Scores for a common standardized college aptitude test are normally distributed with a mean of 496 and a standard deviation of 110. 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)

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

chicken nugget stir fry

step-by-step explanation:

80° is the measure of gh

step-by-step explanation:

circle f is congruent to circle j

we have two circle with center f and j.

both circles are congruent. the radius of the circle would be equal.

see the attachment for the circle.

we are given ∠efd ≅ ∠gjh

therefore, arc (ed) = arc (gh)     (∴ circle f is congruent to circle j)

but ∠dfe=80°=arc (ed)

therefore, arc (gh) = 80°

thus, 80° is the measure of gh