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A coffee shop currently sells 490 lattes a day at $3.00 each. They recently tried raising the by price by $0.25 a latte, and found that they sold 30 less lattes a day. a) Assume that the number of lattes they sell in a day, N, is linearly related to the sale price, p (in dollars). Find an equation for N as a function of p.

 Jan 18, 2020
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A coffee shop currently sells 490 lattes a day at $3.00 each.

They recently tried raising the by price by $0.25 a latte, and found that they sold 30 less lattes a day.

 

a) Assume that the number of lattes they sell in a day, N, is linearly related to the sale price, p (in dollars). Find an equation for N as a function of p.

 

\(\begin{array}{|rcll|} \hline \dfrac{N-460}{p-3.25} &=& \dfrac{490-460}{3-3.25} \\\\ N-460 &=& (p-3.25)*\dfrac{490-460}{3-3.25} \\\\ N &=& 460+(p-3.25)*\dfrac{490-460}{3-3.25} \\\\ N &=& 460+(p-3.25)*\dfrac{30}{-0.25} \\\\ N &=& 460-(p-3.25)*\dfrac{30}{0.25} \\\\ N &=& 460-p*\dfrac{30}{0.25}+3.25*\dfrac{30}{0.25} \\\\ N &=& 460-p*\dfrac{30}{0.25}+390 \\\\ N &=& 850-p*\dfrac{30}{0.25} \\\\ N &=& 850-p*120 \\\\ \mathbf{N} &=& \mathbf{850 - 120p} \\ \hline \end{array}\)

 

laugh

 Jan 18, 2020

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