On a college campus of 10,000 students, a single student relurned to campus infected by a discase. The spread of the disease through the student body is given by y = 10000/1+9999e^0.99t where y is the total number infected at time r (in days). (a) How many are infected after 4 days? (b) The school will shut down if 50% of the students are ill. During what day will it close?

Question

On a college campus of 10,000 students, a single student relurned to campus infected by a discase. The spread of the disease through the student body is given by y = 10000/1+9999e^0.99t where y is the total number infected at time r (in days). (a) How many are infected after 4 days? (b) The school will shut down if 50% of the students are ill. During what day will it close?

Answer 1

well, you have the formula, so plug in your numbers.

But I dislike the formula, since it starts at 1 and then decreases. I suspect you meant
10000/(1+9999e^(-0.99*4))
That starts at 1 and then increases.
(a) 10000/(1+9999e^(-0.99*4)) = 0.019065 = 52
(b) 10000/(1+9999e^(-0.99t)) = 5000
1+9999e^(-0.99t) = 2
e^(-.99t) = 1/9999
-.99t = -ln9999
t = 9.3

Answer 2

To calculate the number of infected students after 4 days, we substitute t = 4 into the given equation y = 10000/(1 + 9999e^(0.99t)):

(a) y = 10000/(1 + 9999e^(0.99*4))

To find out the day when the school will close, we need to determine when 50% (or 0.5) of the students are infected. We can set up the equation y = 0.5 * 10000 and solve for t:

(b) 0.5 * 10000 = 10000/(1 + 9999e^(0.99t))

Let's calculate the values for both parts:

(a) After 4 days:
y = 10000/(1 + 9999e^(0.99*4))
= 10000/(1 + 9999e^3.96)
≈ 192

Therefore, approximately 192 students will be infected after 4 days.

(b) The day when the school will close:
0.5 * 10000 = 10000/(1 + 9999e^(0.99t))

Dividing both sides of the equation by 10000, we get:

0.5 = 1/(1 + 9999e^(0.99t))

Now, rearranging the equation:

1 + 9999e^(0.99t) = 2

Subtracting 1 from both sides:

9999e^(0.99t) = 1

Dividing both sides by 9999:

e^(0.99t) = 1/9999

Now, taking the natural logarithm of both sides:

0.99t = ln(1/9999)

Dividing both sides by 0.99:

t = ln(1/9999)/0.99

Using a calculator, we can determine the value of t:

t ≈ 68.64

Therefore, the school will close on approximately the 69th day.

Answer 3

To find the number of students infected after 4 days, we can plug in t = 4 into the equation given:

y = 10000 / (1 + 9999e^0.99t)

(a) After 4 days:
y = 10000 / (1 + 9999e^0.99(4))
y ≈ 10000 / (1 + 9999e^3.96)
y ≈ 10000 / (1 + 9999 * 53.38)
y ≈ 10000 / (1 + 533745.62)
y ≈ 10000 / 533746.62
y ≈ 0.01875

Approximately 0.01875 students will be infected after 4 days.

To determine when the school will close, we need to find the day when 50% of the students are infected. We can set y = 0.5*10000, simplify the equation, and solve for t.

(b) When 50% of the students are infected:
0.5*10000 = 10000 / (1 + 9999e^0.99t)
5000 = 10000 / (1 + 9999e^0.99t)

Next, we can solve this equation for t by multiplying both sides by (1 + 9999e^0.99t) and simplifying:

5000(1 + 9999e^0.99t) = 10000
1 + 9999e^0.99t = 2
9999e^0.99t = 1

Now, divide both sides by 9999:
e^0.99t = 1/9999

To solve for t, take the natural logarithm (ln) of both sides:
ln(e^0.99t) = ln(1/9999)
0.99t = ln(1/9999)

Finally, divide both sides by 0.99 to find t:
t = ln(1/9999) / 0.99

Use a calculator to evaluate the right side of this equation, and you will find the approximate value of t.