So, let's look at the problem once again.
8%, or 8/100 of the bag were green jelly beans.
Let's see what we can do to find out a number of other jelly beans, not including the green ones.
We know that 92% of jelly-beans are not green.
92% is 11.5 times more than 8%
(92/8)
24 (or 8%) times 11.5 = 276
276 - number of jelly beans that are not green.
276 plus 24 = 300.
Answer: 300 jelly beans were in one bag.
The attached graph represents the solution set of 
<h3>How to determine the graph?</h3>
The complete options are not given.
So, I would plot the graph that represents the solution set
The inequality is given as:
Start by splitting the inequalities as follows:
The above means that we plot the graphs of f(x) and g(x) to represent the solution set
See attachment
Read more about inequalities at:
brainly.com/question/25275758
#SPJ1
Answer:
15x + 40 = 115
We set the equation to 115 because the total cost is $115.
We add the "+ 40" because of the $40 to rent the trailer for the day.
The 15x is because it represents the miles per hour (x) times the number of hours.
The number of hours times the charge per mile, plus the $40 to rent it, will equal $115.
Hope this helps!!
Answer: <em>33 square feet</em>
Step-by-step explanation:
<em>First, let's take the area of the rectangle shown here which is 6ft long and 4ft wide</em>
<em>6x4=24</em>
<em>Now let's take the area of the square which is 3ft long and 3ft wide</em>
<em>3x3=9</em>
<em>Now add 24 and 9 for the total area</em>
<em>24+9=</em><em>33 square feet</em>
the solid is made up of 2 regular octagons, 8 sides, joined up by 8 rectangles, one on each side towards the other octagonal face.
from the figure, we can see that the apothem is 5 for the octagons, and since each side is 3 cm long, the perimeter of one octagon is 3*8 = 24.
the standing up sides are simply rectangles of 8x3.
if we can just get the area of all those ten figures, and sum them up, that'd be the area of the solid.
![\bf \textit{area of a regular polygon}\\\\ A=\cfrac{1}{2}ap~~ \begin{cases} a=apothem\\ p=perimeter\\[-0.5em] \hrulefill\\ a=5\\ p=24 \end{cases}\implies A=\cfrac{1}{2}(5)(24)\implies \stackrel{\textit{just for one octagon}}{A=60} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{two octagon's area}}{2(60)}~~+~~\stackrel{\textit{eight rectangle's area}}{8(3\cdot 8)}\implies 120+192\implies 312](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20regular%20polygon%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B2%7Dap~~%20%5Cbegin%7Bcases%7D%20a%3Dapothem%5C%5C%20p%3Dperimeter%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D5%5C%5C%20p%3D24%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%285%29%2824%29%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bjust%20for%20one%20octagon%7D%7D%7BA%3D60%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Btwo%20octagon%27s%20area%7D%7D%7B2%2860%29%7D~~%2B~~%5Cstackrel%7B%5Ctextit%7Beight%20rectangle%27s%20area%7D%7D%7B8%283%5Ccdot%208%29%7D%5Cimplies%20120%2B192%5Cimplies%20312)
