Answer:
See below. (C*a) + (P*b) = M
Step-by-step explanation:
amount of students = 45
amount of candy per student = 1
amount of popcorn per student = 1
Let "S" be the total amount of students
Let "C" be the total amount of candy
Let "P" be the total amount of popcorn
Let "T" be the total number of snacks
Let "a" be the price per candy
Let "b" be the price per popcorn
Let "M" be the total amount of money spent at the game
The price of the candy is unknown.
The price of the popcorn is unknown.
The total amount of money spent at the game is unknown.
(C*a) + (P*b) = M
I am not sure what you are asking for but, here is what I drew up for you. Hope this helps!
Answer:
The probability of 1 or less children from that group to learn how to swim before 6 years of age is 0.072
Step-by-step explanation:
In this case we need to compute the probability of none of these 12 children learns to swim before 6 years of age. This is given by:
p(0) = (1 - 0.312)^(12) = 0.688^(12) = 0.01124
We now need to calculate the probability that one child learns to swim before 6 years of age.
p(1) = 12*0.312*(1 - 0.312)^(11) = 3.744*(0.688)^(11)
p(1) = 3.744*0.01634
p(1) = 0.0612
The probability of 1 or less children from that group to learn how to swim before 6 years of age is:
p = p(0) + p(1) = 0.01124 + 0.0612 = 0.07244
Answer:
i think that it is oil
Step-by-step explanation:
because it does not change
Answer:
5/6 (i think im not good at math soz)
-Mina
Step-by-step explanation:
Branily stop deleting my awnsers
Answer:
CD = 5
Step-by-step explanation:
AC = 5
BC = 7
∆ACB ≅ ∆DCE, therefore,
AC = CD,
BC = CE, and,
AB = DE
Thus,
AC = CD = 5
CD = 5
