Answer:
w<6 THE SMALL LINE UNDER THE LESS THAN SIGN IS STILL THERE I JUST CANNOT ADD IT
Step-by-step explanation:
first solve the equation to the right.
-3(2w+1) = -6w-3
now solve the whole equation
-33-w< -6w-3
-w+6w < -3+33
5w < 30
divide the 5 from both sides to simplify
5/5w < 30/5
w < 6
THE ANSWER IS w < 6
Hello,
Very nice as problem.
2 solutions:
1 quater,8 dimes, 2 pennies
and
3 quaters,3 dimes, 2 pennies
since
107=( 0, 0, 107) but : 100= 0*25+ 0*10+ 100
107=( 0, 1, 97) but : 100= 0*25+ 1*10+ 90
107=( 0, 2, 87) but : 100= 0*25+ 2*10+ 80
107=( 0, 3, 77) but : 100= 0*25+ 3*10+ 70
107=( 0, 4, 67) but : 100= 0*25+ 4*10+ 60
107=( 0, 5, 57) but : 100= 0*25+ 5*10+ 50
107=( 0, 6, 47) but : 100= 0*25+ 6*10+ 40
107=( 0, 7, 37) but : 100= 0*25+ 7*10+ 30
107=( 0, 8, 27) but : 100= 0*25+ 8*10+ 20
107=( 0, 9, 17) but : 100= 0*25+ 9*10+ 10
107=( 0, 10, 7) but : 100= 0*25+ 10*10+ 0
107=( 1, 0, 82) but : 100= 1*25+ 0*10+ 75
107=( 1, 1, 72) but : 100= 1*25+ 1*10+ 65
107=( 1, 2, 62) but : 100= 1*25+ 2*10+ 55
107=( 1, 3, 52) but : 100= 1*25+ 3*10+ 45
107=( 1, 4, 42) but : 100= 1*25+ 4*10+ 35
107=( 1, 5, 32) but : 100= 1*25+ 5*10+ 25
107=( 1, 6, 22) but : 100= 1*25+ 6*10+ 15
107=( 1, 7, 12) but : 100= 1*25+ 7*10+ 5
107=( 1, 8, 2) is good
107=( 2, 0, 57) but : 100= 2*25+ 0*10+ 50
107=( 2, 1, 47) but : 100= 2*25+ 1*10+ 40
107=( 2, 2, 37) but : 100= 2*25+ 2*10+ 30
107=( 2, 3, 27) but : 100= 2*25+ 3*10+ 20
107=( 2, 4, 17) but : 100= 2*25+ 4*10+ 10
107=( 2, 5, 7) but : 100= 2*25+ 5*10+ 0
107=( 3, 0, 32) but : 100= 3*25+ 0*10+ 25
107=( 3, 1, 22) but : 100= 3*25+ 1*10+ 15
107=( 3, 2, 12) but : 100= 3*25+ 2*10+ 5
107=( 3, 3, 2) is good
107=( 4, 0, 7) but : 100= 4*25+ 0*10+ 0
It is $7. Anything average 5 you need to round up
Answer:
P(A) = 0.39
Step-by-step explanation:
We are given;
P(W|A) = 0.7
P(W|A^c ) = 0.3
We are told that 60% of the respondents said they voted for A. Thus;
P(A|W) = 60% = 0.6
Now, using the principle of drawing lots, we can be able to find the probability of the event that they are willing to participate in the exit poll which is P(W).
Thus;
P(W) = [P(W|A) × P(A)] +[P(W∣A^c) × P(A^c)]
Now, P(A^c) can be expressed as 1 - P(A)
Thus, we now have;
P(W) = [P(W|A) × P(A)] + [P(W∣A^c) × (1 - P(A)]
Plugging in the relevant values gives;
P(W) = 0.7P(A) + 0.3(1 - P(A))
P(W) = 0.7P(A) + 0.3 - 0.3P(A)
P(W) = 0.3 + 0.4P(A)
Now,using Baye's theorem, we can find an expression for P(A|W)
Thus;
P(A|W) = [P(A ∩ W)]/P(W)
This can be further expressed as;
P(A|W) = [P(A) × P(W|A)]/P(W)
Plugging in relevant values, we have;
0.6 = 0.7P(A)/(0.3 + 0.4P(A))
Cross multiply to get;
0.6(0.3 + 0.4P(A)) = 0.7P(A)
0.18 + 0.24P(A) = 0.7P(A)
0.18 = 0.7P(A) - 0.24P(A)
0.46P(A) = 0.18
P(A) = 0.18/0.46
P(A) = 0.39
