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Naya [18.7K]
3 years ago
5

Harold and Millicent are getting married and need to combine their already-full libraries. If Harold, who has 1/2 as many books

as Millicent, brings 1/3 of his books to their new home, then Millicent will have enough room to bring 1/2 of her books to their new home. What fraction of Millicent's old library capacity is the new home's library capacit
Mathematics
1 answer:
slavikrds [6]3 years ago
3 0

Answer:

2/3

Step-by-step explanation:

Let x be the number of books that Millicent has. So the number of books that Harold has is x/2. And the number of book that Harold is bringing to the new home is (x/2)/3 = x/6.

The home has enough room for another half of Millicent's books, which is x/2. So the total books capacity of the new home is

\frac{x}{6} + \frac{x}{2} = \frac{x}{6} + \frac{3x}{6} = \frac{4x}{6} = \frac{2x}{3}

Since x is also the number of books in Millicent's old library capacity, we can conclude that the fraction of Millicent's old library capacity to the new home's library capacity is 2/3, or 2 thirds

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(a) y = 350,000 \times (1 + 0.07132)^t

(b) (i) The population after 8 hours is 607,325

(ii) The population after 24 hours is 1,828,643

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(d) The doubling time of the population is approximately, 10.06 hours

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(a) The initial population of the bacteria, y₁ = a = 350,000

The time the colony grows, t = 12 hours

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The exponential growth model, can be written as follows;

y = a \cdot (1 + r)^t

Plugging in the values, we get;

800,000 = 350,000 \times (1 + r)^{12}

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12·㏑(1 + r) = ㏑(16/7)

㏑(1 + r) = (㏑(16/7))/12

r = e^((㏑(16/7))/12) - 1 ≈ 0.07132

The  model is therefore;

y = 350,000 \times (1 + 0.07132)^t

(b) (i) The population after 8 hours is given as follows;

y = 350,000 × (1 + 0.07132)⁸ ≈ 607,325.82

By rounding down, we have;

The population after 8 hours, y = 607,325

(ii) The population after 24 hours is given as follows;

y = 350,000 × (1 + 0.07132)²⁴ ≈ 1,828,643.92571

By rounding down, we have;

The population after 24 hours, y = 1,828,643

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Initial population = y

Final population = 2·y

The doubling time of the population is therefore;

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Therefore, we have;

2·y/y =2 = (1 + 0.07132)^t

t = ln2/(ln(1 + 0.07132)) ≈ 10.06

The doubling time of the population is approximately, 10.06 hours.

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