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Nutka1998 [239]
3 years ago
8

Paula has $10 and earns $0.50 per lap she runs for her school’s fundraiser. Jackie has $15 and earns $0.25 per lap she runs for

her school’s fundraiser. After how many laps will Paula have more money than Jackie?
Mathematics
1 answer:
algol133 years ago
6 0

Answer: After 20 laps Paula will have more money than Jackie.

Step-by-step explanation:

Let x = Number of laps.

Given, Paula has $10 and earns $0.50 per lap she runs for her school’s fundraiser.

i.e. Total amount Paula has =$ [ 10 +0.50 (Number of laps)]

= $ [10+0.50x]

Jackie has $15 and earns $0.25 per lap she runs for her school’s fundraiser.

i.e. Total amount Jackie has =$ [ 15 +0.25 (Number of laps)]

= $ [15+0.25x]

When Paula has more money , then

15+0.25x< 10+0.50x\\\\\Rightarrow\ 15-10

Hence, after 20 laps Paula will have more money than Jackie.

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Answer:

<u>Exponential model</u>

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  • A = initial value
  • r = rate of growth/decay
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<h3><u>Part (a)</u></h3>

Given:

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Substituting given values into the formula and solving for A:

\begin{aligned}y & =Ae^{rt}\\\implies 100 & = Ae^{r \times 0}\\100 & = Ae^0\\100 & = A(1)\\A & = 100\end{aligned}

<h3><u>Part (b)</u></h3>

Given:

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Substituting the given values into the equation and solving for r:

\begin{aligned}y& =Ae^{rt}\\\\\implies 50 & =100e^{30.17r}\\\\\dfrac{1}{2} & = e^{30.17r}\\\\ln \dfrac{1}{2} & = \ln e^{30.17r}\\\\\ln 1-\ln2 & =30.17r \ln e\\\\0-\ln 2 & =30.17r(1)\\\\-\ln 2 & =30.17r\\\\r & = \dfrac{-\ln 2}{30.17}\end{aligned}

Therefore, the final equation is:

y=100e^{\left(-\dfrac{\ln 2}{30.17}\right)t}

<h3><u>Question 1</u></h3>

<u>Part (a)</u>

Q:   From 100g how much remains in 80 years?

\begin{aligned}t=80 \implies y & =100e^{\left(-\dfrac{\ln 2}{30.17}\right)80}\\& = 15.91389949 \: \sf g\end{aligned}

<u>Part (b)</u>

Q:  How long will it take to have 10% remaining?

10% of 100 g = 10 g

\begin{aligned}y=10 \implies 10 & =100e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\dfrac{1}{10} & =e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\ln \dfrac{1}{10} & =\ln e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\ln 1 - \ln 10 & =\left(-\dfrac{\ln 2}{30.17}\right)t\ln e\\\\0 - \ln 10 & =\left(-\dfrac{\ln 2}{30.17}\right)t(1)\\\\-\ln 10 & =\left(-\dfrac{\ln 2}{30.17}\right)t\\\\t & = \dfrac{- \ln 10}{\left(-\dfrac{\ln 2}{30.17}\right)}\\\\t & = 100.2225706\: \sf years\end{aligned}

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<u>Part (a)</u>

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<u></u>

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Q:   How long to reach 20 g (amount remaining)?

<u></u>\begin{aligned}y=20 \implies 20 & =100e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\dfrac{1}{5} & =e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\ln \dfrac{1}{5} & =\ln e^{\left(-\dfrac{\ln 2}{30.17}\right)t}\\\\\ln 1 - \ln 5 & =\left(-\dfrac{\ln 2}{30.17}\right)t\ln e\\\\0 - \ln 5 & =\left(-\dfrac{\ln 2}{30.17}\right)t(1)\\\\-\ln 5 & =\left(-\dfrac{\ln 2}{30.17}\right)t\\\\t & = \dfrac{- \ln 5}{\left(-\dfrac{\ln 2}{30.17}\right)}\\\\t & = 70.05257062\: \sf years\end{aligned}

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