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Fed [463]
2 years ago
13

Use fundamental theorem of calculus to find derivative of the function LOOK AT PHOTO

Mathematics
1 answer:
kykrilka [37]2 years ago
8 0

Let c > 0. Then split the integral at t = c to write

f(x) = \displaystyle \int_{\ln(x)}^{\frac1x} (t + \sin(t)) \, dt = \int_c^{\frac1x} (t + \sin(t)) \, dt - \int_c^{\ln(x)} (t + \sin(t)) \, dt

By the FTC, the derivative is

\displaystyle \frac{df}{dx} = \left(\frac1x + \sin\left(\frac1x\right)\right) \frac{d}{dx}\left[\frac1x\right] - (\ln(x) + \sin(\ln(x))) \frac{d}{dx}\left[\ln(x)\right] \\\\ = -\frac1{x^2} \left(\frac1x + \sin\left(\frac1x\right)\right) - \frac1x (\ln(x) + \sin(\ln(x))) \\\\ = -\frac1{x^3} - \frac{\sin\left(\frac1x\right)}{x^2} - \frac{\ln(x)}x - \frac{\sin(\ln(x))}x \\\\ = -\frac{1 + x\sin\left(\frac1x\right) + x^2\ln(x) + x^2 \sin(\ln(x))}{x^3}

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Let the sample size of leg strengths to be 7 and the sample mean and sample standard deviation be 630 watts and 32 watts, respec
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Answer:

a. There is_<u><em>sufficient</em></u> evidence that the leg

C. 0.010 < P-value < 0.025

b. Power of test = 1- β=0.2066

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Step-by-step explanation:

We formulate the null and alternative hypotheses as

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Here n= 7 and significance level ∝= 0.005

Critical value for a right tailed test with 6 df is 1.9432

Sample Standard deviation = s= 32

Sample size= n= 7

Sample Mean =x`= 630

Degrees of freedom = df = n-1= 7-1= 6

The test statistic used here is

Z = x- x`/ s/√n

Z= 630-600 / 32 / √7

Z= 2.4797= 2.48

P- value = 0.0023890 > ∝ reject the null hypothesis.

so it lies between 0.010 < P-value < 0.025

b) Power of test if true strength is 610 watts.

For  a right tailed test value of z is = ± 1.645

P (type II error) β= P (Z< Z∝-x- x`/ s/√n)

Z = x- x`/ s/√n

Z= 610-630 / 32 / √7

Z=0.826

P (type II error) β= P (Z< 1.645-0.826)

= P (Z> 0.818)

= 0.7933

Power of test = 1- β=0.2066

(c)

true mean = 610

hypothesis mean = 600

standard deviation= 32

power = β=0.9

Z∝= 1.645

Zβ= 1.282

Sample size needed

n=( (Z∝ +Zβ )*s/ SE)²

n=  ((1.645+1.282) 32/ 10)²

Putting the values  and solving we get 87.69

So the sample size is 88

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