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A study estimates that the cost of tuition at a university will increase by 2.8% each year. The cost of tuition at the University in 2015 was $33,741 the function b(x) , models the estimated tuition cost , where x is the number of years since 2015.
finds the expression that completes the function b(x)​

Respuesta :

so, the cost will increase 2.8% per annum... so that simply means is a compound interest rate, so let's use the compound interest formula for this one.

[tex]\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 &\$33741\\ r=rate\to 2.8\%\to \frac{2.8}{100}\dotfill &0.028\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=\textit{years after 2015}\dotfill &t \end{cases} \\\\\\ A=33741\left(1+\frac{0.028}{1}\right)^{1\cdot t}\implies b(x)=33741(1.028)^x[/tex]

fichoh

The complete expression for the compounding tuition fee function is [tex]b(x) = 33741(1.028) {}^{x} [/tex]

Using the compound interest relation :

[tex] b = P(1 + \frac{r}{n} ) {}^{nx} [/tex]

Where ;

  • b = final amount after x years
  • P = Initial amount = 33741
  • r = rate = 2.8% = 0.028
  • x = number of years since 2015
  • n = number of compounding times per period = 1 (yearly)

The function b(x) can be written as :

[tex]b(x) = 33741(1 + \frac{0.028}{1} ) {}^{x} [/tex]

Therefore, the expression for the function b(x) is :

[tex]b(x) = 33741(1.028) {}^{x} [/tex]

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