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%

step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =)     take the ln   of both sides

ln (1/2)   = )   and we can write

ln(1/2) = ) lne     and lne   = 1     so we have

ln(1/2) = -.1092t     divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)

step-by-step explanation:

the multiplicative inverse of a fraction a/b is b/a. for the multiplicative inverse of a real number, divide 1 by the number. for example, the reciprocal of 5 is one fifth 1/5 or 0.2, and the reciprocal of 0.25 is 1 divided by 0.25, or 4.