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1 year ago

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Julio wants to break his school’s scoring record of 864 points during his 24-game basketball season. During the first 8 games of

the season, he scored a total of 256 points. Which inequality can be used to find x, the number of points Julio must average per game during the rest of the season to break the record?

2 answers:

Scilla [17]1 year ago

8 0

Answer:

x≥38 points

Step-by-step explanation:

Julio wants to break his school’s scoring record of 864 points during his 24-game basketball season. During the first 8 games of the season, he scored a total of 256 points. Which inequality can be used to find x, the number of points Julio must average per game during the rest of the season to break the record?

julio has a 24 game basketball season.

he has played 8, it means there are 16 more games to go

therefore=

he has scored 256 nts, wic means there are still 608 points to go .

864-256=608

x is the points he more score per every game.

16x=608

to break the records ,he must score an extra point 1

so 16x≥608

x≥38

Lyrx [107]1 year ago

7 0

Answer:

256 + 16x > 864

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A sample of 900 college freshmen were randomly selected for a national survey. Among the survey participants, 372 students were

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Answer:

a) $ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423$

b) $0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371$

Step-by-step explanation:

1) Data given and notation

n=900 represent the random sample taken

X=372 represent the students were pursuing liberal arts degrees

$\hat p=\frac{372}{900}=0.413$ estimated proportion of students were pursuing liberal arts degrees

$\alpha=0.01$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

p= population proportion of students were pursuing liberal arts degrees

2) Solution to the problem

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

The margin of error is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If we replace we have:

$ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423$

And replacing into the confidence interval formula we got:

$0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371$

$0.413 + 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.455$

And the 99% confidence interval would be given (0.371;0.455).

We are confident (99%) that about 37.1% to 45.5% of students were pursuing liberal arts degrees.

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Answer:

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Step-by-step explanation:

The hypothesis for the test above will be stated as follows :

The claim to be tested is the alternative hypothesis, which is the negation of the Null hypothesis

H0 : μ < 14

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Data given and notation

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Concepts and formulas to use

We need to conduct a hypothesis in order to check if the two proportions differs , the system of hypothesis would be:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

We need to apply a z test to compare proportions, and the statistic is given by:

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Calculate the statistic

Replacing in formula (1) the values obtained we got this:

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Statistical decision

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