Sounds like a problem in binomial probability. What do you think?
Here the # of experiments is 6, so n = 6. The probability of a baby girl being born is 0.50.
Using my TI-83 Plus calculator:
binompdf(6, 0.5, 2) = 0.234
binompdf(6, 0.5, 3) = 0.313
binompdf(6, 0.5, 4) = 0.234
binompdf(6, 0.5, 5) = 0.094
binompdf(6, 0.5, 6) = 0.016
To get the prob. of at least 2 girls in 6 births, add up the 5 probabilities given above:
P(at least 2 girls in 6 births) = 0. ???
???? Can you be more specific please
Answer:
x=1,y=-1
Step-by-step explanation:
the ans is 1 and -1.
Answer:
d. $298 U.S. dollars
Step-by-step explanation:
$1 Can = $0.85 US
Divide both sides of the equation by $1 Can
1 = ($0.85 US)/($1 Can)
$350 Can * ($0.85 US)/($1 Can) = $297.50 US
Answer: d. $298 U.S. dollars
