There are 91 such ways in whih the volunteers can be assigned if two of them cannot be assigned from 14 volunteers.
Given that a school dance committee has 14 volunteers and each dance requires 3 volunteers at the door, 5 volunteers on the floor and 6 on floaters.
We are required to find the number of ways in which the volunteers can be assigned.
Combinations means finding the ways in which the things can be choosed to make a new thing or to do something else.
n
=n!/r!(n-r)!
Number of ways in which the volunteers can be assigned is equal to the following:
Since 2 have not been assigned so left over volunteers are 14-2=12 volunteers.
Number of ways =14
=14!/12!(14-12)!
=14!/12!*2!
=14*13/2*1
=91 ways
Hence there are 91 such ways in whih the volunteers can be assigned if two of them cannot be assigned.
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Let be "x" the number of colored pencils in the box.
According to the information given in the exercise, there are 84 pencils in the box and 53 of them are regular pencils.
Knowing this information, you can set up the following equation:
Finally, you have to solve for the variable "x" in order to find its value. In order to do this, you can apply the Subtraction property of equality by subtracting 53 from both sides of the equation:
The answer is: There are 31 colored pencils in the box.
Answer:
Step-by-step explanation:
Answer:
37. {-1, -1}.
Step-by-step explanation:
I'll solve the first one . The other can be solved in a similar way. We can use the method of elimination.
x1 - x2 = 0
3x1 - 2x2 = -1
We can multiply the first equation by -2. We then have an equation containing + 2x2 so when we add this to the second equation the 2x2 will be eliminated
So the first equation becomes:
-2x1 + 2x2 = 0 Bring down the second equation:
3x1 - 2x2 = -1 Now adding, we get:
x1 + 0 = -1
so x1 = -1.
Now we substitute this value of x1 in the original first equation:
-1 - x2 = 0
-1 = x2
x2 = -1.
So the solution set is {-1, -1}.
If there are more than 2 equations you can use a combination of substitutions and eliminations.
