Answer: white = $13, double = $14
<u>Step-by-step explanation:</u>
Let white chocolate chip dough = x
Let double chocolate chip dough = y
Mary: 3x + 2y = 67
Darryl: <u> -(3x + 4y = 95)</u>
-2y = -28
<u> ÷-2 </u> <u>÷-2 </u>
y = 14
3x + 2y = 67
3x + 2(14) = 67
3x + 28 = 67
<u> -28 </u> <u>-28</u>
3x = 39
<u>÷3 </u> <u>÷3 </u>
x = 13
Answer:
A. 2-liter bottle;$1.40
Step-by-step explanation:
We have to find out which option has the higher cost per liter. To do this we must find out how much each option would be if it was a liter.
1st set up each equation
2-liter = $2.80 we can divide by 2 on both sides of the equal sign and we get 1 liter = $1.40.
6(1/2) liter bottles = $3.90 this becomes 3 liters = $3.90 then we divide by 3 on both sides of the equal sign and we get $1.30
2nd we compare
$1.40(2 liter bottle) > $1.30 (6 1/2 liter bottles)
In order to make a frequency plot first we need to find the proportion of each outcome.
Total number of results = 15+20+5+5+5 = 50
Frequency of 0 = 15
Proportion of 0 = 15/50 = 0.3
Frequency of 1 = 20
Proportion of 0 = 20/50 = 0.4
Frequency of 2 = 5
Proportion of 2 = 5/50 = 0.1
Frequency of 3 = 5
Proportion of 3 = 5/50 = 0.1
Frequency of 4 = 5
Proportion of 4 = 5/50 = 0.1
Now we need to plot the data on a frequency plot. The x-axis shows the outcomes from 0 to 4 and y-axis shows the frequency of each outcomes. The frequency plot is shown in the figure attached with.
so, let's keep in mind that

so let's make a quick table of those solutions, say A, B, C solutions with x,y,z liters of acid, with an acidity of 0.25, 0.40 and 0.60 respectively.

we know she's using "z" liters and those are 3 times as much as "y" liters, so z = 3y.
![\bf \begin{cases} x+y+3y=78\\ x+4y=78\\[-0.5em] \hrulefill\\ 0.25x+0.4y+0.6(3y)=35.1\\ 0.25x+0.4y=1.8y=35.1\\ 0.25x+2.2y=35.1 \end{cases}\implies \begin{cases} x+4y=78\\\\ 0.25x+2.2y=35.1 \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ x+4y=78\implies \boxed{x}=78-4y \\\\\\ \stackrel{\textit{using substitution on the 2nd equation}}{0.25\left( \boxed{78-4y} \right)+2.2y=35.1}\implies 19.5-y+2.2y=35.1](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20x%2By%2B3y%3D78%5C%5C%20x%2B4y%3D78%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%200.25x%2B0.4y%2B0.6%283y%29%3D35.1%5C%5C%200.25x%2B0.4y%3D1.8y%3D35.1%5C%5C%200.25x%2B2.2y%3D35.1%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Bcases%7D%20x%2B4y%3D78%5C%5C%5C%5C%200.25x%2B2.2y%3D35.1%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20x%2B4y%3D78%5Cimplies%20%5Cboxed%7Bx%7D%3D78-4y%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Busing%20substitution%20on%20the%202nd%20equation%7D%7D%7B0.25%5Cleft%28%20%5Cboxed%7B78-4y%7D%20%5Cright%29%2B2.2y%3D35.1%7D%5Cimplies%2019.5-y%2B2.2y%3D35.1)
![\bf 1.2y=15.6\implies y=\cfrac{15.6}{1.2}\implies \blacktriangleright y=13 \blacktriangleleft \\\\\\ x=78-4y\implies x=78-4(13)\implies \blacktriangleright x=26 \blacktriangleleft \\\\\\ z=3y\implies z=3(13)\implies \blacktriangleright z=39 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{25\%}{26}\qquad \stackrel{40\%}{13}\qquad \stackrel{60\%}{39}~\hfill](https://tex.z-dn.net/?f=%5Cbf%201.2y%3D15.6%5Cimplies%20y%3D%5Ccfrac%7B15.6%7D%7B1.2%7D%5Cimplies%20%5Cblacktriangleright%20y%3D13%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5C%5C%20x%3D78-4y%5Cimplies%20x%3D78-4%2813%29%5Cimplies%20%5Cblacktriangleright%20x%3D26%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5C%5C%20z%3D3y%5Cimplies%20z%3D3%2813%29%5Cimplies%20%5Cblacktriangleright%20z%3D39%20%5Cblacktriangleleft%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7B25%5C%25%7D%7B26%7D%5Cqquad%20%5Cstackrel%7B40%5C%25%7D%7B13%7D%5Cqquad%20%5Cstackrel%7B60%5C%25%7D%7B39%7D~%5Chfill)
Given the vertex of the parabola at point (2, -3):
The quadratic function in vertex form is:
f(x) = (x - 2)^2 - 3
where the vertex (h, k) is the minimum point = (2, -3).