Answer:
410.32
Step-by-step explanation:
Given that the initial quantity, Q= 6200
Decay rate, r = 5.5% per month
So, the value of quantity after 1 month, 

The value of quantity after 2 months, 

From equation (i)

The value of quantity after 3 months, 

From equation (ii)


Similarly, the value of quantity after n months,

As 4 years = 48 months, so puttion n=48 to get the value of quantity after 4 years, we have,

Putting Q=6200 and r=5.5%=0.055, we have

Hence, the value of quantity after 4 years is 410.32.
Answer:
its an enlargment
Step-by-step explanation: ur welcome :)
28.7083333333...
But you could also round it up to 29.
Answer:
2^5 and 3^4
Step-by-step explanation:
you can see there are 5 2's and 4 3's. so it would be 2^5 and 3^4
Answer:
1. 1343 years
2. 9 hours
3. 39 years
Step-by-step explanation:
1. Given, half-life of carbon = 5730 years.
∴ λ = 0.693/half-life of carbon = 0.693/5730 = 0.000121
If N₀ = 100 then N = 85
Formula:- N = N₀*e^(-λt)
∴ 85 = 100 * e^(-0.000121t)
∴㏑(-0.85)=-0.000121t
∴ t = 1343 years
2. Given half-life of aspirin = 12 hours
λ = 0.693/12 = 0.5775
Also N₀ = 100 then 70 will disintegrate and N = 30 will remain disintegrated.
∴ 70 = 100 *e^(-0.05775t)
0.70 = e^(-0.05775t)
㏑(0.70) = -0.05775t
∴ t = 9 hours
3. The population of the birds as as A=A₀*e^(kt)
Given that the population of birds fell from 1400 from 1000, We are asked how much time it will take for the population to drop below 100, let that be x years.
The population is 1400 when f = 0, And it is 1000 when f = 5
We can write the following equation :
1400 = 1000e^(5t).
∴1400/1000 = e^(5k)
∴ k = ㏑(1.4)/5
We need to find x such that 1400/100 = e^(xk)
14 = e^(xk)
∴ x = 39 years