$ According to a study published in a reputable science magazine, about 8 women in 100,000 have cervical cancer (C), so P(C)=0.00008. Suppose the chance that a Pap smear will detect cervical cancer when it is
present is 0.85. Therefore, P(test pos|C)=0.85. What is the probability that a randomly chosen woman who has this test will both have cervical cancer AND test positive for it?
Pſhave C AND test positive)-(Round to six decimal places as needed.)

coin flip

non-answer:

choosing

sorry if this isnt what you're looking for i dont really understand the question

for a it would be 24 children.

for b i don't understand the question. is it saying for each individual person they pick 5 reds to 7 greens or is it the whole group. clarify and i hope this can you.

step-by-step explanation:

x² + 11x + 18 < 0  

[-11 ± sqrt(121 - 72)]/2 = (-11 ± 7)/2  

solution: (-9, -2)

step-by-step explanation:

he is going to fill two shelves and the third shelf will only have five video games