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Waiting period. Upstate University currently has a 6,300-car parking capacity for faculty, staff, and students. This year, the university issued 4,272 parking passes. The number of parking passes issued has been growing at a rate of \( 7 \% \) per year. How long will it be before the university will need to add parking spaces? How long will it be before the university will need to add parking spaces? \( \square \) years (Round to twhtdecimal places.)

Ask by Wyatt Mcfarlane. in the United States
Jan 22,2025

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Tutor-Verified Answer

Answer

It will take approximately 5.74 years before the university needs to add parking spaces.

Solution

To determine how long it will take for Upstate University to exceed its current parking capacity of 6,300 cars, given that the number of parking passes is growing at an annual rate of \(7\%\), we can use the formula for exponential growth: \[ P(t) = P_0 \times (1 + r)^t \] Where: - \(P(t)\) is the future number of parking passes. - \(P_0\) is the current number of parking passes (4,272). - \(r\) is the growth rate (7% or 0.07). - \(t\) is the number of years. We need to find \(t\) when \(P(t) = 6,300\): \[ 6,300 = 4,272 \times (1.07)^t \] First, divide both sides by 4,272: \[ \frac{6,300}{4,272} = (1.07)^t \] \[ 1.475 \approx (1.07)^t \] Next, take the natural logarithm of both sides to solve for \(t\): \[ \ln(1.475) = t \times \ln(1.07) \] \[ t = \frac{\ln(1.475)}{\ln(1.07)} \] Calculating the natural logarithms: \[ \ln(1.475) \approx 0.3887 \] \[ \ln(1.07) \approx 0.0677 \] Thus, \[ t \approx \frac{0.3887}{0.0677} \approx 5.74 \text{ years} \] **Answer:** Approximately **5.74 years** are needed before the university must add parking spaces.

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To determine how long it will take before Upstate University needs to add parking spaces, we can use the formula for exponential growth. The number of parking passes issued is growing at a rate of \( 7\% \) per year. The formula we can use is: \[ P(t) = P_0 \times (1 + r)^t \] Where: - \( P(t) \) is the future number of parking passes, - \( P_0 \) is the current number of parking passes (4,272), - \( r \) is the growth rate (0.07), - \( t \) is the number of years. We want to find \( t \) when \( P(t) = 6,300 \): \[ 6,300 = 4,272 \times (1 + 0.07)^t \] To solve for \( t \), we can rearrange the equation: \[ (1.07)^t = \frac{6,300}{4,272} \] Calculating the right side: \[ \frac{6,300}{4,272} \approx 1.477 \] Now we take the logarithm of both sides: \[ \log((1.07)^t) = \log(1.477) \] Using the power rule of logarithms: \[ t \cdot \log(1.07) = \log(1.477) \] Now we divide both sides by \( \log(1.07) \): \[ t = \frac{\log(1.477)}{\log(1.07)} \] Calculating this gives: \[ t \approx \frac{0.1692}{0.0291} \approx 5.8 \text{ years} \] So, to add parking spaces, it would take approximately \( 5.80 \) years. **Answer:** \( 5.80 \) years

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