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Lelechka [254]
3 years ago
12

The volume of the sphere is 500/3pi cubic units. What is the value of x

Mathematics
1 answer:
Sphinxa [80]3 years ago
5 0

Answer:

Step-by-step explanation:

Formula for the volume of a sphere is V = (4/3) (π) r³

                                                                        3V           4²

and so the cube of the radius, "r," is   r³ = ------------- * -----

                                                                           4           4²

Taking the cube root of both sides, we get        

                                                ∛[3V / 4²]                3V

and so the radius, "r," is   r = ------------------ =  ∛ ( --------- )  = (1/4)*∛(3*v)

                                                      ∛[4³]                   4³

Then

r = (1/4)*∛(3*V), after substituting 500/(3π) for V, becomes:

r = (1/4)*∛[ 3*500/3π ]  =   (1/4)*∛[ 500/π ]

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The manufacturer of an airport baggage scanning machine claims it can handle an average of530 bags per hour.(a-1) At α = .05 in
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Answer:

b. H0: μ < 530. Reject H0 if tcalc > -1.753

Step-by-step explanation:

1) Data given and notation  

\bar X=507 represent the sample mean

s=47 represent the sample standard deviation  

n=16 sample size  

\mu_o =530 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is less than 530 (left tailed tes), the system of hypothesis would be:  

Null hypothesis:\mu \geq 530  

Alternative hypothesis:\mu < 530  

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t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

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We can replace in formula (1) the info given like this:  

t=\frac{507-530}{\frac{47}{\sqrt{16}}}=-1.957

Critical value

Since we are conducting a left tailed test we need to first find the degrees of freedom for the statistic given by:

df=n-1=16-1=15

Now we need to look on the t distribution with 15 degrees of freedom that accumulates 0.05 of the area on the left area. And on this case the critical value would be t_{\alpha}=-1/753.

And we can use the following excel code to find it: "=T.INV(0.05,15)"  

So then the correct rejection zone for H0 would be: Reject H0 if t_{calc}

P-value  

Since is a one side left tailed test the p value would be:  

p_v =P(t_{(15)}  

Conclusion  

If we compare the p value and the significance level for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean is significantly lower than 530 at 5% of significance.  

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Answer:

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Step-by-step explanation:

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