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klemol [59]
3 years ago
13

Jeremy is having his basketball team over for an end-of-season party. He has $80 to spend on pizza and soda. Pizzas cost $13.99

each and cases of soda cost $4.49 each. Write an inequality to represent this scenario. Make sure to define your variables.
Mathematics
1 answer:
Lapatulllka [165]3 years ago
5 0

Answer:

13.99p + 4.49s ≤ 80

Step-by-step explanation:

Pizza is p and soda is s.  If he has only $80 to spend on a combination of both, he cannot obviously spend more than that.  Therefore, the inequality sign is "less than or equal to".  He can spend 80, but he can also spend less than 80.  He cannot spend anything over that, since he doesn't have it to spend!

If pizzas are represented by p, and we know the cost of each one is 13.99, we represent one pizza as 13.99p; if the cost of sida is 4.49 per case, we represent one case as 4.49s.  

The sum of these cannot exceed 80, so the inequality, then, is:

13.99p + 4.49s ≤ 80

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A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro
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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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