The total tickets to be purchased to guarantee the win = 504 tickets
Step-by-step explanation:
Step 1 :
Number of entries in the trifecta race = 9
The win is to select the first finisher, second finisher and third finisher in their proper order.
We need to find the number of tickets to be purchased to guarantee the win
Step 2 :
Number of ways to select the first finisher = 9
Number of ways to select the second finisher = 8 [the first is selected and fixed. So the number of available finishes is reduced by 1]
Number of ways to select the third finisher = 7
Hence the total tickets to be purchased to guarantee the win = 9 × 8 × 7 = 504
Step 3 :
Answer :
The total tickets to be purchased to guarantee the win = 504 tickets
For the answer to the question above the probability of exactly x successes is
P(X=x)=b(x;n,p) = (nCx)(p^x)((1-p)^(n-x))
where nCx is number of combinations of n things taken x at a time, and "^" means exponentiation.
n = 10
p = 0.2
P(X=7) = 10C7*(0.2^7)*(0.8^3) = 0.00079
P(X=8) = 10C8*(0.2^8)*(0.8^2) = 0.00007
P(X=9) = 10C9*(0.2^9)*(0.8^1) = 0.00000
P(X=10) = 10C10*(0.2^10)*(0.8^0) = 0.00000
.00079 + .00007 + 0 + 0 = .00086 = .0009 rounded
Answer:
Given Below
Step-by-step explanation:
v = u+at
u+at = v
at = v-u
t = v-u/a
Hope this helps
Answer:
$95.78
Step-by-step explanation:
f(t) = 300t / (2t² + 8)
t = 0 corresponds to the beginning of August. t = 1 corresponds to the end of August. t = 2 corresponds to the end of September. So on and so forth. So the second semester is from t = 5 to t = 10.
$T₂ = ∫₅¹⁰ 300t / (2t² + 8) dt
$T₂ = ∫₅¹⁰ 150t / (t² + 4) dt
$T₂ = 75 ∫₅¹⁰ 2t / (t² + 4) dt
$T₂ = 75 ln(t² + 4) |₅¹⁰
$T₂ = 75 ln(104) − 75 ln(29)
$T₂ ≈ 95.78
Answer:
26
Step-by-step explanation:
Subtract the mean (84) from each of the four scores, obtaining
2, -3, 3, - 2
These are the "deviations."
Next, square each of these four deviations, obtaining
(2)^2 = 4, and:
(-3)^2 = 9, and:
(3)^2 = 9, and:
(-2)^2 = 4
Then the sum of the squared deviations of the scores from the mean is 26.
