If the number of bacteria in a colony doubled every 207 min and there is currently a population of 1,500 bacteria what will the population be 621 minutes from now

the population would be 11085 in 621 minutes from now

Step-by-step explanation:

We would apply the formula,

y = ab^t

Where

a represents the initial amount of bacteria.

t represents the doubling time

From the information given

a = 1500

t = 207 minutes

Since after 207 minutes, the amount of bacteria doubles, then

y = 2 × 1500 = 3000

Therefore

3000 = 1500 × b^207

Dividing through by 1500, it becomes

2 = b^207

Taking log of both sides of the equation, it becomes

Log 2 = 207 log b

0.301 = 207 log b

Log b = 0.301/207 = 0.00145

Taking inverse log of both sides of the equation, it becomes

10^logb = 10^ 0.00145

b = 1.00334

The equation becomes

y = 1500(1.00334)^t

When t = 621 minutes, the population would be

y = 1500(1.00334)^621

y = 11085

oof tough question

step-by-step explanation:

sorry i can't ya : c gl tho

93.8

step-by-step explanation:

lets say, first number = x

second number = 3x+7

total = 49

(x)+(3x+7)= 49

4x+7 = 49

4x = 42

x = 42/4

x= 10.5

x= 10.5

3x+7 = 3(10.5)+7 = 93.8

hope this