I don’t understand inequalities that much so can someone help fast I will give 10 points since is the max I can give

Mathematics

1 answer:

Tatiana [17]3 years ago

8 0

its either b or a since we don't exactly where x is but the arrow stops at 5 so it must be b

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What is the domain and range?

Rasek [7]

Answer:

Domain:- All real numbers

Range:-[3,infinite)

Step-by-step explanation:

In the function there's no value of x for which it is not defined thus domain is R.

Now a modulus will have always it's minimum value 0 thus minimum value of function is 0+3=3. And Max value of a modulus is infinite so infinite+3=infinite.

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3 years ago

Solve by elimination -8x+8y=19 -4x+4y=8

Harrizon [31]

Answer: the first one :x=y+ −19 /8

the second one: x=y−2

Step-by-step explanation:

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3 years ago

How to evaluate this question

Harlamova29_29 [7]

The answer is 16

Refer image for working :-

Drop comments if any doubt arises.

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3 years ago

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Help me on what you can preferably all

Arturiano [62]

When dividing exponents you subtract the exponents. When multiplying you add them. When raising an exponent by an exponent you multiply them.

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4 years ago

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

Daniel [21]

Part A:

Significant level:

α = 0.05

Null and alternative hypothesis:

h0 : μ = 3 vs h1: μ ≠ 3

Test statistics:

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

P-value:

P(-0.9467) = 0.1719

Since the test is a two-tailed test, p-value = 2(0.1719) = 0.3438

Conclusion:

Since the p-value is greater than the significant level, we fail to reject the null hypothesis and conclude that there is no sufficient evidence that the true mean is different from 3.

Part B:

The power of the test is given by:

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

Therefore, the power of the test if μ = 3.25 is 0.8105.

Part C:

The 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 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 16.

8 0

3 years ago