Question

The Conowingo Reservoir had an original storage capacity of 300,000 acre-feet at the end of 1928, the year in which it was built. Starting in 1929, sediment carried downstream by the Susquehanna River collected in the reservoir and began reducing the reservoir's storage capacity at the approximate rate of 1,700 acre-feet per year. If the reservoir's capacity t years after 1928 was between 290,000 and 292,000 acre-feet, which of the following must be true? a. t < 2 b. 2 < t < 4 c. 4 < t < 6 d. 6 < t < 8

Answer 1

The value of t representing the years after 1928 when the Conowingo Reservoir's capacity is between 290,000 and 292,000 acre-feet is 6 < t < 5.

Step-by-step explanation:

To find the value of t that represents the years after 1928 when the Conowingo Reservoir's capacity is between 290,000 and 292,000 acre-feet, we need to set up an inequality.

The reservoir's capacity after t years can be represented as:

Storage Capacity after t years = Original Storage capacity - Rate of reduction x t

Using the given information:

Original Storage capacity = 300,000 acre-feet

Rate of reduction = 1,700 acre-feet per year

We can set up the inequality:

290,000 < 300,000 - 1,700t < 292,000

Now, let's solve this inequality to find the possible values of t:

Subtract 300,000 from all three parts:

-10,000 < -1,700t < -8,000

Divide all three parts by -1,700 (remember to reverse the direction of the inequality because we are dividing by a negative number):

5.88 > t > 4.71

Round the values to the nearest whole number, since time cannot be a fraction:

6 > t > 5

Therefore, the value of t must be 6 < t < 5. None of the given answer choices align with this, so none of the options presented are true.

Answer 2

The capacity of the Conowingo Reservoir is between 290,000 and 292,000 acre-feet when 6 < t < 8 years after 1928, given the sedimentation rate of 1,700 acre-feet per year.

Step-by-step explanation:

The student is asking about the capacity of the Conowingo Reservoir after a certain number of years due to sediment accumulation. Using the given rate of accumulation, we can solve for the time range during which the reservoir's capacity is between 290,000 and 292,000 acre-feet.

Original capacity (end of 1928): 300,000 acre-feet

Sediment accumulation rate: 1,700 acre-feet per year

Capacity after t years: 300,000 acre-feet - (1,700 acre-feet/year × t)

To find the time range when the capacity is between 290,000 and 292,000 acre-feet, we set up two inequalities:

300,000 - 1,700t ≥ 292,000

300,000 - 1,700t ≤ 290,000

Solving the first inequality for t gives us t ≤ 4.7, and solving the second inequality gives us t ≥ 5.9. Therefore, the correct answer is d. 6 < t < 8