A certain paper suggested that a normal distribution with mean 3,500 grams and a standard deviation of 560 grams is a reasonable model for birth weights of babies born in Canada.
One common medical definition of a large baby is any baby that weighs more than 4,000 grams at birth.
What is the probability that a randomly selected Canadian baby is a large baby?

the probability that a randomly selected Canadian baby is a large baby is 0.19

Step-by-step explanation:

Since the birth weights of babies born in Canada is assumed to be normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = birth weights of babies

µ = mean weight

σ = standard deviation

From the information given,

µ = 3500 grams

σ = 560 grams

We want to find the probability or that a randomly selected Canadian baby is a large baby(weighs more than 4000 grams). It is expressed as

P(x > 4000) = 1 - P(x ≤ 4000)

For x = 4000,

z = (4000 - 3500)/560 = 0.89

Looking at the normal distribution table, the probability corresponding to the z score is 0.81

P(x > 4000) = 1 - 0.81 = 0.19

3=log(n)=2

answer a: product property

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2