4^2 equals 16. 16+25=41
The answer would be 41
The answer is 448 whatever the measurement is
<span>Don't forget S is measured in thousands of units so you are solving for :
100 < 74.5 + 43.75Sin(πt/6)
25.5 < 43.75Sin(πt/6)
Sin(πt/6) >25.5/43.75 = 0.582857
ASrcSin(πt/6) > 0.62224 radians
πt/6 > 0.62224
t > 6 x 0.62224/π = 1.1884 (4dp)
This initial value occurs when the sine value is increasing and it will reach its maximum value of 1 when Sin(πt/6) = Sinπ/2, that is when t = 3.
Consequently, monthly sales exceed 100,000 during the period between t = 1.1884 and 4.8116
[3 - 1.1884 = 1.8116 so the other extreme occurs at 3 + 1.8116]
Note : on the basis of these calculations, January is 0 ≤ t < 1 : February is 1 ≤ t < 2 :....May is 4 ≤ t < 5
So the period when sales exceed 100,000 occurs between Feb 6 and May 25 and annually thereafter.</span>
Answer:
3.9
Step-by-step explanation:
Given the data:
Payout ($) (x) : 0 2 4 8 10
Probability p(x) : 0.35 0.2 0.1 0.2 0.15
The expected winning ; E(X) = Σ(x * p(x))
Σ(x * p(x)) = (0*0.35)+(2*0.2)+(4*0.1)+(8*0.2)+(10*0.15)
= 0 + 0.4 + 0.4 + 1.6 + 1.5
= 3.9
The first term is 138
The difference is 55
The iterative rule for the amount of money Mr Speas has after n weeks is
55/2 n² + 221/2 n
During the first week she has $138 in his bank account. At the end of each week she deposited $55 into her bank account.
The first term will be 138 .
The common difference is 55 because her bank always increase by $55 dollars every week. The sequence will be 138, 193, 248, 303, 358...…
The difference = 193 - 138 = 55.
The iterative rule for the amount of money Mr Speas has after n weeks can be represented below
n = number of weeks
a = first term = 138
d = common difference = 55
Using AP formula,
sₙ = n/2(2a + (n - 1)d)
sₙ = n/2 (2(138)+ (n - 1)55)
sₙ = n /2(276 + 55n -55)
sₙ = n /2(221 + 55n)
sₙ = 55/2 n² + 221/2 n
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