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Arada [10]
2 years ago
7

(Look at photo)

Mathematics
1 answer:
Lorico [155]2 years ago
3 0

Answer: Choice G   \left| \frac{p-n}{n} \right|

============================================================

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 |p| > |n| is a true statement).

I'll go for these values

  • p = 5
  • n = -2

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.

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\\\\

Now onto choice G

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\\\\

Next up is evaluating the expression for choice H.

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\\\\

Then let's compute expression J.

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\\\\

Lastly, compute expression K.

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\\\\

-----------------------------------

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, \left| \frac{p-n}{n} \right| has the largest value when p > 0 and n < 0.

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Normal probability distribution:

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