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alina1380 [7]
3 years ago
15

John has four more nickels than dimes in his pocket, for a total of $1.25. Which equation could be used to determine the numbet

of dimes, x, in his pocket?
Mathematics
1 answer:
ahrayia [7]3 years ago
7 0
So, you have to have your equation in terms of dimes. What you said about what he has: four more nickels than dimes in his pocket. Will help with our equation.
We know what a nickel and dime is worth. A nickel is .05 and a dime is .1
We don't know how many dimes are in his pocket, since we're trying to solve it.
.1x+(4+x).05=1.25; 
In the parenthesis, it shows how there are four more nickels. Let's solve it now.
.1x + (4+x).05=1.25
.1x + .2+ .05x = 1.25; Let's add like terms.
.15x + .2 = 1.25; Subtract .2 from both sides.
.15x= 1.05
1.05÷.15= 7 =x
There are 7 dimes in his pocket, let's check our answer.
We now know there are 11 nickels, since there are four more nickels than dimes.
11(.05) +.1(7) = 1.25
.55+ .7 = 1.25
Now that we've tried it, we know there are 7 dimes in his pocket.
Tell me if this helps!
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Answer:

t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606    

The degrees of freedom are given by:

df=n-1=5-1=4  

The p value wuld be given by:

p_v =2*P(t_{(4)}>2.606)=0.060  

For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40

The distribution with the critical values are in the figure attached

Step-by-step explanation:

Information given

48, 41, 40, 51, and 50

The sample mean and deviation can be calculated with these formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=46 represent the mean height for the sample  

s=5.148 represent the sample standard deviation

n=5 sample size  

\mu_o =40 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

Hypothesis to test

We want to test if the true mean for this case is equal to 40, the system of hypothesis would be:  

Null hypothesis:\mu = 40  

Alternative hypothesis:\mu \neq 40  

The statistic is given by:

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

Replacing we got:

t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606    

The degrees of freedom are given by:

df=n-1=5-1=4  

The p value wuld be given by:

p_v =2*P(t_{(4)}>2.606)=0.060  

For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40

The distribution with the critical values are in the figure attached

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3 years ago
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