in a school with 1500 students , one student returned from vacation with a contagious flu virus on 1st January. the spread of the virus through student body is given by y=1500/1+1499e^-0.85t where y is the total number of students infected after t days. the school would cancel classes if 40% or more of the students are ill.

Question

in a school with 1500 students , one student returned from vacation with a contagious flu virus on 1st January. the spread of the virus through student body is given by y=1500/1+1499e^-0.85t where y is the total number of students infected after t days. the school would cancel classes if 40% or more of the students are ill.

1) how many students would be infected after 6 days?
2)from which date onwards will the school have to cancel classes?

Answer 1

1) t=6

1500
2) January 6th = because January 1st is first day

Answer 2

I think you meant:

y = 1500/(1+1499e^-0.85t)

when t = 6 , I get
y = 148 , to the nearest whole number

40% of 1500 = 600

so 1500/(1+1499e^-0.85t) = 600
600(1+1499e^-0.85t) = 1500
1+1499e^-0.85t = 2.5
1499e^-0.85t = 1.5
e^-.85t = .001000667
-.85t = -6.907
t = 8.126

It would take 8 days to hit 400 students infected.

Answer 3

To find the number of students infected after 6 days, we can substitute t = 6 into the given equation y = 1500 / (1 + 1499e^(-0.85t)):

y = 1500 / (1 + 1499e^(-0.85*6))

Now, we can calculate this using a calculator or a computer program:

y = 1500 / (1 + 1499e^(-5.1))
≈ 1500 / (1 + 1499 * 0.00677575)
≈ 1500 / (1 + 10.16311581)
≈ 1500 / 11.16311581
≈ 134.61

So, after 6 days, approximately 135 students would be infected.

To determine from which date onwards the school will have to cancel classes, we need to find when 40% or more of the students are ill. In other words, we need to find when y is equal to or greater than 40% of the total number of students (40% of 1500 is 600).

We can set up the equation:

y = 1500 / (1 + 1499e^(-0.85t))

Setting y to 600, we have:

600 = 1500 / (1 + 1499e^(-0.85t))

Multiplying both sides by (1 + 1499e^(-0.85t)) gives us:

600(1 + 1499e^(-0.85t)) = 1500

Dividing both sides by 1500:

(1 + 1499e^(-0.85t)) = 1500/600
1 + 1499e^(-0.85t) = 2.5

Subtracting 1 from both sides:

1499e^(-0.85t) = 1.5

Dividing both sides by 1499:

e^(-0.85t) ≈ 0.001000667

To solve for t, we need to take the natural logarithm of both sides:

ln(e^(-0.85t)) ≈ ln(0.001000667)
-0.85t ≈ ln(0.001000667)

Finally, divide by -0.85 to isolate t:

t ≈ ln(0.001000667) / -0.85
t ≈ -6.908142
t ≈ 8.124

So, from approximately 8.124 days onwards, the school will have to cancel classes due to 40% or more of the students being ill.