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Lemur [1.5K]
3 years ago
13

A type of organism's cells split every minute. If the organism has 4 cells when the cells begin to split, how many cells will th

e organism have after 6 minutes?
Mathematics
1 answer:
snow_lady [41]3 years ago
4 0

Answer:

The total of cells will the organism have after 6 minutes is 252

Step-by-step explanation:

* Lets consider this problem as a geometric series

- There is a constant ratio between each two consecutive numbers

Ex:

# 5  ,  10  ,  20  ,  40  ,  80  ,  ………………………. (×2)

# 5000  ,  1000  ,  200  ,  40  ,  …………………………(÷5)

* General term (nth term) of a Geometric series:

∵ U1 = a  ,  U2  = ar  ,  U3  = ar^2  ,  U4 = ar^3  ,  U5 = ar^4

∴ Un = ar^n-1

- Where r is the ratio between each two consecutive terms and

 n is the position of the number in the series

* The sum of first n terms of a Geometric Progression is calculate from

 Sn = a(1 - r^n)/1 - r  

- In the problem the cell split every minute

- The organism has 4 cells when the cells begin to split

∴ a = 4

# 4 , 8 , 16 , 32 , .......

∴ r = 2

- The total of cells will the organism have after 6 minutes can

  calculate from the rule of the sum

∴ S6 = 4(1 - 2^6)/1 - 2 = 4(1 - 64)/-1 = 4(-63)/-1 = 252 cells

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Brian invests ?1900 into a savings account. The bank gives 3.5% compound interest for the first 2 years and 4.9% thereafter. How
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let's check how much is it after 2 years firstly.


\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &1900\\ r=rate\to 3.5\%\to \frac{3.5}{100}\dotfill &0.035\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases} \\\\\\ A=1900\left(1+\frac{0.035}{1}\right)^{1\cdot 2}\implies A=1900(1.035)^2\implies A=2035.3275


Brian invested the money for 6 years, so now let's check how much is that for the remaining 4 years.


\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &2035.3275\\ r=rate\to 4.9\%\to \frac{4.9}{100}\dotfill &0.049\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}


\bf A=2035.3275\left(1+\frac{0.049}{1}\right)^{1\cdot 4}\implies A=2035.3275(1.049)^4 \\\\\\ A\approx 2464.54\implies \boxed{\stackrel{\textit{rounded up }}{A=2465}}

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