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Radda [10]
3 years ago
7

Jane is organizing a fundraiser to buy a ping-pong table for the community center. The table costs $500.00. Jane is asking contr

ibutors to pay for an equal share of the cost of the table. She already has five contributors lined up. What function models the cost per share, C, with respect to the number of new contributors, n? How many new contributors must Jane recruit in order for the cost of each share to be $20.00?

Mathematics
1 answer:
kogti [31]3 years ago
4 0

Answer:

Function: C = 500 / n

Jane must recruit 20 more contributors or 25 contributors in total for the cost of each share to be $20.

Step-by-step explanation:

1. Cost per share, C, with respect to the number of contributors, n is

C = 500 / n

So, with 5 contributors, the price is $100 per share. $100 = 500/5

2. C= $20

20= 500 / n

20 * n = 500 * n

20n = 500

n= 25

She already has 5 contributors, so she needs 20 more.



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A lathe is set to cut bars of steel into lengths of 6 cm. The lathe is considered to be in perfect adjustment if the average len
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Answer:

t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723    

E. -0.723

df=n-1=93-1=92  

p_v =2*P(t_{(92)}  

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Step-by-step explanation:

Information provided

\bar X=5.97 represent the sample mean for the length

s=0.4 represent the sample standard deviation

n=93 sample size  

\mu_o =6 represent the value that we want to test

\alpha=0.05 represent the significance level

t would represent the statistic  

p_v represent the p value for the test

System of hypothesis

We need to conduct a hypothesis in order to check if the lathe is in perfect adjustment (6cm), then the system of hypothesis would be:  

Null hypothesis:\mu = 6  

Alternative hypothesis:\mu \neq 6  

since we don't know the population deviation the statistic is:

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

Replacing in formula (1) we got:

t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723    

E. -0.723

P value

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df=n-1=93-1=92  

Since is a two tailed test the p value would be:  

p_v =2*P(t_{(92)}  

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

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