All categories

3 years ago

Part A

Mathematics

2 answers:

Trava [24]3 years ago

7 0

Answer:

The total water usage at the park this year, in gallons, U = 16,206 + 22.2·x

Step-by-step explanation:

The given parameters are;

The increase in water usage for both regular days and special event for the current year = 20%

The water usage for last year regular days = 13,505 gallons

The water usage per special event = 18.5 gallons

The total water usage = 13,616 gallons

Therefore, we have;

The water usage for regular days on the current year = 13,505 gallons + (20/100)×13,505 gallons = 16,206 gallons

The water usage per special event for the current year, x = 18.5 gallons + (20/100) × 18.5 gallons = 22.2 gallons

The total water usage, U, at the park this year in terms of 'x', the number of special events the park hosted is given by the sum of the amount of water used on regular days and the product of 'x', and the amount of water used per special event this year as follows;

U = 16,206 gallons + x × 22.2 gallons

∴ The total water usage at the park this year, U = 16,206 + 22.2·x

leva [86]3 years ago

7 0

Answer:

U=16,206+22.2 ^. x

Step-by-step explanation:

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