Birth weights in norway are normally distributed with a mean of 3570 g and a standard deviation of 500 g. if the ulleval university hospital officials plan to require special treatment for the lightest 3% of newborn babies, what birth weight separates those requiring special treatment from those who do not?

We are given that:

μ = mean = 3570

σ = standard deviation = 500

From the normal probability table, find the value of z which corresponds to the largest 3%.

at P = 0.03, z = 1.88

The formula for z score is:

z = (x - μ) / σ

Finding for the sample value x:

1.88 =(x-3570)/500

x=(1.88)(500)+3570

x = 4510

The birth weight that separates those who do and do not require is 4510 g

x=17

step-by-step explanation:

solve 3x-24=x+10

then that makes it 2x=34 because you subtract an x from both sides to get rid of the one on the right, and then you add 24 to both sides to get rid of the 24 on the left

divide by both sides by 2 to get your answer so that you can get your answer

the first one, all those points on the line just make a strait line.