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Deffense [45]
2 years ago
7

How do I solve the equation 6=-w/8

Mathematics
2 answers:
MrRa [10]2 years ago
4 0

Answer:

-48

Step-by-step explanation:

Multiply both sides of the equation by -8

-8(6)  = \frac{-w}{8}(-8)

-48 = w

Ivenika [448]2 years ago
3 0

Answer:

-48

Step-by-step explanation:

i got the rught anser on my test

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Evaluate using integration by parts ​
PolarNik [594]

Rather than carrying out IBP several times, let's establish a more general result. Let

I(n)=\displaystyle\int x^ne^x\,\mathrm dx

One round of IBP, setting

u=x^n\implies\mathrm du=nx^{n-1}\,\mathrm dx

\mathrm dv=e^x\,\mathrm dx\implies v=e^x

gives

\displaystyle I(n)=x^ne^x-n\int x^{n-1}e^x\,\mathrm dx

I(n)=x^ne^x-nI(n-1)

This is called a power-reduction formula. We could try solving for I(n) explicitly, but no need. n=5 is small enough to just expand I(5) as much as we need to.

I(5)=x^5e^x-5I(4)

I(5)=x^5e^x-5(x^4e^x-4I(3))=(x-5)x^4e^x+20I(3)

I(5)=(x-5)x^4e^x+20(x^3e^x-3I(2))=(x^2-5x+20)x^3e^x-60I(2)

I(5)=(x^2-5x+20)x^3e^x-60(x^2e^x-2I(1))=(x^3-5x^2+20x-60)x^2e^x+120I(1)

I(5)=(x^3-5x^2+20x-60)x^2e^x+120(xe^x-I(0))

Finally,

I(0)=\displaystyle\int e^x\,\mathrm dx=e^x+C

so we end up with

I(5)=(x^4-5x^3+20x^2-60x+120)xe^x-120e^x+C

I(5)=(x^5-5x^4+20x^3-60x^2+120x-120)e^x+C

and the antiderivative is

\displaystyle\int2x^5e^x\,\mathrm dx=(2x^5-10x^4+40x^3-120x^2+240x-240)e^x+C

8 0
3 years ago
What is the simplified form of the expressions?<br> a. 2−3<br> b. (5.5) 0
Semmy [17]
A= 1/8
B=1
HOPE THIS HELPS
3 0
3 years ago
Please Help! Multiple Choice!
Katyanochek1 [597]

Using the z-distribution, the p-value would be given as follows:

b) 0.0086.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis we test if the means are equal, hence:

H_0: \mu_D - \mu_C = 0

At the alternative hypothesis, it is tested if they are different, hence:

H_1: \mu_D - \mu_C \neq 0

<h3>What are the mean and the standard error for the distribution of differences?</h3>

For each sample, they are given as follows:

  • \mu_D = 12, s_D = \frac{5.2}{\sqrt{73}} = 0.6086
  • \mu_C = 14, s_C = \frac{4.1}{\sqrt{81}} = 0.4556

Hence, for the distribution of differences, they are given by:

  • \overline{x} = 12 - 14 = -2.
  • s = \sqrt{0.6086^2 + 0.4556^2} = 0.76

<h3>What is the test statistic?</h3>

The test statistic is given by:

z = \frac{\overline{x} - \mu}{s}

In which \mu = 0 is the value tested at the null hypothesis.

Hence:

z = \frac{\overline{x} - \mu}{s}

z = \frac{-2 - 0}{0.76}

z = -2.63.

Using a z-distribution calculator, for a two-tailed test, with z = -2.63, the p-value is of 0.0086.

Hence option B is correct.

More can be learned about the z-distribution at brainly.com/question/13873630

#SPJ1

4 0
2 years ago
you have 33 math and science problems for homework. you have 7 more math problems than science problems. how many problems do yo
Verdich [7]
You have 33 math problems and 26 science.
3 0
3 years ago
Read 2 more answers
PLEASE HELP!! brainliest if correct!!
Sonja [21]
Graph #1: No
Graph #2: Yes
Graph #3: No
Graph #4: No
Graph #5: No
Graph #6: Yes

Reasoning:

The vertical line test is a test that determines wether a graph is a function or a relation. The vertical line test shows that if you construct a vertical line through any point on the graph, then the vertical line should only intercept the graph once for it to be a function.
6 0
2 years ago
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