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BlackZzzverrR [31]
2 years ago
14

In a certain Algebra 2 class of 26 students, 5 of them play basketball and 12 of them play baseball. There are 12 students who p

lay neither sport. What is the probability that a student chosen randomly from the class plays basketball or baseball?
Mathematics
1 answer:
loris [4]2 years ago
6 0

Answer:\dfrac{7}{13}

Step-by-step explanation:

There are only two games basketball and baseball.

Any student who plays could play basketball or baseball.

Given that there are 26 students in total.

Given that there are 12 students who don't play any game at all.

So,there are 26-12=14 students who play play some baseball or basketball.

Probability=\frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}

The required probability is \frac{14}{26}=\frac{7}{13}

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-Dominant- [34]

Answer:

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

Step-by-step explanation:

Data given and notation  

\bar X=1.6 represent the sample mean

s=0.46 represent the sample deviation

n=32 sample size  

\mu_o =1.35 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 true mean is different from 1.35 in/year, the system of hypothesis would be:  

Null hypothesis:\mu =1.35  

Alternative hypothesis:\mu \neq 1.35  

The statistic is given by:

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

Calculate the statistic  

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

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

P-value  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

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3 years ago
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Answer:

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6. 4

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