Question

score on last try: 3.33 of 5 pts. see details for more. at least one scored part is incorrect. jump to first changable incorrect part. get a similar question you can retry this question below utah, the land of milk and honey where math students have courage and freedom rings true, is one of the fastest growing states in america. the state population was 2.942 million people in the year 2014, and 3.273 million people in the year 2020. for the prompts below, you may approximate this as an exponential function with a constant growth rate. what is the annual percentage growth rate between 2014 and 2020, rounded to two decimals? growth rate = 1.78 % at the current growth rate, what is the expected state population in the year 2036, rounded to two decimals? expected population of utah in 2036 = million people enter an integer or decimal number more.. at the current growth rate, during what year (round down to a whole number) is the population expected to reach 4.6 million people? utah expects to reach 4.6 million people in the year: 2039 hint: remember to not round off any decimals during calculations! question help: video

Explanation:

Step1: Identify the exponential growth formula

The exponential growth formula is \( P(t) = P_0(1 + r)^t \), where \( P(t) \) is the population at time \( t \), \( P_0 \) is the initial population, \( r \) is the annual growth rate (in decimal), and \( t \) is the time in years.

We know \( P_0 = 2.942 \) million (2014 population), \( P(t) = 3.273 \) million (2020 population), and \( t = 2020 - 2014 = 6 \) years.

Step2: Solve for the growth rate \( r \)

Substitute the known values into the formula:
\( 3.273 = 2.942(1 + r)^6 \)

First, divide both sides by \( 2.942 \):
\( \frac{3.273}{2.942} = (1 + r)^6 \)

Calculate \( \frac{3.273}{2.942} \approx 1.1125 \)

Then, take the 6th root of both sides:
\( 1 + r = 1.1125^{\frac{1}{6}} \)

Calculate \( 1.1125^{\frac{1}{6}} \approx 1.0178 \) (since \( 1.0178^6 \approx 1.1125 \))

So, \( r \approx 1.0178 - 1 = 0.0178 \), which is \( 1.78\% \) (this matches the given growth rate, so we can use \( r = 0.0178 \) for the next parts).

Step3: Find the expected population in 2036

First, find the time \( t \) from 2014 to 2036: \( t = 2036 - 2014 = 22 \) years.

Use the exponential growth formula again with \( P_0 = 2.942 \), \( r = 0.0178 \), and \( t = 22 \):
\( P(22) = 2.942(1 + 0.0178)^{22} \)

Calculate \( (1 + 0.0178)^{22} \approx (1.0178)^{22} \). Let's compute this:
\( 1.0178^{22} \approx e^{22 \times \ln(1.0178)} \) (using the natural exponent approximation, or direct calculation)
\( \ln(1.0178) \approx 0.0176 \), so \( 22 \times 0.0176 \approx 0.3872 \)
\( e^{0.3872} \approx 1.472 \) (alternatively, direct calculation: \( 1.0178^{22} \approx 1.472 \))

Then, \( P(22) = 2.942 \times 1.472 \approx 4.33 \) million (rounded to two decimals). Wait, but let's check the time from 2020 to 2036: \( 2036 - 2020 = 16 \) years. Wait, maybe I made a mistake in the initial time. Wait, 2014 to 2036 is 22 years, but if we use 2020 as a reference: 2020 population is 3.273 million, time to 2036 is \( 2036 - 2020 = 16 \) years.

Let's recalculate using 2020 as \( P_0 = 3.273 \), \( t = 16 \), \( r = 0.0178 \):

\( P(16) = 3.273(1 + 0.0178)^{16} \)

Calculate \( (1.0178)^{16} \approx e^{16 \times 0.0178} \approx e^{0.2848} \approx 1.329 \)

Then, \( P(16) = 3.273 \times 1.329 \approx 4.35 \)? Wait, maybe my initial time was wrong. Wait, the problem says "at the current growth rate, what is the expected state population in the year 2036". Let's use the correct time from 2014 to 2036: 2036 - 2014 = 22 years.

Using \( P_0 = 2.942 \), \( t = 22 \), \( r = 0.0178 \):

\( (1 + 0.0178)^{22} \approx 1.0178^{22} \). Let's calculate this more accurately:

\( 1.0178^1 = 1.0178 \)

\( 1.0178^2 = 1.0178 \times 1.0178 \approx 1.0359 \)

\( 1.0178^4 = (1.0359)^2 \approx 1.0731 \)

\( 1.0178^8 = (1.0731)^2 \approx 1.1515 \)

\( 1.0178^{16} = (1.1515)^2 \approx 1.326 \)

\( 1.0178^{22} = 1.0178^{16} \times 1.0178^4 \times 1.0178^2 \approx 1.326 \times 1.0731 \times 1.0359 \approx 1.326 \times 1.112 \approx 1.475 \)

Then, \( P(22) = 2.942 \times 1.475 \approx 4.34 \) million. Wait, but maybe the problem expects using the growth rate to calculate from 2014 to 2036. Alternatively, maybe the first part (finding the growth rate) was given as 1.78%, so we can use that.

Wait, the problem says "At the current growth rate, what is the expected state population in the year 2036, rounded to two decimals?"

So, \( P_0 = 2.942 \) (2014), \( t = 2036 - 2014 = 22 \), \( r = 0.0178 \)

\( P(t) = 2.942(1 + 0.0178)^{22} \)

Calculate \( (1.0178)^{22} \):

Using a calculator, \( 1.0178^{22} \approx e^{22 \times \ln(1.0178)} \)

\( \ln(1.0178) \approx 0.01763 \)

\( 22 \times 0.01763 \approx 0.38786 \)

\( e^{0.38786} \approx 1.472 \)

So, \( 2.942 \times 1.472 \approx 4.33 \) million. Wait, but maybe the time is from 2020 to 2036: 2036 - 2020 = 16 years.

\( P_0 = 3.273 \) (2020), \( t = 16 \), \( r = 0.0178 \)

\( P(16) = 3.273(1.0178)^{16} \)

\( (1.0178)^{16} \approx e^{16 \times 0.0178} \approx e^{0.2848} \approx 1.329 \)

\( 3.273 \times 1.329 \approx 4.35 \) million. Hmm, maybe there's a miscalculation. Wait, let's use the formula correctly.

Alternatively, maybe the growth rate is 1.78% per year, so the formula is \( P(t) = P_0 e^{rt} \) (continuous growth), but the problem says "exponential function with a constant growth rate", which can be either discrete or continuous. But since we found the growth rate as 1.78% (discrete), let's stick to discrete.

Wait, the first part: from 2014 (2.942) to 2020 (3.273) is 6 years, so \( 3.273 = 2.942(1 + r)^6 \), so \( r = (\frac{3.273}{2.942})^{1/6} - 1 \approx 0.0178 \), so that's correct.

Now, for 2036, which is 2036 - 2014 = 22 years later.

So, \( P(22) = 2.942(1 + 0.0178)^{22} \)

Calculate \( (1.0178)^{22} \):

Using a calculator, \( 1.0178^{22} \approx 1.0178 \times 1.0178 \times ... \) (22 times)

Alternatively, use the formula for compound interest:

\( A = P(1 + r)^t \)

So, \( A = 2.942(1.0178)^{22} \)

Let's compute \( 1.0178^{10} \approx 1.0178^{10} \approx 1.190 \) (since \( 1.0178^5 \approx 1.091 \), then square it: \( 1.091^2 \approx 1.190 \))

\( 1.0178^{20} \approx (1.190)^2 \approx 1.416 \)

\( 1.0178^{22} = 1.0178^{20} \times 1.0178^2 \approx 1.416 \times 1.0359 \approx 1.467 \)

Then, \( 2.942 \times 1.467 \approx 4.31 \) million? Wait, this is confusing. Wait, maybe the problem has a typo, or maybe I made a mistake. Wait, the given growth rate is 1.78%, so let's use that.

Wait, maybe the time from 2014 to 2036 is 22 years, so:

\( P = 2.942(1 + 0.0178)^{22} \)

Using a calculator for \( (1.0178)^{22} \):

\( \ln(1.0178) = 0.01763 \)

\( 22 \times 0.01763 = 0.38786 \)

\( e^{0.38786} = 1.472 \) (if we use continuous growth, but the problem says exponential with constant growth rate, which can be discrete or continuous. But since we found the discrete rate as 1.78%, maybe the problem expects discrete. However, the difference between discrete and continuous for small rates is minimal.

Wait, maybe the answer is approximately 4.33 million (rounded to two decimals). But let's check with the given numbers.

Alternatively, maybe the problem is using the growth rate to calculate from 2020 to 2036. 2020 population is 3.273 million, time to 2036 is 16 years.

So, \( P = 3.273(1 + 0.0178)^{16} \)

Calculate \( (1.0178)^{16} \):

\( \ln(1.0178) = 0.01763 \)

\( 16 \times 0.01763 = 0.2821 \)

\( e^{0.2821} = 1.326 \)

\( 3.273 \times 1.326 = 4.34 \) million.

But the problem says "rounded to two decimals", so maybe 4.33 or 4.34. Wait, maybe the correct answer is 4.33 million. But let's check with the initial numbers.

Wait, the first part: 2014 to 2020 is 6 years, population from 2.942 to 3.273.

\( 3.273 / 2.942 = 1.1125 \)

\( 1.1125^{1/6} = 1.0178 \), so growth rate is 1.78% per year.

Then, 2036 is 22 years after 2014, so:

\( 2.942 \times (1.0178)^{22} \)

Using a calculator, \( (1.0178)^{22} \approx 1.0178^{22} \approx 1.472 \)

\( 2.942 \times 1.472 \approx 4.33 \) million.

So, the expected population in 2036 is approximately 4.33 million (rounded to two decimals).

Step4: Find the year when population reaches 4.6 million

We need to find \( t \) such that \( 4.6 = 2.942(1 + 0.0178)^t \)

Divide both sides by 2.942:

\( \frac{4.6}{2.942} = (1.0178)^t \)

Calculate \( \frac{4.6}{2.942} \approx 1.5636 \)

Take the natural logarithm of both sides:

\( \ln(1.5636) = t \ln(1.0178) \)

Calculate \( \ln(1.5636) \approx 0.447 \)

\( \ln(1.0178) \approx 0.0176 \)

Then, \( t = \frac{0.447}{0.0176} \approx 25.4 \) years.

Since we start in 2014, the year is \( 2014 + 25.4 \approx 2039.4 \), so we round down to 2039 (as the problem says to round down to a whole number).

Answer:

The expected population of Utah in 2036 is approximately \(\boxed{4.33}\) million people.

(For the year when it reaches 4.6 million, it's 2039, as given in the hint, but the question here is about the population in 2036, so the answer is 4.33 (or similar, depending on precise calculation).)