For Questions 1 — 2, use the following. Scooters are often used in European and Asian cities because of their ability to negotiate crowded city streets. The number of scooters (in thousands) sold each year in India can be approximated by the function N = 61.86t^2 — 237.43t + 943.51 where f is the number of years since 1990. 1. Find the vertical intercept. What is the practical meaning of the vertical intercept in this situation? 2. Use a numerical method to find the year when the number of scooters sold reaches 1 million. (Note that 1 million is 1,000 thousand, so N = 1000) Show three rows of the table you used to support your answer and write a clear answer to the problem.

Question
Modeling
asked 2020-10-18
For Questions 1 — 2, use the following. Scooters are often used in European and Asian cities because of their ability to negotiate crowded city streets. The number of scooters (in thousands) sold each year in India can be approximated by the function \(N = 61.86t^2 — 237.43t + 943.51\) where f is the number of years since 1990. 1. Find the vertical intercept. What is the practical meaning of the vertical intercept in this situation? 2. Use a numerical method to find the year when the number of scooters sold reaches 1 million. (Note that 1 million is 1,000 thousand, so N = 1000) Show three rows of the table you used to support your answer and write a clear answer to the problem.

Answers (1)

2020-10-19
Step 1 Given function \(N = 61.86t^2 − 237.43t + 943.51\) 1) For finding the vertical intercept, put \(t = 0\) \(N = 61.86(0)^2 − 237.43(0) + 943.51\)
\(N = 943.51\) The practical meaning of the vertical intercept in this situation is: When\(t = 0,\) that is in 1990, the number of scooters sold in India = 943 (approximately). 2) When N = 1000, the equation becomes, \(61.86t^2 − 237.43t + 943.51 = 1000\) Solving using bisection method, \(f(t) = 61.86t^2 − 237.43t + 943.51 − 1000\) \(= 61.86t^2 − 237.43t − 56.49\) Step 2 \(f(−1) = 61.86+237.43−56.49 = 242.8\)
\(f(0) = −56.49\)
\(f(1) = 61.86 − 237.43 − 56.49 =−232.06\) Since \(f(−1)>0 and f(0)<0\)</span>, zero of the function lies between -1 and 0. Take \(t1=\frac{−1 + 0}{2}} =\frac{−1}{2} = −0.5\) \(f(−0.5) = 15.465 + 118.715 − 56.49\)
\(=77.69 > 0\) Thus root lies between -0.5 and 0 Take \(t_2 = \frac{−0.5 + 0}{2} = −0.25\)
\(f(−0.25) = 3.86 + 59.36 − 56.49\)
\(=6.73 > 0\) Step 3 Thus root lies between -0.25 and 0 Take \(t_3 = \frac{−0.25 + 0}{2} = −0.125\)
\(f(−0.125) = 0.96+29.69−56.49\)
\(=−25.84<0\)</span> \(\begin{array}{|c|c|}n & a (negative) & b (positive) & t=a+b2 & f(t)\ 1 & -1 & 0 & -0.5 & 77.69\\ 2 & -0.5 & 0 & -0.25 & 6.73\\ 3 & -0.25 & 0 & -0.125 & -25.84\end{array}\) Root here lies between -0.25 and -0.125. Thus root lies between \(\frac{−0.25−0.125}{2} = −0.1875\).
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The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations,
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\(\displaystyle{0},{025}<{P}-\text{value}<{0},{050}\)
\(\displaystyle{0},{005}<{P}-\text{value}<{0},{025}\)
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P.vaiue Pevgiue
P-value f P-value
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