Answer:
The median is 50.
Step-by-step explanation:
First you put all the numbers in order. (37, 44, 46, 54, 62, 88) then you look for the middle number. Since there is two middle numbers, you add them together then divide it by 2.
So 46+54= 100
100/2= 50
Check the picture below. so, that'd be the triangle's sides hmmm so let's use Heron's Area formula for it.
![~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{15}) ~\hfill a=\sqrt{[ 15- 10]^2 + [ 15- 5]^2} \\\\\\ ~\hfill \boxed{a=\sqrt{125}} \\\\\\ (\stackrel{x_1}{15}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{30}~,~\stackrel{y_2}{9}) ~\hfill b=\sqrt{[ 30- 15]^2 + [ 9- 15]^2} \\\\\\ ~\hfill \boxed{b=\sqrt{261}}](https://tex.z-dn.net/?f=~%5Chfill%20%5Cstackrel%7B%5Ctextit%7B%5Clarge%20distance%20between%202%20points%7D%7D%7Bd%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%7D~%5Chfill~%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B10%7D~%2C~%5Cstackrel%7By_1%7D%7B5%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B15%7D~%2C~%5Cstackrel%7By_2%7D%7B15%7D%29%20~%5Chfill%20a%3D%5Csqrt%7B%5B%2015-%2010%5D%5E2%20%2B%20%5B%2015-%205%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20~%5Chfill%20%5Cboxed%7Ba%3D%5Csqrt%7B125%7D%7D%20%5C%5C%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B15%7D~%2C~%5Cstackrel%7By_1%7D%7B15%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B30%7D~%2C~%5Cstackrel%7By_2%7D%7B9%7D%29%20~%5Chfill%20b%3D%5Csqrt%7B%5B%2030-%2015%5D%5E2%20%2B%20%5B%209-%2015%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20~%5Chfill%20%5Cboxed%7Bb%3D%5Csqrt%7B261%7D%7D)
![(\stackrel{x_1}{30}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{5}) ~\hfill c=\sqrt{[ 10- 30]^2 + [ 5- 9]^2} \\\\\\ ~\hfill \boxed{c=\sqrt{416}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%28%5Cstackrel%7Bx_1%7D%7B30%7D~%2C~%5Cstackrel%7By_1%7D%7B9%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B10%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%20~%5Chfill%20c%3D%5Csqrt%7B%5B%2010-%2030%5D%5E2%20%2B%20%5B%205-%209%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20~%5Chfill%20%5Cboxed%7Bc%3D%5Csqrt%7B416%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{125}\\ b=\sqrt{261}\\ c=\sqrt{416}\\ s\approx 23.87 \end{cases} \\\\\\ A\approx\sqrt{23.87(23.87-\sqrt{125})(23.87-\sqrt{261})(23.87-\sqrt{416})}\implies \boxed{A\approx 90}](https://tex.z-dn.net/?f=%5Cqquad%20%5Ctextit%7BHeron%27s%20area%20formula%7D%20%5C%5C%5C%5C%20A%3D%5Csqrt%7Bs%28s-a%29%28s-b%29%28s-c%29%7D%5Cqquad%20%5Cbegin%7Bcases%7D%20s%3D%5Cfrac%7Ba%2Bb%2Bc%7D%7B2%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D%5Csqrt%7B125%7D%5C%5C%20b%3D%5Csqrt%7B261%7D%5C%5C%20c%3D%5Csqrt%7B416%7D%5C%5C%20s%5Capprox%2023.87%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%5Capprox%5Csqrt%7B23.87%2823.87-%5Csqrt%7B125%7D%29%2823.87-%5Csqrt%7B261%7D%29%2823.87-%5Csqrt%7B416%7D%29%7D%5Cimplies%20%5Cboxed%7BA%5Capprox%2090%7D)
Answer:
He needs 16 ounces of raisins and 8 ounces of nuts
Step-by-step explanation:
Let
x -----> the number of ounces of raisins
y -----> the number of ounces of nuts
we know that
x+y=24 -----> equation A
x=2y -----> equation B
Solve the system by substitution
Substitute equation B in equation A and solve for y
(2y)+y=24
3y=24
y=8 ounces of nuts
Find the value of x
x=2(8)=16 ounces of raisins
therefore
He needs 16 ounces of raisins and 8 ounces of nuts
