Question

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 谢谢...

Answer 1

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

Answer 2

1、准备一个初始投资万美元是感激8年注册一个帐号,合成seminannually挣9%的兴趣。找的钱的数量在账户末端的时期。

2、某种放射性同位素有相当于大约1100多年。有多少年到最近的一年需要一定数量的同位素衰减到30%的金额吗?