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. ???
(x,y) or
(x,f(x))
so just plug the first value in and see if yo get the 2nd value
A. when x=-2, then is f(x)=48?
remember, x^-m=1/(x^m)
3(1/4)^-2=3/((1/4)^2)=3/(1/16)=3*16=48
true
B. x^0=1
so 3(1/4)^0=3(1)=3
true
C. when x=2
3(1/4)^2, 3(1/16), 3/16
true
D. 12
das is obviously false but anyway
3(1/4)^12=3(1/big number)=3/big number=small number but not zero
answer is A,B,C
Answer:
<em> i got -380</em>
Step-by-step explanation:
sorry if not helpful
Answer:
The equation that can be used to determine the number of rounds that Emma golfs is:
x=(c-7.50)/6.25
Step-by-step explanation:
From the information given, the total amount Emma pays to go miniature golfing is equal to the price of the admission ticket plus the result of multiplying the price per round of golf for the number of rounds and according to this, the equation would be:
c=7.50+6.25x, where:
c is the total cost for Emma to go miniature golfing
x is the number of rounds
Now, you can solve for x to determine the equation that can be used to find the number of rounds that Emma golfs:
x=(c-7.50)/6.25
You can replace c with 26.25:
x=(26.25-7.50)/6.25
x=18.75/6.25
x=3
