Let's analyze each part of the problem step by step:
### Given:
- Initial population [tex]\( P_0 = 1,000,000 \)[/tex]
- Decay rate [tex]\( r = 0.04 \)[/tex]
- The population function is [tex]\( P(t) = P_0 e^{-0.04t} \)[/tex]
### Part (a)
Find the population at time [tex]\( t = 1 \)[/tex] year to the nearest thousand.
The population function is:
[tex]\[ P(t) = P_0 e^{-0.04t} \][/tex]
Substitute [tex]\( t = 1 \)[/tex] and [tex]\( P_0 = 1,000,000 \)[/tex]:
[tex]\[ P(1) = 1,000,000 \cdot e^{-0.04 \cdot 1} \][/tex]
[tex]\[ P(1) = 1,000,000 \cdot e^{-0.04} \][/tex]
[tex]\[ P(1) \approx 960,789.44 \][/tex]
To the nearest thousand:
[tex]\[ \boxed{961,000} \][/tex]
### Part (b)
How long, to the nearest tenth of a year, will it take for the population to decline to 750,000?
We need to solve for [tex]\( t \)[/tex] when [tex]\( P(t) = 750,000 \)[/tex]:
[tex]\[ 750,000 = 1,000,000 \cdot e^{-0.04t} \][/tex]
Divide both sides by 1,000,000:
[tex]\[ 0.75 = e^{-0.04t} \][/tex]
Take the natural logarithm of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ \ln(0.75) = -0.04t \][/tex]
Solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(0.75)}{-0.04} \][/tex]
[tex]\[ t \approx 7.192 \][/tex]
To the nearest tenth of a year:
[tex]\[ \boxed{7.2} \][/tex]
### Part (c)
How long, to the nearest tenth of a year, will it take for the population to decline to half the initial number?
We need to solve for [tex]\( t \)[/tex] when [tex]\( P(t) = \frac{P_0}{2} = \frac{1,000,000}{2} = 500,000 \)[/tex]:
[tex]\[ 500,000 = 1,000,000 \cdot e^{-0.04t} \][/tex]
Divide both sides by 1,000,000:
[tex]\[ 0.5 = e^{-0.04t} \][/tex]
Take the natural logarithm of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ \ln(0.5) = -0.04t \][/tex]
Solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(0.5)}{-0.04} \][/tex]
[tex]\[ t \approx 17.329 \][/tex]
To the nearest tenth of a year:
[tex]\[ \boxed{17.3} \][/tex]
So the answers to the given problems are:
- (a) 961,000
- (b) 7.2 years
- (c) 17.3 years
