Acandy distributor needs to mix a 40% fat-content chocolate with a 60% fat-content chocolate to create 100 kilograms of a 42% fat-content chocolate. how many kilograms of each kind of chocolate must they use?

Step-by-step explanation:

The easiest way to solve something like this is to make a table.  The nice thing about the table is that is consistent among problems like this.  It never changes.  It can be used for this type of mixture problem and it can be used to find out how much salt or HCl is in a solution.  Same table.  The number of gallons or liters or kilograms or pounds, etc. goes in the first column.  The percent per gallon or liter or kilograms or pounds, etc. OR the cost per whatever the unit is goes in the second column.  The third column is the product of the first 2.  Our table looks like this:

                           # kg        x          % fat            =            kg fat

40% fat choc

60% fat choc

42% fat choc

That's our template for all these types of problems and we only change the columns to match our problem.  We know the percent fat of each type of chocolate so we will put that into the table (in decimal form), and we know that the number of kg we want of the 42% fat chocolate is 100.

                           # kg          x          % fat          =          kg fat

40% fat choc                                     .40

60% fat choc                                     .60

42% fat choc         100        x            .42           =

If we are to make 100 kg total, then we have x kg of the 40% choc and 100-x kg of the 60% chocolate.  For example, if we had 55 kg of the 40%, we can only have 45 kg of the 60% because 55 + 45 = 100 and we are only making 100.  Filling that in:

                            # kg          x          % fat          =          kg fat

40% fat choc            x           x            .40    

60% fat choc      100-x          x            .60    

42% fat choc        100            x           .42

Now we can do the multiplication to get the last column.

                            # kg           x           % fat          =          kg fat

40% fat choc          x              x             .40           =           .40x

60% fat choc     100-x            x             .60          =        .6(100-x)

42% fat choc        100            x              .42         =            42

Now we have to add the 40% fat value to the 60% fat value and set them equal to 42.  

.4x + 60 - .6x = 42 and

-.2x = -18 so

x = 90

That means that we need 90 kg of the 40% and 10 kg of the 60% to get what we need.

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

a framing bias

step-by-step explanation:

rodrigo is subject to a framing bias i believe.