Consider an individual selected at random from a sample of 750 married women (see the data in the file P05_61.xlsx) in answering each of the following questions. Round your answers to three decimal places, if necessary. A. What is the probability that this woman does not work outside the home, given that she has at least one child
B. What is the probability that this woman has no children, given that she works part time?
C. What is the probability that this woman has at least two children, given that she does not work full time?

P(A) = 86 / 265

P(B) = 39 / 153

P(C) = 243 / 326

Note:

Missing data as follow

Number of children    No job      Part-time      Full-time     Total

0                                      16                39                98            153

1                                       86               59                126           271

2                                      163              80                83             326

Total                                265             178               307           750

Computation:

P(A) = 86 / 265

P(B) = 39 / 153

P(C) = (163 + 80) / 326 = 243 / 326

not sure but it may be b, as i said i'm not sure.

6^2 or 36

step-by-step explanation:

6000n

6×1000n

6×10^3 n

so we need 6n need to be a perfect cube

so n has to be 6^2 so that 6n is 6^3

what i'm saying is if you let n be 6^2 you have 6^3×10^3 or 60^3 which is a perfect cube