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Marianna [84]
3 years ago
14

Find r(x + 1) if r(x) = x3 + x + 1

Mathematics
2 answers:
suter [353]3 years ago
7 0

Substitute x + 1 as x in the equation of the function:

r(x)=x^3+x+1\\\\r(x+1)=(x+1)^3+(x+1)+1=x^3+3(x^2)(1)+3(x)(1^2)+1^3+x+1+1\\\\=x^3+3x^2+3x+x+2=x^3+3x^2+4x+2

Used

(a+b)^3=a^3+3a^3b+3ab^3+b^3

Ad libitum [116K]3 years ago
4 0

Answer:

<em>r(x + 1) = x^3 + 3x^2 + 4x + 3</em>

Step-by-step explanation:

r(x) = x3 + x + 1

Replace x with x + 1 is r(x) and simplify.

r(x + 1) = (x + 1)^3 + x + 1 + 1

= (x^2 + 2x + 1)(x + 1) + x + 2

= x^3 + x^2 + 2x^2 + 2x + x + 1 + x + 2

r(x + 1) = x^3 + 3x^2 + 4x + 3

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Step-by-step explanation:

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A manufacturer produces crankshafts for an automobile engine. the crankshafts wear after 100,000 miles (0.0001 inch) is of inter
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Part A:

Significant level:

<span>α = 0.05

Null and alternative hypothesis:

</span><span>h0 : μ = 3 vs h1: μ ≠ 3

Test statistics:

z= \frac{\bar{x}-\mu}{\sigma/\sqrt{n}}  \\  \\ = \frac{2.78-3}{0.9/\sqrt{15}}  \\  \\ = \frac{-0.22}{0.2324} =-0.9467

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Part B:

The power of the test is given by:

\beta=\phi\left(Z_{0.025}+ \frac{3-3.25}{0.9/\sqrt{15}}\right) -\phi\left(-Z_{0.025}+ \frac{3-3.25}{0.9/\sqrt{15}}\right) \\  \\ =\phi\left(1.96+ \frac{-0.25}{0.2324} \right)-\phi\left(-1.96+ \frac{-0.25}{0.2324} \right)=\phi(0.8842)-\phi(-3.0358) \\  \\ =0.8117-0.0012=0.8105

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Part C:

</span>The <span>sample size that would be required to detect a true mean of 3.75 if we wanted the power to be at least 0.9 is obtained as follows:

1-0.9=\phi\left(Z_{0.025}+ \frac{3-3.75}{0.9/\sqrt{n}}\right) -\phi\left(-Z_{0.025}+ \frac{3-3.75}{0.9/\sqrt{n}}\right) \\ \\ \Rightarrow0.1=\phi\left(1.96+ \frac{-0.75}{0.9/\sqrt{n}}\right)-\phi\left(-1.96+ \frac{-0.75}{0.9/\sqrt{n}} \right) \\  \\ =\phi\left(1.96+(-3.2415)\right)-\phi\left(1.96+(-3.2415)\right) \\  \\ \Rightarrow\frac{-0.75}{0.9/\sqrt{n}}=-3.2415 \\ \\ \Rightarrow\frac{0.9}{\sqrt{n}}=\frac{-0.75}{-3.2415}=0.2314 \\  \\ \Rightarrow\sqrt{n}=\frac{0.9}{0.2314}=3.8898

\Rightarrow n=(3.8898)^2=15.13

Therefore, the </span>s<span>ample size that would be required to detect a true mean of 3.75 if we wanted the power to be at least 0.9 is 16.</span>
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