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Rudiy27
3 years ago
6

Scores for a common standardized college aptitude test are normally distributed with a mean of 496 and a standard deviation of 1

10. Randomly selected men are given a Test Preparation Course before taking this test. Assume, for sake of argument, that the preparation course has no effect. If 1 of the men is randomly selected, find the probability that his score is at least 545.7. P(X > 545.7)
Mathematics
1 answer:
Sloan [31]3 years ago
5 0

Answer:

0.3257

Step-by-step explanation:

Given an approximately normal distribution:

Z = (x - mean) / standard deviation

Mean = 496 ; Standard deviation = 110

For :

P(x > 545.7)

We obtain the standardized, Z score

Z = (545.7 - 496) / 110

Z = 49.7 / 110

Z = 0.4518

Hence, using the Z distribution table :

P(Z > 0.4518) = 1 - P(Z < 0.4518)

P(Z > 0.4518) = 1 - 0.67429

P(Z > 0.4518) = 0.32571

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4 0
3 years ago
Solve:<br><br><img src="https://tex.z-dn.net/?f=%5Cunderset%7Bx%5Crightarrow~3%7D%7B%5Clim%7D~%5Cdfrac%7B2x%5E2-18%7D%7Bx%5E2-3x
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Hello, please consider the following.

\displaystyle \lim_{x\rightarrow3}~\dfrac{2x^2-18}{x^2-3x} \\ \\ \\ =\lim_{x\rightarrow3}~\dfrac{2(x^2-3^2)}{x(x-3)} \\ \\ \\ =\lim_{x\rightarrow3}~\dfrac{2(x-3)(x+3)}{x(x-3)} \\ \\ \\ =\lim_{x\rightarrow3}~\dfrac{2(x+3)}{x} \\ \\ \\=\dfrac{2(3+3)}{3}\\ \\ \\=\dfrac{2*3*2}{3} =\Large \boxed{\sf \bf \ 4 \ }

Thank you

8 0
3 years ago
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