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3 years ago

-33 < -7n - 12 < -26

1 answer:

3 0

-33 < -7n - 12 < -26
First step is to add 12 to both sides.
-33 < -7n < -26-21 < -7n < -14

Now divide both sides by -7.
3 < n < 2

Because you divided by a negative number though, you need to switch up the signs. Both the < will turn into >.
3 > n > 2

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An NBA fan named Mark claims that there are more fouls called on his team 1 point

inn [45]

Answer:

$t=\frac{12.2-11.5}{\frac{1.6}{\sqrt{34}}}=2.551$

$df = n-1=34-1=33$

$p_v =P(t_{(33)}>2.551)=0.008$

Since the p value is less than the significance level of 0.05 we have enough evidence to reject the null hypothesis in favor of the claim

And the best conclusion for this case would be:

b)The p-value is 0.008, indicating sufficient evidence for his claim.

Step-by-step explanation:

Information provided

$\bar X=12.2$ represent the sample mean fould against

$s=1.6$ represent the sample standard deviation

$n=34$ sample size

represent the value that we want to test

$\alpha=0.05$ 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)

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean is higher than 11.5 fouls per game:

Null hypothesis: $\mu \leq 11.5$

Alternative hypothesis: $\mu > 11.5$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

The statistic is given by:

$t=\frac{12.2-11.5}{\frac{1.6}{\sqrt{34}}}=2.551$

P value

The degreed of freedom are given by:

$df = n-1=34-1=33$

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

$p_v =P(t_{(33)}>2.551)=0.008$

Since the p value is less than the significance level of 0.05 we have enough evidence to reject the null hypothesis in favor of the claim

And the best conclusion for this case would be:

b)The p-value is 0.008, indicating sufficient evidence for his claim.

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A new car is purchased for 15500 dollars. The value of the car depreciates at 7.5% per year. What will the value of the car be,

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Answer:

$\$9,709.17$

Step-by-step explanation:

we know that

The formula to calculate the depreciated value is equal to

$V=P(1-r)^{x}$

where

V is the depreciated value

P is the original value

r is the rate of depreciation in decimal

x is Number of Time Periods

in this problem we have

$P=\$15,500\\r=7.5\%=0.075\\x=6\ years$

substitute

$V=\$15,500(1-0.075)^{6}=\$9,709.17$

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