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katen-ka-za [31]
3 years ago
6

A clothing vendor estimates that 78 out of every 100 of its online customers do not live within 50 miles of one of its physical

stores. Using this​ estimate, what is the probability that a randomly selected online customer does not live within 50 miles of a physical​ store?
Mathematics
1 answer:
Korolek [52]3 years ago
4 0

Answer:

78% probability that a randomly selected online customer does not live within 50 miles of a physical​ store.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this problem, we have that:

Total outcomes:

100 customers

Desired outcomes:

A clothing vendor estimates that 78 out of every 100 of its online customers do not live within 50 miles of one of its physical stores. So the number of desired outcomes is 78 customers.

Using this​ estimate, what is the probability that a randomly selected online customer does not live within 50 miles of a physical​ store?

p = \frac{78}{100} = 0.78

78% probability that a randomly selected online customer does not live within 50 miles of a physical​ store.

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Answer:

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

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

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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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Therefore, we get;

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

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When t = 20, H = 12

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

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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}}

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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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