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# Tutor profile: Daniel T.

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Daniel T.
College Mathematics Instrutor for Four Years
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## Questions

### Subject:Trigonometry

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Question:

Find the exact value for cos⁡(-3π/8).

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Daniel T.

Using the half-angle identity for the Cosine function, we have that cos⁡(-3π/8)=cos⁡(3π/8)=cos⁡((3π/4)/2)=√((1+cos⁡(3π/4))/2)=√((1-√2/2)/2)=√(((2-√2)/2)/2)=√((2-√2)/4) cos⁡(-3π/8)=√(2-√2) /2

### Subject:Calculus

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Question:

Household electricity is supplied in the form of alternating current that varies from -155 V to 155 V with a frequency of 60 cycles per second (Hz). The voltage is thus given by the equation E(t)=155 sin⁡(120πt) where t is the time in seconds. Voltmeters read the RMS (root-mean-square) voltage, which is the square root of the average value of [E(t)]^2 over one cycle. a. Calculate the RMS voltage of household current. b. Many electric stoves require an RMS voltage of 220 V. Find the corresponding amplitude A needed for the voltage E(t)=A sin⁡(120πt).

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Daniel T.

Since the frequency is 60 cycles per second, one cycle is 1/60 of a second. Therefore, the RMS voltage of household current is √(1/(1/60-0) ∫_0^(1/60)▒〖[155 sin⁡(120πt) ]^2 dt〗)=√(1,441,500∫_0^(1/60)▒〖sin^2⁡(120πt)dt〗)=√(1,441,500∫_0^(1/60)▒〖(1-cos⁡(240πt))/2 dt〗)=√(720,750∫_0^(1/60)▒(1-cos⁡(240πt) )dt)=√(720,450[t-sin⁡(240πt)/240π]_0^(1/60) )=√(720,450(1/60) )≈109.6 V Therefore, household current has an RMS voltage of about 110 V. Now, replacing 155 in the integral above with A, we have that √(A^2/2)=220⇒A≈311 Therefore, the amplitude needs to be about 310 for an electric stove requiring 220 V.

### Subject:Statistics

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Question:

A scientist estimates that the mean carbon dioxide emissions per country in a recent year are greater than 150 megatons. A random sample of 42 countries has a mean amount of carbon dioxide emissions of 77.3 megatons and a standard deviation of 259.8 megatons. Assume the population standard deviation is 816 megatons. At α=0.06, can you support the scientist’s claim?

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Daniel T.

The claim is given by μ>150. This is not a statement of equality, so it represents the alternate hypothesis. Therefore, we have that H_0:μ≤150 H_a:μ>150 (claim) Note that this is a right-tailed test by the alternate hypothesis. Since σ is known, a standard z-test for μ is used. For α=0.06, the positive critical z-score is to the right of 94% of the data. This is the only critical value necessary since the test is right-tailed. The critical value is (using the calculator command invNorm) invNorm(0.94,0,1)≈1.555. Therefore, the rejection region is z>1.555. z=(x-μ)/(σ/√n)=(77.3-150)/(816/√42)≈-0.58 Since -0.58 is not greater than 1.555, it does not lie in the rejection region. Therefore, we fail to reject to null hypothesis. Therefore, there is not enough evidence to support the claim at the 6% level that μ>150.

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