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anygoal [31]
3 years ago
13

11) The Cost of maintaing a

Mathematics
1 answer:
dezoksy [38]3 years ago
3 0

Answer:

a). Cost of 44 pupils = $14265

b). Least number of pupils = 31

Step-by-step explanation:

The given question is incomplete; here is the complete question.

The cost of maintaining a school is partly constant and partly varies as the number of pupils. With 50 pupils, the cost is $15,705.00 and with 40 pupils, it is $13,305.00.

(a) Find the cost when there are 44 pupils.

(b) If the fee per pupil is $360.00, what is the least number of pupils for which the school can run without a loss?

Let the equation representing the total cost of maintaining a school is,

C = ax + b

Where C = Total cost of maintaining a school

a = Fee per pupil

b = Fixed running cost

x = number of pupils

a). Cost of 50 pupils = $15705

    Equation will be,

    15705 = 50a + b -------(1)

    Cost of 40 pupils = $13305

    Equation will be,

    13305 = 40a + b --------(2)

    By subtracting equation (2) from equation (1),

    15705 - 13305 = (50a + b) - (40a + b)

    2400 = 10a

    a = 240

    From equation (1),

    b = 3705

    Equation representing the total cost will be,

    C = 240x + 3705

    If x = 44

    C = 240(44) + 3705

    C = $14265

b). If the fee per pupil 'a' = $360

    Let the number of pupils = p

    Total fee of 'p' pupils = $360p

    Total cost to run the school will be = 3705 + 240p

    For the school not to be in the loss,

    360p ≥ 3705 + 240p

    360p - 240p ≥ 3705

    120p ≥ 3705

    p ≥ \frac{3705}{120}

    p ≥ 30.875

    Therefore, to run the school without loss, number pupils should be at least 31.

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Step-by-step explanation:

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6 0
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A ship embarked on a long voyage. At the start of the voyage, there were 300 ants in the cargo hold of the ship. One week into t
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Answer:

It takes 1 week for the ant population to double

It takes 1.58 weeks for the ant population to triple

It takes 5.06 weeks for the ant population to reach 10000

It takes 5.09 weeks for the ant population to be 200 times the ant eater population.

Step-by-step explanation:

It takes only 1 week for the population to double from 300 to 600

We can model the population of ant (or anteater) as the following:

p = ae^{kt}

Where a = 300 is the initial population at t = 0

When t = 1, P = 600

600 = 300e^{1k}

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t = 1.1/k = 1.1/0.693 = 1.58 weeks.

When there are 10000 ants on board, p = 10000:

300e^{kt} = 10000

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kt = ln33.33 = 3.51

t = 3.51 / k = 3.51 / 0.693 = 5.06 weeks.

Similarly for anteater, at t = 0 there are 17 of them so A = 17. We can solve for their K parameter if the population doubled after 3.2 weeks

e^{3.2K} = P/A = 2

3.2K = ln2

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At the time there are 200 ants per anteater

p = 200P

300e^{kt} = 200*17e^{Kt}

e^{kt - Kt} = 200*17/300

e^{0.693t - 0.2166t} = 11.33

e^{0.4765t} = 11.33

0.4765t = ln11.33

t = ln11.33/0.4765 = 5.09 weeks

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