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
Answer:
8
Step-by-step explanation:
I got 8 in a short version. The full answer is not complete. It is 8.18812697...
But if you shorten it, 8 is the best estimation.
Answer:
- 2
Step-by-step explanation:
The average rate of change of h(x) in the closed interval [a, b ] is
Here [a, b ] = [1, 8 ], thus
f(b) = f(8) = - 8² + 7(8) + 14 = - 64 + 56 + 14 = 6
f(a) = f(1) = - 1² + 7(1) + 14 = - 1 + 7 + 14 = 20
average rate of change = 
= 
= - 2
