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ludmilkaskok [199]
3 years ago
13

You have quarters and dimes that total $2.80. your friend says it is possible that the number of quarters is 8 more than the num

ber of dimes. is your friend correct? explain. let dd represent the number of dimes.an equation that models this situation is?
Mathematics
1 answer:
ss7ja [257]3 years ago
4 0
To answer this question, you need to know that a quarter is worth $0.25 and dimes worth $0.10. In this case, the total number of money that made from quarters and dimes is $2.80
With Q=quarter and D=dime, then the equation should be:
2.80= 0.25 Q + 0.10 D

The condition that needs to be fulfilled in this case is that the quarter should be 8 more than dimes. Then you need to find out the highest possible number of the quarter and lowest possible number of the dimes.  
If you divide 2.8 with 0.25 you will found it will be 11.2 but 11 quarter will result with 0.05 value which you cannot remove, so the highest possible of the quarter should be 10.
If the highest possible of the quarter is 10, then the lowest possible of dimes should be:
2.8= 0.25(10) + 0.10D
0.10D= 0.3
D=3

Since 10-3 is 7, then it is not possible for the quarter to be 8 more than the dimes.
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z=\frac{3.1-3}{\frac{2.449}{\sqrt{47}}}=0.280    

p_v =P(z>0.280)=0.390  

If we compare the p value and the significance level assumed \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its NOT significant higher than 0.3 at 1% of signficance.  

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Data given and notation  

\bar X=3.1 represent the sample mean

\sigma=\sqrt{6} represent the sample standard deviation for the sample

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\mu_o =3 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)  

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We need to conduct a hypothesis in order to check if the mean is higher than 3, the system of hypothesis would be:  

Null hypothesis:\mu \leq 3  

Alternative hypothesis:\mu > 3  

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

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

z=\frac{3.1-3}{\frac{2.449}{\sqrt{47}}}=0.280    

P-value

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

p_v =P(z>0.280)=0.390  

Conclusion  

If we compare the p value and the significance level assumed \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its NOT significant higher than 0.3 at 1% of signficance.  

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