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11. The health department of a large city has developed an air pollution index that MeaSureS the level ol several air pollutants that cISC' respiratory dlistre...

Question

11. The health department of a large city has developed an air pollution index that MeaSureS the level ol several air pollutants that cISC' respiratory dlistress in Iaus , The following table gives the pollution indlex (On scale of to 10. with 10 being the worst ) for randomly selected SCT days aId the mnber of patients with acute respiratory problems admitted t0 the emergency rOoms of the city s hospitals. Air pollution index 15 6,7 82 50 4,6 6 3.0 Emergency aclmissions 53 82 102 60 39 Tak

11. The health department of a large city has developed an air pollution index that MeaSureS the level ol several air pollutants that cISC' respiratory dlistress in Iaus , The following table gives the pollution indlex (On scale of to 10. with 10 being the worst ) for randomly selected SCT days aId the mnber of patients with acute respiratory problems admitted t0 the emergency rOoms of the city s hospitals. Air pollution index 15 6,7 82 50 4,6 6 3.0 Emergency aclmissions 53 82 102 60 39 Taking the air pollution index as an indlependent variable $ and the Im- ber o emergency admissions a5 # dependent variable We ASSMMC probabilistie linear regression moclel given by v ="+6r + 6 (3 points) Compute the least squares estimates a and b for and respectively: points) Compute SSE. i2 the Hinimm achieved SUm of squares of errors



Answers

Pollution. In a large city, the amount of sulfur dioxide pollutant released into the atmosphere due to the burning of coal and oil for heating purposes varies seasonally. Suppose that the number of tons of pollutant released into the atmosphere during the $n$ th week after January 1 is given approximately by $$ P(n)=1+\cos \frac{\pi n}{26} \quad 0 \leq n \leq 104 $$ (A) What is the rate of change of pollutant $n$ weeks after the first of the year? (B) What is the rate of change of pollutant 13 weeks after the first of the year? 26 weeks after the first of the year? 30 weeks after the first of the year? (C) Find all local maxima and minima for $0<t<104$. (D) Find the absolute maximum and minimum for$0 \leq t \leq 104$ (E) Repeat part ( $\mathrm{C}$ ), using a graphing calculator.

In part A. We want to find the rate of the change of the pollution, which is just the derivative of the pollution function. So to take the relatives to the pollution function. We use a chain rule. The derivative of pile over 26 is just a pie over 26. And the derivative of the cold Sine function is a negative sine function enhance its derivative, adjust. My nurse pi over 26 times. Sign pi over 26 for part B. We just evaluate the derivative 13 26. Young 30. And their function values miners 0.1210 and the 0.56 respectively. For percy we want to find the local maximum and local minimum. So we first need to find the critical points and we just let the relative equals zero. And it gives us three critical points. Reach 26 52 78. And to determine whether they are local maximum on local minimum. Official use the second order derivative. So were you vary it the second other derivative. These three points and their signs are positive, negative and positive respectively. And hence their local minimum. Local maximum and local minimum respectively. For per day. We want to find the absolute maximum minimum between the injured grow from 0 to 100 and four. All these very critical numbers are visiting this interval and hence we just evaluate their function values and this five point the function values are 2020 And to respect enhanced the other absolute maximum, absolute minimum, absolute maximum absolute minimum, an absolute maximum respectively. And here is a graph of the pollution function. You can also use this graph to verify the answer in party and party

Yes of exercise 70 pts are pollution function for parade We need to evaluate the function value at zero 39 52 and 65. Co sign zero echoes one and has P zero vehicles to 39 pi over 26 as three. Power to and co sign serve. Our two equals zero. Hence, p 39 equals one mhm 52 pi over 26 Because do pine and co sign two pi equals what hands p 52 You close to 65 pi over 26 equals five. Power to and co sign firepower to equal zero. Hence pie 65 equals What for Barbie. We evaluated the function value at 10 and 95. Here we have no and medical solutions. Hence videos a calculator to find the numerical approximation. Pitching is approximately 1.355 m p 95 years approximately 1.465 They are the amount of pollution released into the atmosphere at the 10th and 95th week. This is Percy. The graph of this pollution function

Yeah, Here the pollution function is given which is one plus Kozulin party over 26. And we want to evaluate its value. Yeah. Zero 39 52 and 65. So which is a substitute T. Busies values. And my nose at its coastline. zero equals one call sign three power two equals zero coastline to pi equals minus one and co sign firepower draw so equals zero and hence function values 2121 respect for part B. We want to evaluate the pollution at 10 and 95. Therefore we're just a substitute tv these two values and then they grew up calculator. We can find the approximate value which are 1.355 And one of 465 respect.

Hello. We have question number 58 related to the optimization. Okay, Um, we have been provided with a model 80 which is 136 divided by one place, 0.25 T minus 4.5. Holds choir plus 28. Where t is between 0 to 11. And this is the okay. This is the amount of nitrogen dioxide. Yeah, it's an 80 standard and XT million hours. Okay, no problem. We have to find the time of the day when the pollution is at its highest level. So we have to optimize this. So for, uh, to be at highest level one can easily see that for this expression to be maximum expression to be maximum, this denominator has to be minimum. So we have to minimize this denominated. We should write fall 80 to be max the denominator. One plus 0.25 t minus 4.5. Whole square has to be minimum. And for this expression to be minimum this because since the simpler science, so anything and this whole square, so anything here other than zero will get it to the value greater than what we require. So for this to be minimum, we have t minus 4.5 to be equal to zero. If this expression becomes zero, then only it can be minimized. So t equal to 4.5 Ours now 4.5 hours means seven plus 4.5. That is 11.5 11.5 a m. So this is the required time where pollution level is maximum. Thank you.


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