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Anastaziya [24]
3 years ago
13

A jar is filled with only dimes and quarters. The jar contains a total of 60 coins and the value of the jar is $11.25. How many

quarters are in the jar?
Mathematics
1 answer:
inessss [21]3 years ago
3 0

Answer:

35 quarters

Step-by-step explanation:

This situation has two unknowns - the total number of dimes and the total number of quarters. Because we have two unknowns, we will write a system of equations with two equations using the two unknowns.

  • d+q=60 is an equation representing the total number of coins
  • 0.10d+0.25q=11.25 is an equation representing the total value in money based on the number of coin. 0.10 and 0.25 come from the value of a dime and quarter individually.

We write the first equation in terms of q by subtracting it across the equal sign to get d=60-q. We now substitute this for d in the second equation.

0.10(60-q)+0.25q=11.25\\6-0.10q+0.25q=11.25\\6+0.15q=11.25

After simplifying, we subtract 6 across and divide by the coefficient of q.

6+0.15q=11.25\\0.15q=5.25\\q=35

We now know of the 60 coins that 35 are quarters. To find the total value of the quarters, we multiply 35 by 0.25 and find 8.75.

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Answer:

t=\frac{51.6-48}{\frac{1.1}{\sqrt{10}}}=10.349      

p_v =P(t_9>10.349)=1.34x10^{-6}  

If we compare the p value and the significance level given for example \alpha=0.05 we see that p_v so we can conclude that we reject the null hypothesis, and the the actual mean is significantly higher than 48 MPa at 5% of significance.

Step-by-step explanation:

1) Data given and notation      

\bar X=51.6 represent the sample mean

s=1.1 represent the standard deviation for the sample      

n=10 sample size      

\mu_o =48 represent the value that we want to test    

\alpha=0.05 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 determine if the mean score is higher than 48, the system of hypothesis would be:      

Null hypothesis:\mu \geq 48      

Alternative hypothesis:\mu > 48      

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:      

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

t-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:      

t=\frac{51.6-48}{\frac{1.1}{\sqrt{10}}}=10.349      

Calculate the P-value      

First we find the degrees of freedom:

df=n-1=10-1=9

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

p_v =P(t_9>10.349)=1.34x10^{-6}  

Conclusion      

If we compare the p value and the significance level given for example \alpha=0.05 we see that p_v so we can conclude that we reject the null hypothesis, and the the actual mean is significantly higher than 48 MPa at 5% of significance.        

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