The spread of a flu virus on a college campus is modeled by y= 5000/(1+4999e^-0.8t), where y is the number of students infected after t days. 1) how many students are infected after 8 days? 2) how many days will it take before half of the student population is infected?

We apply the use of algebraic equations and natural logarithm to solve the above question.

Natural logarithm is represented in short form as "In". It is the logarithm to the base constant of e.

The number of students affected after 8 days is 537 students

It will take 11 days for half the student population to be affected.

From the question, we are  given the equation:

y= 5000/(1+4999e^-0.8t)

where

y =  number of students infected after t days.

t = 8

Substituting 8 for t in the above equation,

y= 5000/(1+4999e^-0.8t)

y= 5000/(1+4999e^-0.8 x 8)

y =537.28 students

Approximately to the nearest whole number,

y = 537 students

Therefore, 537 students will be affected after 8 days.

Half the student population = 5000/2 = 2500

Hence, y = 2500

2500 =  5000 / (1+4999e^-0.8t)

Cross Multiply

2500 (1+4999e^-0.8t) = 5000

Divide both sides by 2500

(1+4999e^-0.8t) = 5000/2500

(1+4999e^-0.8t) = 2

Collect like terms

(4999e^-0.8t) = 2-1

(4999e^-0.8t) = 1

Divide both sides by 4999

(4999 / 4999e^-0.8t)=1/4999

e^-0.8t=1/4999

Take the natural logarithm of both sides

In (e^-0.8t) = In(1/4999)

-0.8t = -8.52

Divide both sides by -0.8

-0.8/-8t = -8.52/-0.8

t = 10.6462 days

Approximately to the nearest whole number

= 11 days

Therefore, it will take 11 days for half the student population to be infected.

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537 students

11 days

Step-by-step explanation:

       5000

y=  

    (1+4999e^-0.8t)

a)  after 8 days  means t =8

       5000

y=  

    (1+4999e^-0.8*8)

       5000

y=  

    (1+4999e^-6.4)

       5000

y=  

    (1+8.306)

       5000

y=  

    (9.306)

y =537.28

Rounding to the nearest student

y = 537 students

b)  1/2 the student population  means y =2500  (The 5000 is the student population)

             5000

2500=  

        (1+4999e^-0.8t)

Multiply each side by  (1+4999e^-0.8t)

2500 (1+4999e^-0.8t) = 5000

Divide each side by 2500

(1+4999e^-0.8t) = 5000/2500

(1+4999e^-0.8t) = 2

Subtract 1

(4999e^-0.8t) = 2-1

(4999e^-0.8t)=1

Divide by 4999

(4999/4999e^-0.8t)=1/4999

e^-0.8t=1/4999

Take the natural log of each side

ln(e^-0.8t)=ln(1/4999)

-.8t = ln(1/4999)

Divide by -.8

-.8/-8t = -1/.8 *ln(1/4999)

t = -1/.8 *ln(1/4999)

t≈10.6462

Rounding, it will take 11 days

by definition, an arc subtended by a central angle of some given measure has the same measure. so .

i think it would be 42cm