The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds. the probability of a player weighing less than 250 pounds is closest to:

Using the normal distribution, it is found that there is a 0.9772 = 97.72% probability of a player weighing less than 250 pounds.

In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:

In this problem:

The probability of a player weighing less than 250 pounds is the p-value of Z when X = 250, hence:

has a p-value of 0.9772.

0.9772 = 97.72% probability of a player weighing less than 250 pounds.

More can be learned about the normal distribution at link

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116

answer:

step-by-step explanation:

here, the total dishes = 10

dishes having meat = 3 ( chicken, meatloaf and hamburger )

thus, the probability of having meat on the chosen dishes = number of dishes having meat / total dishes

=

therefore, the probability of not having meat = 1- the probability of having meat on the chosen dishes

=

=

=

⇒ option b is correct.