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3 years ago

Everett made 3/5 of the basket he shot supposed he shot 60 baskets how many did he make

Mathematics

2 answers:

vlada-n [284]3 years ago

6 0

You multiply 4 * 15 to get 60 so then you would multiply 3 * 15 so he made 45 out of the 60 shots.

olga nikolaevna [1]3 years ago

5 0

So your denominator was 5 but now your turning it into 60. So 5x12=60.
What you did to the denominator you have to do to the numerator. So 3x12=36. Your answer would be 36/60 or 36 out of 60.

Hope this helps.

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a) The expression for the height, 'H', of the plant after 't' day is;

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Step-by-step explanation:

The given maximum theoretical height of the plant = 30 in.

The height of the plant at the beginning of the experiment = 5 in.

a) The logistic differential equation can be written as follows;

$\dfrac{dH}{dt} = K \cdot H \cdot \left( M - {P} \right)$

Using the solution for the logistic differential equation, we get;

$H = \dfrac{M}{1 + A\cdot e^{-(M\cdot k) \cdot t}}$

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M = The maximum height = 30 in.

Therefore, we get;

$5 = \dfrac{30}{1 + A\cdot e^{-(30\cdot k) \cdot 0}}$

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$12 = \dfrac{30}{1 + 5\cdot e^{-(30\cdot k) \cdot 20}}$

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$5\cdot e^{-(30\cdot k) \cdot 20} = 2.5 - 1 = 1.5$

∴ -(30·k)·20 = ㏑(1.5)

k = ㏑(1.5)/(30 × 20) ≈ 6·7577518 × 10⁻⁴

k ≈ 6·7577518 × 10⁻⁴

Therefore, the expression for the height, 'H', of the plant after 't' day is given as follows

$H = \dfrac{30}{1 + 5\cdot e^{-(30\times 6.7577518 \times 10^{-4}) \cdot t}} = \dfrac{30}{1 + 5\cdot e^{-(2.02732554 \times 10^{-3}) \cdot t}}$

b) The height of the plant after 30 days is given as follows

$H = \dfrac{30}{1 + 5\cdot e^{-(2.02732554 \times 10^{-3}) \cdot t}}$

At t = 30, we have;

$H = \dfrac{30}{1 + 5\cdot e^{-(2.02732554 \times 10^{-3}) \times 30}} \approx 19.4258866473$

The height of the plant after 30 days, H ≈ 19.426 in.

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