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marissa [1.9K]
3 years ago
6

Use integration by parts to find the integrals in Exercise. ∫(4x-12)e-8x dx.

Mathematics
1 answer:
stepladder [879]3 years ago
5 0

Answer:

e(2x^{2} -12x)-4x^{2}+C

Step-by-step explanation:

We have been given an indefinite integral as \int \left(4x-12\right)e-8x\:dx. We are asked to find the given integral.

Let us solve our given problem.

\int \left(4x-12\right)e\:dx-\int 8x\:dx

Take out constant:

e\int \left(4x-12\right)\:dx-8\int x\:dx

e(\int 4x\:dx -\int 12\right\:dx)-8\int x\:dx

e(\frac{4x^{1+1}}{2} -12x)-8*\frac{x^{1+1}}{1+1}+C

e(\frac{4x^{2}}{2} -12x)-8*\frac{x^{2}}{2}+C

e(2x^{2} -12x)-4x^{2}+C

Therefore, our required integral would be e(2x^{2} -12x)-4x^{2}+C.

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Answer:

B is True

A, C. D are false

Step-by-step explanation:

Given :

Sample size, n = 120

Mean diameter, m = 10

Standard deviation, s = 0.24

Confidence level, Zcritical ; Z0.05/2 = Z0.025 = 1.96

The confidence interval represents how the true mean value compares to a set of values around the mean computed from a set of sample drawn from the population.

The population here is N = 10000

To obtain

Confidence interval (C. I) :

Mean ± margin of error

Margin of Error = Zcritical * s/sqrt(n)

Margin of Error = 1.96 * 0.24/sqrt(120)

Confidence interval for the 10,000 ball bearing :

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Hence. The confidence interval defined as :

10 ± 1.96 * (0.24) / sqrt(120) is the 95% confidence interval for the mean diameter of the 10,000 bearings in the box.

4 0
3 years ago
Simplify the expression. -2/3c -9/5 + 14c + 3/10
Andrew [12]

Answer:

13 \frac{1}{3} c - 1 \frac{1}{2}

Step-by-step explanation:

We want to combine alike terms so

-2/3c+14c & -9/5+3/10

14c-2/3

\frac{14}{1} c -  \frac{2}{3} c

Multiply 14/1 by 3/3 to get the denominators the same

\frac{42}{3} c -  \frac{2}{3} c

Subtract

\frac{40}{3} c

simplify

13 \frac{1}{3} c

now

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Multiply 9/5 by 2/2

\frac{3}{10}  -  \frac{ 18}{10}

Subtract

- \frac{15}{10}

Simplify

- 1 \frac{1}{2}

put both together

13 \frac{1}{3}c  - 1 \frac{1}{2}

Hope this helps! If you have any questions on how I got my answer feel free to ask. Stay safe!

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3 years ago
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