Answer: Choice G [tex]\left| \frac{p-n}{n} \right|[/tex]
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Explanation:
Pick a positive number for p, and a negative value for n. Make sure that p is further away from 0 than n is (so that [tex]|p| > |n|[/tex] is a true statement).
I'll go for these values
Replace every copy of p with 5, and replace every copy of n with -2. Then evaluate or simplify. Use your calculator if needed.
Let's start with choice F.
[tex]F = \left| \frac{p-n}{p} \right|\\\\F = \left| \frac{5-(-2)}{5} \right|\\\\F = \left| \frac{5+2}{5} \right|\\\\F = \left| \frac{7}{5} \right|\\\\F = \left| 1.4 \right|\\\\F = 1.4\\\\[/tex]
Now onto choice G
[tex]G = \left| \frac{p-n}{n} \right|\\\\G = \left| \frac{5-(-2)}{-2} \right|\\\\G = \left| \frac{5+2}{-2} \right|\\\\G = \left| \frac{7}{-2} \right|\\\\G = \left| -3.5 \right|\\\\G = 3.5\\\\[/tex]
Next up is evaluating the expression for choice H.
[tex]H = \left| \frac{p+n}{p-n} \right|\\\\H = \left| \frac{5+(-2)}{5-(-2)} \right|\\\\H = \left| \frac{5-2}{5+2} \right|\\\\H = \left| \frac{3}{7} \right|\\\\H \approx \left| 0.4286\right|\\\\H \approx 0.4286\\\\[/tex]
Then let's compute expression J.
[tex]J = \left| \frac{p+n}{p} \right|\\\\J = \left| \frac{5+(-2)}{5} \right|\\\\J = \left| \frac{5-2}{5} \right|\\\\J = \left| \frac{3}{5} \right|\\\\J = \left| 0.6 \right|\\\\J = 0.6\\\\[/tex]
Lastly, compute expression K.
[tex]K = \left| \frac{p+n}{n} \right|\\\\K = \left| \frac{5+(-2)}{-2} \right|\\\\K = \left| \frac{5-2}{-2} \right|\\\\K = \left| \frac{3}{-2} \right|\\\\K = \left| -1.5 \right|\\\\K = 1.5\\\\[/tex]
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To summarize, when plugging p = 5 and n = -2 into each expression, we got these results:
- F = 1.4
- G = 3.5
- H = 0.4286 (approximate)
- J = 0.6
- K = 1.5
Expression G leads to the largest output.
Therefore, [tex]\left| \frac{p-n}{n} \right|[/tex] has the largest value when p > 0 and n < 0.
