Question

crandall wants to order keychains to market his business. the company he is ordering from has a deal that says the more you buy, the lower the unit price of each keychain. therefore, the unit price of each keychain varies inversely with the number of keychains. when he submitted his last order for 220 keychains, the unit price for each was $1.15. what should crandall expect the unit price to be if he orders 270 keychains? round your answer to the nearest cent. (1 point) $\square$

Explanation:

Step1: Recall inverse variation formula

For inverse variation, the relationship is \( y = \frac{k}{x} \) (or \( xy = k \)), where \( k \) is the constant of variation. Let \( p \) be the unit price and \( n \) be the number of keychains. So \( p = \frac{k}{n} \), which can be rewritten as \( k = pn \).

Step2: Find the constant \( k \)

We know that when \( n = 220 \), \( p = 1.15 \). Substitute these values into \( k = pn \):
\( k = 1.15\times220 \)
\( k = 253 \)

Step3: Find the new unit price

Now we need to find \( p \) when \( n = 270 \) and \( k = 253 \). Using \( p = \frac{k}{n} \):
\( p = \frac{253}{270} \)
\( p \approx 0.937 \)
Rounding to the nearest cent (hundredth place), we get \( p \approx 0.94 \).

Answer:

\( \$0.94 \)