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kolezko [41]
3 years ago
15

How do i find the area between the two curves

Mathematics
1 answer:
viktelen [127]3 years ago
4 0
A=22/10

A=integral(a,b) [f(x)-g(x)]dx

Since the function is even (the function is mirrored over the y axis) we can evaluate the integral from 0 to 1 and then multiply our answer by 2 since we have the same area on each side of the y axis.

We get A=2*int.(0, 1)[(x^2)-(-2x^4)]dx

Now we can integrate by term.

2*[int.(0, 1)[x^2]dx+int(0, 1)[2x^4]dx]

Now factor out constants.

2*[int(0,1)[x^2]dx+2int(0,1)[x^4]dx]

Now integrate.

2*[(x^3/3)|(0,1) + 2*(x^5/5)|(0,1)]

Now solve.

2*[(1/3)+2*(1/5)]

=22/10

Hope you can decipher what I wrote!
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Step-by-step explanation:

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Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were 3 mill
abruzzese [7]

Answer:

V' = -0.11552 *V\\= -0.11552(1.8)\\ \\=-0.20794 million per year

Step-by-step explanation:

Given that oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were 3 million barrels of oil in the well; six years later 1,500,000 barrels remain.

i.e. if V stands for volume of oil, then

V' =kV\\dv/V= kdt\\ln V = kt+C\\V = Ae^{kt}

To find A and k

V(0) = A = 3 million

Hence V = 3e^{kt}

V(6) = 1.5

i.e. 1.5 = 3e^{6k}\\ln 1.5 =ln 3 +6k\\k = -0.11552

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a) Using the above value of k , we have

V' = -0.11552 *V\\= -0.11552(1.8)\\ \\=-0.20794 million per year.

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3 years ago
Solve the quadratic equation for x. What is one of the roots? (x − 5)2 = 9
Sonja [21]

Answer:

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7 0
3 years ago
Read 2 more answers
"A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation is 20 percen
Marizza181 [45]

Answer:

The company should take a sample of 148 boxes.

Step-by-step explanation:

Hello!

The cable TV company whats to know what sample size to take to estimate the proportion/percentage of cable boxes in use during an evening hour.

They estimated a "pilot" proportion of p'=0.20

And using a 90% confidence level the CI should have a margin of error of 2% (0.02).

The CI for the population proportion is made using an approximation of the standard normal distribution, and its structure is "point estimation" ± "margin of error"

[p' ± Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }]

Where

p' is the sample proportion/point estimator of the population proportion

Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} } is the margin of error (d) of the confidence interval.

Z_{1-\alpha /2} = Z_{1-0.05} = Z_{0.95}= 1.648

So

d= Z_{1-\alpha /2} * \sqrt{\frac{p'(1-p')}{n} }

d *Z_{1-\alpha /2}= \sqrt{\frac{p'(1-p')}{n} }

(d*Z_{1-\alpha /2})^2= \frac{p'(1-p')}{n}

n*(d*Z_{1-\alpha /2})^2= p'(1-p')

n= \frac{p'(1-p')}{(d*Z_{1-\alpha /2})^2}

n= \frac{0.2(1-0.2)}{(0.02*1.648)^2}

n= 147.28 ≅ 148 boxes.

I hope it helps!

3 0
3 years ago
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