It’s is 1800
Hope this helps
Have a great day/night
Answer:1,200 meters per minute
Step-by-step explanation:
Answer:
95%: (3278.354 ; 3270.083)
99% : (3221.646 ; 3278.354)
Step-by-step explanation:
Given :
Sample size, n = 12
Mean, xbar = 3250
Sample standard deviation = √1000
The 95% confidence interval :
Mean ± Margin of error
Margin of Error = Tcritical * s/√n
Tcritical at 0.05, df=12-1 = 11 ;
Tcritical at 95% = 2.20
Hence,
Margin of Error = (2.20 * √1000/√12) = 20.083
Confidence interval : 3250 ± 20.083
Lower boundary = 3250 - 20.083 = 3229.917
Upper boundary = 3250 + 20.083 = 3270.083
2.)
The 99% confidence interval :
Mean ± Margin of error
Margin of Error = Tcritical * s/√n
Tcritical at 0.01, df=12-1 = 11 ;
Tcritical at 99% = 3.106
Hence,
Margin of Error = (3.106 * √1000/√12) = 28.354
Confidence interval : 3250 ± 28.354
Lower boundary = 3250 - 28.354 = 3221.646
Upper boundary = 3250 + 28.354 = 3278.354
![\bf f(x)=y=2x+sin(x) \\\\\\ inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x) \\\\\\ \textit{now, the "y" in the inverse, is really just g(x)} \\\\\\ \textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\ -----------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%3Dy%3D2x%2Bsin%28x%29%0A%5C%5C%5C%5C%5C%5C%0Ainverse%5Cimplies%20x%3D2y%2Bsin%28y%29%5Cleftarrow%20f%5E%7B-1%7D%28x%29%5Cleftarrow%20g%28x%29%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bnow%2C%20the%20%22y%22%20in%20the%20inverse%2C%20is%20really%20just%20g%28x%29%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bso%2C%20we%20can%20write%20it%20as%20%7Dx%3D2g%28x%29%2Bsin%5Bg%28x%29%5D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C)
![\bf \textit{let's use implicit differentiation}\\\\ 1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor} \\\\\\ 1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\ -----------------------------\\\\ g'(2)=\cfrac{1}{2+cos[g(2)]}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Blet%27s%20use%20implicit%20differentiation%7D%5C%5C%5C%5C%0A1%3D2%5Ccfrac%7Bdg%28x%29%7D%7Bdx%7D%2Bcos%5Bg%28x%29%5D%5Ccdot%20%5Ccfrac%7Bdg%28x%29%7D%7Bdx%7D%5Cimpliedby%20%5Ctextit%7Bcommon%20factor%7D%0A%5C%5C%5C%5C%5C%5C%0A1%3D%5Ccfrac%7Bdg%28x%29%7D%7Bdx%7D%5B2%2Bcos%5Bg%28x%29%5D%5D%5Cimplies%20%5Ccfrac%7B1%7D%7B%5B2%2Bcos%5Bg%28x%29%5D%5D%7D%3D%5Ccfrac%7Bdg%28x%29%7D%7Bdx%7D%3Dg%27%28x%29%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Ag%27%282%29%3D%5Ccfrac%7B1%7D%7B2%2Bcos%5Bg%282%29%5D%7D)
now, if we just knew what g(2) is, we'd be golden, however, we dunno
BUT, recall, g(x) is the inverse of f(x), meaning, all domain for f(x) is really the range of g(x) and, the range for f(x), is the domain for g(x)
for inverse expressions, the domain and range is the same as the original, just switched over
so, g(2) = some range value
that means if we use that value in f(x), f( some range value) = 2
so... in short, instead of getting the range from g(2), let's get the domain of f(x) IF the range is 2
thus 2 = 2x+sin(x)
![\bf 2=2x+sin(x)\implies 0=2x+sin(x)-2 \\\\\\ -----------------------------\\\\ g'(2)=\cfrac{1}{2+cos[g(2)]}\implies g'(2)=\cfrac{1}{2+cos[2x+sin(x)-2]}](https://tex.z-dn.net/?f=%5Cbf%202%3D2x%2Bsin%28x%29%5Cimplies%200%3D2x%2Bsin%28x%29-2%0A%5C%5C%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Ag%27%282%29%3D%5Ccfrac%7B1%7D%7B2%2Bcos%5Bg%282%29%5D%7D%5Cimplies%20g%27%282%29%3D%5Ccfrac%7B1%7D%7B2%2Bcos%5B2x%2Bsin%28x%29-2%5D%7D)
hmmm I was looking for some constant value... but hmm, not sure there is one, so I think that'd be it
Using equations of linear model function, the number of hours Jeremy wants to skate is calculated as 3.
<h3>How to Write the Equation of a Linear Model Function?</h3>
The equation that can represent a linear model function is, y = mx + b, where m is the unit rate and b is the initial value.
Equation for Rink A:
Unit rate (m) = (35 - 19)/(5 - 1) = 16/4 = 4
Substitute (x, y) = (1, 19) and m = 4 into y = mx + b to find b:
19 = 4(1) + b
19 - 4 = b
b = 15
Substitute m = 4 and b = 15 into y = mx + b:
y = 4x + 15 [equation for Rink A]
Equation for Rink B:
Unit rate (m) = (39 - 15)/(5 - 1) = 24/4 = 6
Substitute (x, y) = (1, 15) and m = 6 into y = mx + b to find b:
15 = 6(1) + b
15 - 6 = b
b = 9
Substitute m = 6 and b = 9 into y = mx + b:
y = 6x + 9 [equation for Rink B]
To find how many hours (x) both would cost the same (y), make both equation equal to each other
4x + 15 = 6x + 9
4x - 6x = -15 + 9
-2x = -6
x = 3
The hours Jeremy wants to skate is 3.
Learn more about linear model function on:
brainly.com/question/15602982
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