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o-na [289]
3 years ago
13

It takes three identical water pumps 8 hours to fill a pool. c How long would it take four of these same pumps to fill the pool,

assuming they all pump at the same rate?
Mathematics
1 answer:
olchik [2.2K]3 years ago
3 0

Answer:

6 hours

Step-by-step explanation:

We can think of this problem as a "work" problem and use the formula:

work = rate x time

Let p be the rate of a single pump.  So the total rate of 3 pumps is 3p. And the total time is 8 hours, so the work needed to fill a pool is:

work = 3p x 8 = 24p

We need 24p to fill up a pool.

So what happens when you have 4 pumps? That is a rate of 4p.  So how much time is needed to fill up a pool that requires 24p of work?

Solve by using the work = rate x time equation:

24p = 4p x t

6 = t

6 hours.

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6x^2 + 8x + 28

Step-by-step explanation:

2 times the length plus 2 times the width gives the perimeter

2(3x^2 + 3x + 6) + 2(x + 8)

6x^2 + 6x + 12 + 2x + 16

6x^2 + 8x + 28

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Thane Company is interested in establishing the relationship between electricity costs and machine hours. Data have been collect
mars1129 [50]

Answer:

If the controller uses the high-low method to estimate costs, the cost equation for electricity costs is "Total cost = $5,010 + ($6.45 * Machine hours)".

Step-by-step explanation:

Note: The data needed to estimate costs using the high-low method is merged together. It is therefore sorted before answering the question as follows:

Month          Machine Hours          Electricity Costs

January              2,000                           $18,950

February            2,400                           $22,100

March                 1,400                            $14,050

April                   2,600                            $24,100

May                    3,300                            $28,800

June                   2,800                           $23,100

July                    3,600                            $25,300

August               3,000                            $23,300

September         1,500                             $16,600

October             3,200                             $27,100

November         4,200                             $32,100

December         3,700                             $28,300

The explanation of the answer is now provided as follows:

Step 1: Calculation of variable cost per hour

From the data above, the highest Machine Hours and Electricity Costs occur in November, while the lowest occur in March. Therefore, we have:

Variable cost per hour = (Highest Electricity Costs - Lowest Electricity Costs) / (Highest Machine Hours - Lowest Machine Hours = ($32,100 - $14,050) / (4,200 – 1,400) = $18,050 / $2,800 = 6.44642857142857

Rounding to 2 decimal places as required, we have:

Variable cost per hour = $6.45

Therefore, the variable-cost components using the high-low method is $6.45.

Step 2: Calculation of total fixed cost

The formula for calculating the total cost is given as follows:

Total cost = Total Fixed Cost + Total Variable Cost ................. (1)

Where;

Total Variable Cost = Variable cost per hour * Machine hours at a particular Electricity Costs

Using highest levels of activity and substitute into equation (1), we have:

$32,100 = Total Fixed Cost + ($6.45 * 4,200)

Total Fixed Cost = $32,100 - ($6.45 * 4,200) = $32,100 - $27,090 = $5,010

Therefore, the fixed-cost components using the high-low method is $5,010.

Step 3: Derivation of the cost equation for electricity costs

The cost equation for electricity costs can be obtained based on the total cost function given in equation (1) above, where:

Total Fixed Cost = $5,010

Total Variable Cost = Variable cost per hour * Machine hours = $6.45 * Machine hours

Substituting the values into equation (1), we have:

Total cost = $5,010 + ($6.45 * Machine hours)

Therefore, if the controller uses the high-low method to estimate costs, the cost equation for electricity costs is "Total cost = $5,010 + ($6.45 * Machine hours)".

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