Total: 73
dogs: x
cats: 2x+10
2x+10+x=73
3x+10=73
3x=63
x=21
dogs: x
dogs: 21
cats: 2x+10
cats: 52
Check answer:
cats+dogs=73
52+21=73
There's no need for a system of linear equations if you were to solve a real life problem like this. In calculus, they don't give a rat about HOW you get your answer to an easy algebra 1 equation, as long as you used one of the correct methods and got the right answer.
Best of luck my friend. :)
Which data set has an outlier? 25, 36, 44, 51, 62, 77 3, 3, 3, 7, 9, 9, 10, 14 8, 17, 18, 20, 20, 21, 23, 26, 31, 39 63, 65, 66,
umka21 [38]
It's hard to tell where one set ends and the next starts. I think it's
A. 25, 36, 44, 51, 62, 77
B. 3, 3, 3, 7, 9, 9, 10, 14
C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39
D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82
Let's go through them.
A. 25, 36, 44, 51, 62, 77
That looks OK, standard deviation around 20, mean around 50, points with 2 standard deviations of the mean.
B. 3, 3, 3, 7, 9, 9, 10, 14
Average around 7, sigma around 4, within 2 sigma, seems ok.
C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39
Average around 20, sigma around 8, that 39 is hanging out there past two sigma. Let's reserve judgement and compare to the next one.
D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82
Average around 74, sigma 8, seems very tight.
I guess we conclude C has the outlier 39. That one doesn't seem like much of an outlier to me; I was looking for a lone point hanging out at five or six sigma.
L = 4 + w
perimeter = 38
38 = w + 4 + w
2w = 34
w = 17
L = 4 + 17
L = 21
i am a mathematics teacher. if anything to ask please pm me
