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san4es73 [151]
3 years ago
11

A quantity with an initial value of 240 grows exponentially at a rate of 9.5% every 2 decades. What is the value of the quantity

after 52 years, to the nearest hundredth?
Mathematics
1 answer:
postnew [5]3 years ago
8 0

Answer:  303.870087962229

Simplify ≈303.87

Step-by-step explanation:

Grows 9.5%→r=0.095 Divide by 100

Grows every 2 decades: exponent of t2

(where t is in decades)

Write a function:

f(t)=240(1+0.095) ^t/2 Percent change every 2 decades

52 years→52/10→5.2 decades There are 10 years in a decade

Plug in t=5.2

f(5.2)=240(1+0.095) 5.2/2

Answer:  303.870087962229

≈303.87

Round to the nearest hundredth

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Thus the expected value is given by

E(x)=3\cdot6\cdot9p(3, 0, 0)+6\cdot3\cdot9p(0, 3, 0)+6\cdot6\cdot6p(0, 0, 3) \\ 4\cdot5\cdot9p(2, 1, 0)+6\cdot4\cdot8p(0, 2, 1)+5\cdot6\cdot7p(1, 0, 2)+4\cdot6\cdot8p(2, 0, 1) \\ +5\cdot4\cdot9p(1, 2, 0)+6\cdot5\cdot7p(0, 1, 2)+5\cdot5\cdot8(1, 1, 1) \\  \\ = \frac{1}{ ^{21}C_3} (162\cdot{ ^6C_3}\cdot{ ^6C_0}\cdot{ ^9C_0}+162\cdot{ ^6C_0}\cdot{ ^6C_3}\cdot{ ^9C_0}+216\cdot{ ^6C_0}\cdot{ ^6C_0}\cdot{ ^9C_3} \\ \\ +180\cdot{ ^6C_2}\cdot{ ^6C_1}\cdot{ ^9C_0}+192\cdot{ ^6C_0}\cdot{ ^6C_2}\cdot{ ^9C_1}

+210\cdot{ ^6C_1}\cdot{ ^6C_0}\cdot{ ^9C_2}+192\cdot{ ^6C_2}\cdot{ ^6C_0}\cdot{ ^9C_1}+180\cdot{ ^6C_1}\cdot{ ^6C_2}\cdot{ ^9C_0} \\  \\ +210\cdot{ ^6C_0}\cdot{ ^6C_1}\cdot{ ^9C_2}+200\cdot{ ^6C_1}\cdot{ ^6C_1}\cdot{ ^9C_1} \\  \\ =\frac{1}{1,330}(324\cdot20+216\cdot84+360\cdot90+384\cdot135+420\cdot216+200\cdot324) \\  \\ =\frac{1}{1,330}(6,480+18,144+32,400+51,840+90,720+64,800) \\  \\ =\frac{1}{1,330}(264,384) \\  \\ =\bold{198.78}
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