According to government data, the probability that a woman between the ages of 25 and 29 was never married is 40%. in a random survey of 10 women in this age group, what is the probability that at least eight were married?

This is a case of binomial distribution. The formula used in calculations for binomial probability is:

P = nCr p^r (1-p)^(n-r) 

Where,

P = probability

nCr = combinations of r from n possibilities

p = success rate = 40% = 0.40

n = sample size = 10

1st: Let us calculate for nCr for r = 8 to 10. Formula is:

nCr = n! / r! (n-r)! 

10C8 = 10! / 8! 2! = 45

10C9 = 10! / 9! 1! = 10

10C10 = 10! / 10! 0! = 1

Calculating for probabilities when r = 8 to 10:

P (r=8) = 45 * 0.4^8 (0.6)^2 = 0.0106

P (r=9) = 10 * 0.4^9 (0.6)^1 = 0.0016

P (r=10) = 1 * 0.4^10 (0.6)^0 = 0.0001

Total probability that at least 8 were married = 0.0106 + 0.0016 + 0.0001

Total probability that at least 8 were married = 0.0123

the camry sales+accord sale=828,000. to make it so the two numbers are 18 thousand apart and still equal 828,000, you first need to find out what half of 828,000 is. 828,000/2=414,000. for the two numbers to equal 828,000, you have to add the same amount as you subtract from the middle. in this case, the number is 9,000 because it's half of 18,000. to equal 828,000, you have to add 9,000 to 414,000 for the camry and subtract 9,000 from 414,000 for the accord. 414,000+9,000=423,000.

414,000-9,000=405,000.

423,000+405,000=828,000.

423,000-405,000=18,

hope it

answer: 12

step-by-step explanation:

because yes