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Diano4ka-milaya [45]
2 years ago
8

A family of 2 children and 1 adult visited an amusement park and paid an entry fee of $90. Another family of 3 children and 2 ad

ults visited the same amusement park and paid an entry fee of $155. What is the entry fee for a child at the amusement park?
Mathematics
1 answer:
stiks02 [169]2 years ago
7 0

\underline{\underline{\large\bf{Solution:-}}}\\

\leadstoLet us assume entry fee for child be x

\leadstoLet us assume entry fee for adult be y

\\

\longrightarrow<u>Since Entry fees for 2 children and 1 adult is $90 an equation for entry fees will be formed as -</u>

\begin{gathered}\\\implies\quad \sf 2\times x + 1 \times y = 90 \\\end{gathered}

\begin{gathered}\\\implies\quad \sf 2 x + y = 90 \\\end{gathered} \_\_\_\_\_(1)

\longrightarrow<u>Similarly, Entry fees for 3 children and 2 adult is </u><u>$</u><u>1</u><u>5</u><u>5</u><u> , so an equation for entry fees will be formed as -</u>

\begin{gathered}\\\implies\quad \sf 3\times x + 2\times y = 155 \\\end{gathered}

\begin{gathered}\\\implies\quad \sf 3x + 2y = 155 \\\end{gathered} \_\_\_\_\_(2)

<u>Multiplying eq (1) by </u><u>2</u><u> </u><u>it will be - </u>

\begin{gathered}\\\implies\quad \sf 4x + 2y = 180\\\end{gathered}\_\_\_\_\_(3)

<u>Subtracting</u><u> eq(</u><u>2</u><u>) </u><u>from</u><u> eq(</u><u>3</u><u>) - </u>

\begin{gathered}\\\implies\quad \sf  4x+2y - (3x+2y) = 180-155\\\end{gathered}

\begin{gathered}\\\implies\quad \sf  4x + 2y - 3x - 2y = 35\\\end{gathered}

\begin{gathered}\\\implies\quad \sf  4x  - 3x +2y- 2y = 35\\\end{gathered}

\begin{gathered}\\\implies\quad \bf x = 35\\\end{gathered}

Thus , entry fee for a child at the amusement park = $ 35

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The number of months until the insect population reaches 40 thousand is 14.29 months and the limiting factor on the insect population as time progresses is 250 thousands.

Given that population P(t) (in thousands) of insects in t months after being transplanted is P(t)=(50(1+0.05t))/(2+0.01t).

(a) Firstly, we will find the number of months until the insect population reaches 40 thousand by equating the given population expression with 40, we get

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Apply the distributive property a(b+c)=ab+ac, we get

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50+2.5t-0.4t-50=80+0.4t-0.4t-50

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(b) Now, we will find the limiting factor on the insect population as time progresses by taking limit on both sides with t→∞, we get

\begin{aligned}\lim_{t \rightarrow \infty}P(t)&=\lim_{t \rightarrow \infty}\frac{50(1+0.05t)}{2+0.01t}\\ &=\lim_{t \rightarrow \infty}\frac{50(\frac{1}{t}+0.05)}{\frac{2}{t}+0.01}\\ &=50\times \frac{0.05}{0.01}\\ &=250\end

(c) Further, we will sketch the graph of the function using the window 0≤t≤700 and 0≤p(t)≤700 as shown in the figure.

Hence, when the population P(t) (in thousands) of insects in t months after being transplanted by P(t)=(50(1+0.05t))/(2+0.01t) then the number of months until the insect population reaches 40 thousand 14.29 months and the limiting factor on the insect population is 250 thousand and the graph is shown in the figure.

Learn more about limiting factor from here brainly.com/question/18415071.

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