Question

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Explanation:

Part (a)

Step 1: Determine the value of \( t \) for the year 2002

The year 2002 is \( 2002 - 1980 = 22 \) years after 1980, so \( t = 22 \).

Step 2: Substitute \( t = 22 \) into the population model \( P(t)=\frac{280}{4 + 66e^{-0.0211t}} \)

First, calculate the exponent: \( - 0.0211\times22=- 0.4642 \)
Then, calculate \( e^{-0.4642}\approx e^{-0.4642}\approx0.630 \) (using a calculator for the exponential function)
Next, calculate the denominator: \( 4+66\times0.630 = 4 + 41.58=45.58 \)
Finally, calculate \( P(22)=\frac{280}{45.58}\approx6.14 \) (rounded to two decimal places, and since we need to round to the nearest hundred million, we check the decimal part. \( 0.14 \) billion is 140 million, so \( 6.14 \) billion is approximately \( 6.1 \) billion? Wait, no, the problem says "round answers to the nearest hundred million people". A hundred million is \( 0.1 \) billion. So \( 6.14 \) billion, the hundred million place is the first decimal place. \( 0.14 \) is closer to \( 0.1 \) or \( 0.2 \)? Wait, \( 6.14 \) billion: the number is \( 6.14 \), so to the nearest hundred million (which is \( 0.1 \) billion increment), \( 6.14 \) is \( 6.1 \) billion? Wait, no, \( 6.14 \) billion is 6 billion 140 million. The nearest hundred million: 140 million is closer to 100 million or 200 million? 140 - 100 = 40, 200 - 140 = 60, so closer to 100 million. So \( 6.1 \) billion? Wait, maybe I made a mistake in the calculation of \( e^{-0.4642} \). Let's recalculate \( e^{-0.4642} \) more accurately. Using a calculator, \( e^{-0.4642}\approx e^{-0.4642}\approx0.630 \) (more accurately, using a calculator: \( e^{-0.4642}\approx0.630 \)). Wait, maybe my initial calculation of the exponential is wrong. Let's use a calculator for \( e^{-0.0211\times22} \). \( 0.0211\times22 = 0.4642 \), so \( e^{-0.4642}\approx e^{-0.4642}\approx0.630 \). Then \( 66\times0.630 = 41.58 \), \( 4 + 41.58 = 45.58 \), \( 280\div45.58\approx6.14 \). So 6.14 billion, which is 6 billion 140 million. Rounding to the nearest hundred million: 140 million is closer to 100 million (since 140 - 100 = 40 and 200 - 140 = 60), so \( 6.1 \) billion? Wait, no, maybe I messed up the rounding. Wait, the problem says "nearest hundred million people". A hundred million is \( 0.1 \) billion. So \( 6.14 \) billion: the tenths place is 1 (representing 100 million) and the hundredths place is 4 (representing 40 million). So when rounding to the nearest hundred million, we look at the hundredths place. Since 4 < 5, we round down, so \( 6.1 \) billion? Wait, but maybe my calculation of the exponential is incorrect. Let's use a more accurate value of \( e^{-0.4642} \). Using a calculator, \( e^{-0.4642}\approx0.630 \) (actually, using a calculator: \( e^{-0.4642}\approx e^{-0.4642}\approx0.630 \)). Alternatively, maybe I should use a calculator for the entire expression. Let's do it step by step with a calculator:

\( t = 22 \)

\( -0.0211\times22=-0.4642 \)

\( e^{-0.4642}\approx e^{-0.4642}\approx0.630 \) (using calculator: \( e^{-0.4642}\approx0.630 \))

\( 66\times e^{-0.4642}\approx66\times0.630 = 41.58 \)

\( 4 + 41.58 = 45.58 \)

\( 280\div45.58\approx6.14 \)

So \( P(22)\approx6.14 \) billion. Rounding to the nearest hundred million (0.1 billion), we get \( 6.1 \) billion? Wait, but maybe the answer is expected to be around 6.1 or 6.2? Wait, perhaps I made a mistake in the exponential calculation. Let's use a more accurate value of \( e^{-0.4642} \). Let's calculate \( e^{-0.4642} \) using a calculator: \( e^{-0.4642}\approx0.630 \) (actually, using a calculator, \( e^{-0.4642}\approx0.630 \)). Alternatively, maybe the problem expects us to use a calculator for the entire expression. Let's check with a calculator:

\( P(22)=\frac{280}{4 + 66e^{-0.0211\times22}} \)

First, calculate \( -0.0211\times22=-0.4642 \)

\( e^{-0.4642}\approx0.630 \)

\( 66\times0.630 = 41.58 \)

\( 4 + 41.58 = 45.58 \)

\( 280\div45.58\approx6.14 \), so approximately 6.1 billion (rounded to the nearest hundred million).

Part (b)

Step 1: Determine the value of \( t \) for the year 2012

The year 2012 is \( 2012 - 1980 = 32 \) years after 1980, so \( t = 32 \).

Step 2: Substitute \( t = 32 \) into the population model \( P(t)=\frac{280}{4 + 66e^{-0.0211t}} \)

First, calculate the exponent: \( -0.0211\times32=-0.6752 \)
Then, calculate \( e^{-0.6752}\approx e^{-0.6752}\approx0.509 \) (using a calculator for the exponential function)
Next, calculate the denominator: \( 4 + 66\times0.509=4 + 33.594 = 37.594 \)
Finally, calculate \( P(32)=\frac{280}{37.594}\approx7.45 \) (rounded to two decimal places). Now, rounding to the nearest hundred million: 7.45 billion is 7 billion 450 million. The hundred million place is the first decimal place (4, representing 400 million) and the hundredths place is 5 (representing 50 million). Since 5 = 5, we round up, so \( 7.5 \) billion? Wait, let's check the calculation again.

Wait, \( e^{-0.6752} \): using a calculator, \( e^{-0.6752}\approx0.509 \) (more accurately, \( e^{-0.6752}\approx0.509 \)). Then \( 66\times0.509 = 33.594 \), \( 4+33.594 = 37.594 \), \( 280\div37.594\approx7.45 \). So 7.45 billion, which is 7 billion 450 million. Rounding to the nearest hundred million: 450 million is closer to 500 million (since 450 - 400 = 50 and 500 - 450 = 50? Wait, no, hundred million increments: 7.45 billion is 7.45, so the tenths place is 4 (400 million) and the hundredths place is 5 (50 million). When rounding to the nearest hundred million, we look at the hundredths place. Since 5 = 5, we round up the tenths place: \( 7.5 \) billion.

Part (c)

Step 1: Analyze the limit as \( t\to\infty \)

We are given that as \( t \) becomes very large (greater than 500), \( e^{-0.0211t}\approx0 \). So we substitute \( e^{-0.0211t} = 0 \) into the population model \( P(t)=\frac{280}{4 + 66e^{-0.0211t}} \).

Step 2: Calculate the limit of \( P(t) \) as \( t\to\infty \)

Substituting \( e^{-0.0211t}=0 \) into the denominator, we get \( 4+66\times0 = 4 \). Then \( P(t)=\frac{280}{4}=70 \). So as \( t \) becomes very large, the population approaches 70 billion. This suggests that the maximum population the Earth can support is 70 billion people.

Answers:

(a) \(\boxed{6.1}\) (or more accurately, if we re - calculate with more precise \( e^{-0.4642}\approx0.630 \), we might get a slightly different value, but following the steps, the answer is approximately 6.1 billion)
(b) \(\boxed{7.5}\) (similarly, the precise calculation gives approximately 7.5 billion)
(c) The maximum population the Earth can support is 70 billion people. So the correct option is: The maximum population the Earth can support is 70 billion people.

Answer:

s:
(a) \(\boxed{6.1}\) (or more accurately, if we re - calculate with more precise \( e^{-0.4642}\approx0.630 \), we might get a slightly different value, but following the steps, the answer is approximately 6.1 billion)
(b) \(\boxed{7.5}\) (similarly, the precise calculation gives approximately 7.5 billion)
(c) The maximum population the Earth can support is 70 billion people. So the correct option is: The maximum population the Earth can support is 70 billion people.