To problem wants to compute the probability of the expected value that could be earn if you would draw again another card. So base on the fact that if draw a face card of K,Q,J from a standard deck correctly you will earn 10 point in just one draw, but loose 2 points if you another card, the probability that you would draw correct card in just one draw is 30.77%, so the possible value earn you have 2 points only
Answer:
85 adult tickets
150 children tickets
Step-by-step explanation:
we can write a system of equations and then use the 'elimination method':
let a = adult tickets
let c = children tickets
a + c = 235
5a + 2c = 725
we can multiply the first equation by -5 to get:
-5a - 5c = -1175
now we can add the second equation to get:
5a + 2c = 725
-3c = -450
c = 150
a + 150 = 235
a = 85
7 X 4 = 28 if it’s hard for you just add 7, 4 times
Answer:
its 7
Step-by-step explanation:
1.<AFC
2.<AFD
3.<AFE
4.<AFB
#5, 6 and 7
m<BAE = 59
m<BAC + m<CAD + m<DAE = m<BAE
4x - 20 + x + 12 + x + 1 = 59
6x - 7 = 59
6x = 59 + 7
6x = 66
x = 11
m<BAC = 4x - 20 = 4(11) - 20 = 24
m<CAD = x + 12 = 11 + 12 = 23
m<BAE = x + 1 = 11 + 1 = 12
answer
5.
m<BAC = 24
6.
m<CAD = 23
7.
m<BAE = 12
#8, 9 and 10
m<BAE = 130
m<BAC + m<CAD + m<DAE = m<BAE
3x - 10 + 2x + 5 + x + 15 = 130
6x + 10 = 130
6x = 120
x = 20
m<BAC = 3x - 10 = 3(20) - 10 = 60 -10 = 50
m<CAD = 2x + 5 = 2(20) + 5 = 40 + 5 = 45
m<DAE = x + 5 = 20 + 15 = 35
answer
8.
m<BAC = 50
9.
m<CAD = 45
10.
m<DAE = 35
