Birth weights in norway are normally distributed with a mean of 3570 g and a standard deviation of 500 g. find the probability that 100 randomly selected birth weights have a mean between 3500 g and 3600 g. a) 0.7257 b) 0.3551 c) 0.6449 d) 0.0796

 0.6449

(The one above is wrong)

D) 0.0796

Step-by-step explanation:

Birth weights in Norway are normally distributed with a mean of 3570 g and a standard deviation of 500 g.

mean = 3570  and SD = 500

We need to find P(3500<x<3600)

P(3500<x<3600)= P(x=3600)- P(x=3500)

to find P(x=3600) we find z-score

Now use z-score table . z-score = 0.5239

P(x=3600)=0.6179

to find P(x=3500) we find z-score

Now use z-score table . z-score = 0.4443

P(x=3600)=0.4443

P(3500<x<3600)= P(x=3600)- P(x=3500)

P(3500<x<3600)=0.5239-0.4443 = 0.0796

this is false ^.^ lol

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

~