Use the following information for the question. Male players at the high school, college and professional ranks use a regulation basketball that weighs 22.0 ounces with a standard deviation of 1.0 ounce. Assume that the weights of basketballs are approximately normally distributed. Roughly what percentage of regulation basketballs weigh more than 23.1 ounces

13.567%

Step-by-step explanation:

We solve the above question, using z score formula

z = (x-μ)/σ, where

x is the raw score = 23.1 ounces

μ is the population mean = 22.0 ounces

σ is the population standard deviation = 1.0 ounce

More than = Greater than with the sign = >

Hence, for x > 23.1 ounces

z = 23.1 - 22.0/1.0

= 1.1

Probability value from Z-Table:

P(x<23.1) = 0.86433

P(x>23.1) = 1 - P(x<23.1)

P(x>23.1) = 1 - 0.86433

P(x>23.1) = 0.13567

Converting to percentage

= 0.13567 × 100

= 13.567%

Therefore, the percentage of regulation basketballs that weigh more than 23.1 ounces is 13.567%

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so:

ion

step-by-step explanation:

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