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Tha fcllowing chart shows Ine Ware_Job: Rochester per OJO working age adults over Year 1997 1998 999 2080 2C01 Jobs 645 700 750 785year Perod_What - the average rat...

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Tha fcllowing chart shows Ine Ware_Job: Rochester per OJO working age adults over Year 1997 1998 999 2080 2C01 Jobs 645 700 750 785year Perod_What - the average rate of change the number (winr waze job: Irom 1997 to 19997 Jobs/ Year What = the dverage rate of change the numbe [ivne wage jobs from 1999 to 20017 Jobs / YearQuestion Help: @video @message instructor

Tha fcllowing chart shows Ine Ware_Job: Rochester per OJO working age adults over Year 1997 1998 999 2080 2C01 Jobs 645 700 750 785 year Perod_ What - the average rate of change the number (winr waze job: Irom 1997 to 19997 Jobs/ Year What = the dverage rate of change the numbe [ivne wage jobs from 1999 to 20017 Jobs / Year Question Help: @video @message instructor



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The graph here shows the average salary for a public school high school principal for several years beginning with $1990 . x$ represents the number of years after 1990 so that $x=0$ represents $1990, x=1$ represents $1991,$ and so on. Let $y$ represent the average salary of a high school principal.
a) Write a linear equation to model this data. Use the data points for 1990 and $2001 .$ (Round the slope to the nearest whole number.) b) Explain the meaning of the slope in the context of the problem. c) Use the equation to estimate the average salary in 1993 d) If the current trend continues, find the average salary in 2005

Party of this one. It wants us to notice the general trend. So what I notice is, uh, the first gap is 11 years, and it went up about 11,000 uh, dollars, and then the next It was in two years, went up about $2000 the last was one year when about $1000. Approximately. So this is not linear, but it's close. So, um, I want to make sure you say not linear. It's closer than some of the others we worked with. Part B wants us to create a function since 1989 is with time since 1989 is the independent variable? Um, so we could dio medium incomes. Let's do a capital I of tea equals funder slopes for him since, but things were going up, it's gonna be a positive slow. And so the difference in our funds from 2003 to 1989 is 44,003. 68 by this 30,056. So it's gonna be 14. 3 12 14,312 and the number of years it's been since 1989 is four team. So 14 gree 12 to bed about 14,022 120.29 So that represents, like, how much it's gonna per year. And then in 1989 it was at 30,056. So there's our equation for our income and in part C. It wants us to find what year? The median sigh of sorry before the 6000 somebody on the next line. So we have 46,000 equals slow whenever we just had 10 1022 and clean and sense times time plus 30,056. So the first step is to subtract 30,000. 56. 46,000 and it's 30,056 which is 15,944 That's gonna equal are $1022.29 times time. So is divide both sides by that and we'll have tea is equal to put this in the calculator. Uh, and 15.6 15.6 years. It will be sometime. Dearing 15.6 years after 1989 would be 2004. So sometime during 2000 for near the second half, uh, the median income will reach that 46,000 according to this model. Okay, thank you very much.

Okay, We are discussing graphs and functions and in this case, particularly the midpoint formula. So remember that the midpoint formula you will add the two numbers have been divide by two. So this will be X one plus x to over to why one plus. Why, too, over to Alright for our particular problem. We have enrollments in public colleges and we assume a linear relationship with me, which means that we can use the midpoint formula. So we need Thio find the poverty cut off in 1987. So between 1970 and 2000 and four, that's 34 years, half of 34 17. So that's in between. So 1987 is the midpoint of 1970 and 2000 and four. So we have 39 68 plus 19,157 over to so in 1987 the income. Excuse me. We're talking about income, not public colleges, and much just of read a seconds of the next problem. So in 1987 the total income is 11,563 approximately

So in 1990 we are approximately 4.9. Oh, no, no. We're at about four point seven, I would say. And at 1985 we are at 4.1, So we're gonna put dollar amounts on top beers on the bottom 4.7 minus 4.1 is just going to be use your 0.6 and then this is just a difference of five years. So your 0.6 divided by five is going to be about a 12 sent increase per year.

The given problem we have a model involving the median age of um us population over the decades. So it can be given by the following model which is F. F. T. Equals um negative zero point 2176. Um -0.2176. And that's going to be t cubed. Mhm Plus 1.962. teaspoons -2.8 33 T. That's 29.4 Between zero and 5. Yeah. So this is the model that we're dealing with you. And we see that since T equals zero corresponds to 1960. We can find the median age And your 2000. If you let this equal fft And then we have for example X. and 40 would be the median age in 2000. We see if that's going to be undefined but we can see the rate at which it was changing. Um Since this is uh this should be to 50. So we see that the rate at which this is the median age and the rated which is changing. So it's decreasing.


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