<span>To produce 1000L of mixture the factory will need 657 liters of grade A and 343 liters of grade B. To determine this you have to figure out the percentage of each grade in the mixture. The ratio is 2.3 liters to 1.2 liters. Therefor in this scenario the unit equaling 100% is 3.5 liters. To find the percentage of grade A you divide the amount used by the total amount of the unit 100%:
2.3 divided by 3.5 = .657 (multiply the answer by 100 to get 65.7%).
To find the percentage of grade B you divide the amount used by the total amount of the unit 100%:
1.2 divided by 3.5 = .343 (multiply the answer by 100 to get 34.3%)
To test your answer make sure both percentages add up to 100%:
65.75 plus 34.35 = 100%
To determine how much of grade A and grade B is needed for a set amount of liters you multiply the percentage by the liters needed.
For this situation you multiply 65.7% (when multiplying percentages you need to multiply in decimal form).
For grade A you multiply .657 (65.7%) by 1000 liters = 657 liters
For grade B you multiply .343 (34.3%) by 1000 liters = 343 liters
To test your answer you can use the same addition as you did to test the percentages:
657 liters plus 343 liters = 1000 liters</span>
Take 3 quarters minus one eighth and you’ll get ur answer
Let
x--------> the amount of gummy candy in pounds
y--------> the amount of jelly beans in pounds
z--------> the amount of hard candy in pounds
we know that
--------> equation 
--------> equation 
--------> equation 
Substitute equation in equation and equation
![[2y]+y+z=9](https://tex.z-dn.net/?f=%5B2y%5D%2By%2Bz%3D9)
-----> 
------> equation 
![2[2y]+3y+3z=23](https://tex.z-dn.net/?f=2%5B2y%5D%2B3y%2B3z%3D23)
-----> 
------> equation 
using a graphing tool ------> Solve the system of equations
see the attached figure
the solution is the point 
<u>Find the value of x</u>
therefore
<u>the answer is</u>
the amount of gummy candy is 
the amount of jelly beans is 
the amount of hard candy is 
Answer:
B
Step-by-step explanation:
If you arrange a statistics in order from least to greatest, the middle number is the median.
If there is even number of numbers in a list, let it be n numbers, then we take the average of n/2th and n/2 + 1 th terms.
Here, all of them have 6 numbers, so 6/2 = 3 and 4th, we take average of 3rd and 4th number to find the median.
Since they are arrange in order we check each:
Jon = average of 6 and 7, (6+7)/2 = 6.5
Leroy = average of 6 and 8, (6+8)/2 = 7
Simon = average of 5 and 6, (5+6)/2 = 5.5
Leroy's median is the greatest (7).
