Math  /  Algebra

Questionnomework11.1: Problem / (1 point)
A 30-second commercial during Super Bowl XLII in 2008 cost advertisers 2.7 million. For the first Super Bowl in 1967, an advertiser could have purchased approximately 26.19 minutes of advertising time for the same amount of money. (a) Assuming that advertising cost is proportional to its length of time, find the cost of advertising, in dollars/second, during the 2008 Super Bowl. cost == \square dollars/second. (round to nearest cent and do not enter commas) (b) Assuming that advertising cost is proportional to its length of time, find the cost of advertising, in dollars/second, during the 1967 Super Bowl. cost == \square dollars/second. (round to nearest cent and do not enter commas) (c) How many times more expensive was Super Bowl advertising in 2008 than in 1967? \square times more expensive (round to nearest whole number)

Studdy Solution
Compare the costs to determine how many times more expensive the 2008 Super Bowl was compared to 1967.
Times more expensive=Cost per second in 2008Cost per second in 1967 \text{Times more expensive} = \frac{\text{Cost per second in 2008}}{\text{Cost per second in 1967}}
Times more expensive=90,0001718.98 \text{Times more expensive} = \frac{90,000}{1718.98}
Calculate:
Times more expensive52 \text{Times more expensive} \approx 52
The cost of advertising during the 2008 Super Bowl was approximately 52 \boxed{52} times more expensive than in 1967.

View Full Solution - Free
Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord