Let a=price of adult ticket
let c=price of a child's ticket
start out by writing the following system of equations:
3a+4c=132
2a+3c=94
then, multiply the first equation by 2, and the second equation by 3 to get the following system of equations:
6a+8c=264
6a+9c=282
subtract the like terms to get the following equation:
-c=-18
divide both sides by -1 to get rid of the negative to get the price of a child's ticket to be $18. to find the price of an adult ticket, pick one of the original equations to substitute the 18 in for c to find a. for example:
2a+3c=94
2a+3(18)=94
2a+54=94
-54 -54
2a=40
2 2
a=20

or if you decide to use the other equation:
3a+4c=132
3a+4(18)=132
3a+72=132
-72 -72
3a=60
3 3
a=20
either way, you still get an adults ticket to be $20 and a child's ticket to be $18.

8 0

3 years ago

Choose the true statement. -2 < -3 -2 > -3 -2 = -3Choose the true statement.

Marrrta [24]

-2 < -3 (I think so anyway)

7 0

2 years ago

Two cards are drawn in succession and without replacement from a standard deck of 52 cards. Find the probability that the second

Svetllana [295]

Answer:

The probability that the second card is a face card if it’s known that the first card was a face card is 0.0497

Step-by-step explanation:

Total number of face cards = 12

Total cards = 52

Probability of getting face card on first draw= $\frac{12}{52}$

Remaining no. of face cards = 11

Remaining number of total cards = 51

Probability of getting face card on second draw= $\frac{11}{51}$

The probability that the second card is a face card if it’s known that the first card was a face card = $\frac{12}{52} \times \frac{11}{51}= \frac{12}{52} \times \frac{11}{51}=0.0497$