it up

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Nvm i got it thank you to everyone who helped :)
torisob [31]

It's always good to help fellow brainly users, we all must acheive A's lolol


6 0
3 years ago
The distribution of passenger vehicle speeds traveling on a certain freeway in California is nearly normal with a mean of 73.7 m
umka2103 [35]

Answer:

a)  0.2347, 0.0007

b) 1.30

c) 0.63

Step-by-step explanation:

Given –  

Mean = 73.4

Sigma = 4.7

Probability that the speed is greater than 70 is  

P (X>70) = P (z> (70-73.4)/4.7) = P (z>-0.72) = 0.7653

a) The probability that a car is not speeding = 1 - 0.7653 = 0.2347

The probability that all 5 cars are not speeding = (0.2347)5 = 0.0007

b) E (X) = 1/p = 1/0.7653 = 1.30  

c) Standard deviation = Sqrt (1-p/p) = (1-0.7653)/0.7653 = 0.63

5 0
3 years ago
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