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2 years ago

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A mold professional uses an antimicrobial chemical in an attic to combat a colony of mold. If the initial number of mold spores

in the colony is
6,200 spores, and the number of spores decreases by 20% every 5 hours,approximately how many hours after will it take for the colony of mold to be
less than 500 spores?

1 answer:

kykrilka [37]2 years ago

4 0

It would take about 57 hours for the number of spores to be less than 500 spores.

<h3>Exponential Function</h3>

An exponential function is in the form:

y = abˣ

Where y,x are variables, a is the initial value of y and b is the multiplication factor.

Let y is the number of pores after x hours.

a = 6200, b = 100% - 20% = 0.8, hence to be less than 500 spores:

$6200(0.8)^\frac{x}{5} < 500$

(x/5) * ln(0.8) < ln(0.08)

x > 56.4 hours

It would take about 57 hours for the number of spores to be less than 500 spores.

Find out more on Exponential Function at: brainly.com/question/12940982

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Step-by-step explanation:

From the question we are told that

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Generally the new probability of companies A being chosen as the sole supplier this year given that supplier E goes out of business is mathematically represented as below according to Bayes theorem

$P(a | e') = \frac{P (a \ and \ e')}{P(e')}$

$P(a | e') = \frac{P (a)}{P(e')}$

$P(a | e') = \frac{P (a)}{1- P(e)}$

=> $P(a | e') = \frac{ 0.20}{1- 0.10}$

=> $P(a | e') = 0.22$

Generally the new probability of companies B being chosen as the sole supplier this year given that supplier E goes out of business is mathematically represented as below according to Bayes theorem

$P(b | e') = \frac{P (b \ and \ e')}{P(e')}$

$P(b | e') = \frac{P (b)}{P(e')}$

$P(b | e') = \frac{P (b)}{1- P(e)}$

=> $P(b | e') = \frac{ 0.25}{1- 0.10}$

=> $P(b | e') = 0.28$

Generally the new probability of companies C being chosen as the sole supplier this year given that supplier E goes out of business is mathematically represented as below according to Bayes theorem

$P(c | e') = \frac{P (c \ and \ e')}{P(e')}$

$P(c | e') = \frac{P (c)}{P(e')}$

$P(c | e') = \frac{P (c)}{1- P(e)}$

=> $P(c | e') = \frac{ 0.15}{1- 0.10}$

=> $P(c | e') = 0.17$

Generally the new probability of companies D being chosen as the sole supplier this year given that supplier E goes out of business is mathematically represented as below according to Bayes theorem

$P(d | e') = \frac{P (d \ and \ e')}{P(e')}$

$P(d | e') = \frac{P (d)}{P(e')}$

$P(d | e') = \frac{P (d)}{1- P(e)}$

=> $P(d | e') = \frac{ 0.30}{1- 0.10}$

=> $P(c | e') = 0.33$

Generally the probability that B, D , E are not chosen this year is mathematically represented as

$P(N) = 1 - [P(e) +P(b) + P(d) ]$

=> $P(N) = 1 - [0.10 +0.25 +0.30 ]$

=> $P(N) = 0.35$

Generally the probability that A is chosen given that E , D , B are rejected this year is mathematically represented as

$P(a | e' , d' , b') = \frac{P(a)}{P(N)}$

=> $P(a | e' , d' , b') = \frac{0.20 }{0.35 }$

=> $P(a | e' , d' , b') = 0.57$

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