英文和数学好的请进1.an initial investment of $14,000 is appreciated f
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英文和数学好的请进
1.an initial investment of $14,000 is appreciated for 8 years in an account that earns 9% interest,compounded seminannually.Find the amount of money in the account at the end of the period.
2.A certain radioactive isotope has a half-life of approximately 1100 years.How many years to the nearest year would be required for a given amount of this isotope to decay to 30% of that amount?
3.140=45+20ln(30t+1) 求t
主要是要步骤 ..
1.an initial investment of $14,000 is appreciated for 8 years in an account that earns 9% interest,compounded seminannually.Find the amount of money in the account at the end of the period.
2.A certain radioactive isotope has a half-life of approximately 1100 years.How many years to the nearest year would be required for a given amount of this isotope to decay to 30% of that amount?
3.140=45+20ln(30t+1) 求t
主要是要步骤 ..
![英文和数学好的请进1.an initial investment of $14,000 is appreciated f](/uploads/image/z/16919552-56-2.jpg?t=%E8%8B%B1%E6%96%87%E5%92%8C%E6%95%B0%E5%AD%A6%E5%A5%BD%E7%9A%84%E8%AF%B7%E8%BF%9B1.an+initial+investment+of+%2414%2C000+is+appreciated+f)
1.投资额$14,000,增值期8年,年利率9%,求结束后的总利息
$14,000*(1+9%)^8 - $14,000 = $13,895.88
2.半衰期1100y,求衰减至30%所需时间
N(t)=Ne^(-λt)
1/2 = e^(-1100λ)
30% = e^(-λt)
t = 1100*(ln30%)/ln(1/2) = 1910.66y
3. ln(30t+1) = (140-45)/20
30t+1 = e^[(140-45)/20]
t = {e^[(140-45)/20]-1}/30 = 3.82
$14,000*(1+9%)^8 - $14,000 = $13,895.88
2.半衰期1100y,求衰减至30%所需时间
N(t)=Ne^(-λt)
1/2 = e^(-1100λ)
30% = e^(-λt)
t = 1100*(ln30%)/ln(1/2) = 1910.66y
3. ln(30t+1) = (140-45)/20
30t+1 = e^[(140-45)/20]
t = {e^[(140-45)/20]-1}/30 = 3.82
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