Question

Algebra 2B

Answer 1

The spread of the disease is an illustration of exponential equations, where the variables are exponents.

The initial number of people affected

The function is given as:

$P(t) = \frac{10000}{1 +e^{3-t}}$

At the initial time, the value of t is 0.

So, we have:

$P(0) = \frac{10000}{1 +e^{3-0}}$

$P(0) = \frac{10000}{1 +e^{3}}$

Estimate the quotient

$P(0) = 474$

Hence, the number of people infected initially is 474

The infected people after three hours

The function is given as:

$P(t) = \frac{10000}{1 +e^{3-t}}$

After three hours, the value of t is 3.

So, we have:

$P(3) = \frac{10000}{1 +e^{3-3}}$

$P(3) = \frac{10000}{1 +e^0}$

Estimate the quotient

$P(3) = 5000$

Hence, the number of people infected after three hours is 5000

The maximum number of people infected

The function is given as:

$P(t) = \frac{10000}{1 +e^{3-t}}$

As t approaches infinity, $e^{3-t}$ approaches 0

So, we have:

$P_{Max} = \frac{10000}{1 +0}$

Estimate the quotient

$P_{Max} = 10000$

Hence, the maximum number of people infected is 10000

The maximum number of people infected is 10000 because the denominator $1 + e^{3-t}$ cannot exceed 1

Time to warn its spread to over 800 people

The function is given as:

$P(t) = \frac{10000}{1 +e^{3-t}}$

When if affects 800 people, we have:

$\frac{10000}{1 +e^{3-t}} > 800$

Divide both sides by 10000

$\frac{1}{1 +e^{3-t}} > 0.08$

Take the reciprocal of both sides

$1 +e^{3-t} > 12.5$

Subtract 1 from both sides

$e^{3-t} > 11.5$

Take the natural logarithm of both sides

$3-t > \ln(11.5)$

Solve for t

$t < 3 - \ln(11.5)$

$t < 0.558$

Convert to minutes

$t < 0.558 * 60$

$t < 33.5$

Hence, the stadium should inform the guests before 33.5 minutes

Read more about exponential equations at:

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