Quation that show an estimate of 7,203+299

Mathematics

2 answers:

Dmitry [639]2 years ago

8 0

The two numbers are very near 7200 and 300. The sum of those is a good estimate of your sum:

... 7200 + 300 = 7500

_____

If you're taught methods of addition that include "make a ten" or "make a hundred", you know you can "borrow" a 1 from 7203 (making it 7202) and loan that 1 to 299 to make it 300. Now, you know your sum is exactly 7202+300 = 7502, with no estimation necessary.

tensa zangetsu [6.8K]2 years ago

7 0

Round each number, I rounded each to nearest hundred.

7,200+300=7,500 estimation

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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

Scorpion4ik [409]

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.

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

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