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vazorg [7]
4 years ago
13

A positive integer is 15 less than another. If 4 times the reciprocal of the smaller integer is subtracted from the reciprocal o

f the larger integer, then the result is − 3/4 . Find all pairs of integers that satisfy this condition.
Mathematics
1 answer:
NemiM [27]4 years ago
7 0
Translate the info given into equations:

<span>A positive integer is 15 less than another   -----> x = y - 15

</span><span>If 4 times the reciprocal of the smaller integer is subtracted from the reciprocal of the larger integer, then the result is − 3/4
</span>------> (1/y) - (4/x) = -3/4

Use substitution, replace 'x' with 'y-15' in 2nd equation

\frac{1}{y} - \frac{4}{y-15} = -\frac{3}{4} \\  \\ \frac{ (y-15)-4y}{y(y-15)}=-\frac{3}{4} \\  \\ 4(-3y-15) = -3y(y-15) \\  \\ -12y -60 = -3y^2+45y \\  \\ 3y^2-57y-60 = 0 \\  \\ y^2-19y -20 =0 \\  \\ (y-20)(y+1) = 0 \\  \\ y = 20,y=-1

We know that both x and y must be positive integers, so y =-1 is not a solution.

If y=20, then x = 20-15 = 5

Final answer:
The pair of integers is (5,20) 
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Answer:

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Step-by-step explanation:

Data given and notation  

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Alternative hypothesis:\mu > 12  

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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".  

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

z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9  

P-value  

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

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Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 5% of signficance.  

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