Set up a system of equations and say green route is x and blue route is y
This will give you
Monday 6x + 5y= 52
Tuesday 12x + 13y = 119
Solve by multiplying the top equation by two to get
12x + 10y= 104
Since you have 12x in both equations, these cancel out when you subtract the bottom from the top. This leaves you with
-3y=-15
Y=5
The blue route is 5 miles, but they want to know the green route. Fill in the y value in either one of your equations. Let’s use the one from Tuesday
12x + 13 (5) = 119
12x= 54
X= 4.5
This means that the answer is 4.5 miles
Answer:
![\hat p_1 -\hat p_2 \pm z_{\alpha/2} SE](https://tex.z-dn.net/?f=%5Chat%20p_1%20-%5Chat%20p_2%20%5Cpm%20z_%7B%5Calpha%2F2%7D%20SE)
And for this case the confidence interval is given by:
![0.11 \leq p_1 -p_2 \leq 0.39](https://tex.z-dn.net/?f=0.11%20%5Cleq%20p_1%20-p_2%20%5Cleq%200.39)
Since the confidenc einterval not contains the value 0 we can conclude that we have significant difference between the two population proportion of interest 1% of significance given. So then we can't conclude that the two proportions are equal
Step-by-step explanation:
Let p1 and p2 the population proportions of interest and let
and
the estimators for the proportions we know that the confidence interval for the difference of proportions is given by this formula:
![\hat p_1 -\hat p_2 \pm z_{\alpha/2} SE](https://tex.z-dn.net/?f=%5Chat%20p_1%20-%5Chat%20p_2%20%5Cpm%20z_%7B%5Calpha%2F2%7D%20SE)
And for this case the confidence interval is given by:
![0.11 \leq p_1 -p_2 \leq 0.39](https://tex.z-dn.net/?f=0.11%20%5Cleq%20p_1%20-p_2%20%5Cleq%200.39)
Since the confidence interval not contains the value 0 we can conclude that we have significant difference between the two population proportion of interest 1% of significance given. So then we can't conclude that the two proportions are equal
![\bf \textit{Law of Cosines}\\ \quad \\ c^2 = {{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)\implies c = \sqrt{{{ a}}^2+{{ b}}^2-(2{{ a}}{{ b}})cos(C)}\\\\ -----------------------------\\\\ \textit{so, let's find the missing side} \\\\\\ c=\sqrt{2.7^2+3.4^2-2(2.7)(3.4)cos(40^o)} \\\\\\ c=\sqrt{4.78542402433556327368}\implies c\approx 2.187561204706182220](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7BLaw%20of%20Cosines%7D%5C%5C%20%5Cquad%20%5C%5C%0Ac%5E2%20%3D%20%7B%7B%20a%7D%7D%5E2%2B%7B%7B%20b%7D%7D%5E2-%282%7B%7B%20a%7D%7D%7B%7B%20b%7D%7D%29cos%28C%29%5Cimplies%20%0Ac%20%3D%20%5Csqrt%7B%7B%7B%20a%7D%7D%5E2%2B%7B%7B%20b%7D%7D%5E2-%282%7B%7B%20a%7D%7D%7B%7B%20b%7D%7D%29cos%28C%29%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%5Ctextit%7Bso%2C%20let%27s%20find%20the%20missing%20side%7D%0A%5C%5C%5C%5C%5C%5C%0Ac%3D%5Csqrt%7B2.7%5E2%2B3.4%5E2-2%282.7%29%283.4%29cos%2840%5Eo%29%7D%0A%5C%5C%5C%5C%5C%5C%0Ac%3D%5Csqrt%7B4.78542402433556327368%7D%5Cimplies%20c%5Capprox%202.187561204706182220)
or we can round it, to say c = 2.19, so hmm that's the missing side
now, we use Heron's Formula, which uses all 3 sides only
![\bf \textit{Heron's Area Formula}\\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} a=2.7\\ b=3.4\\ c\approx 2.19\\\\ s=\cfrac{a+b+c}{2} \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7BHeron%27s%20Area%20Formula%7D%5C%5C%5C%5C%0AA%3D%5Csqrt%7Bs%28s-a%29%28s-b%29%28s-c%29%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Aa%3D2.7%5C%5C%0Ab%3D3.4%5C%5C%0Ac%5Capprox%202.19%5C%5C%5C%5C%0As%3D%5Ccfrac%7Ba%2Bb%2Bc%7D%7B2%7D%0A%5Cend%7Bcases%7D)
and that'd be the area of it
The answer is 23.......................................
Radius
The answer is radius lol