A food server examines the amount of money earned in tips after working an 8-hour shift. The server has a total of $133 in denominations of $1, $5, $10, and $20 bills. The total number of paper bills is 39. The number of $5 bills is 4 times the number of $10 bills, and the number of $1 bills is 1 less than twice the number of $5 bills. Write a system of linear equations to represent the situation. (Assume

x = number of $1 bills,

y = number of $5 bills,

z = number of $10 bills,

and

w = number of $20 bills.)

x + y + z + w = 133
x + y + z + w = 39
y − z = 0
x − y = −1

Use matrices to find the number of each denomination.

x = $1 bill(s)
y = $5 bill(s)
z = $10 bill(s)
w = 20 bills

x = 23 $1 bills

y = 12 $5 bills

z = 3 $10 bills

w = 1 $20 bill

Step-by-step explanation:

First of all we have to setup linear equations with the help of given information.

x = number of $1 bills

y = number of $5 bills

z = number of $10 bills  

w = number of $20 bills

The server has a total of $133 in denominations of $1, $5, $10, and $20 bills.

x + 5y + 10z + 20w = 133 eq. 1

The total number of paper bills is 39.

x + y + z + w = 39 eq. 2

The number of $5 bills is 4 times the number of $10 bills

y = 4z eq. 3

The number of $1 bills is 1 less than twice the number of $5 bills

x = 2y - 1 eq. 4

Now we have got 4 equations and four unknowns x, y, z, w

x + 5y + 10z + 20w = 133 eq. 1

x + y + z + w = 39 eq. 2

0 + y -4z + 0 = 0 eq. 3

x - 2y + 0 + 0 = -1 eq. 4

We have four equations and four unknowns x, y, z, w

We can solve this system of linear equations by various linear algebra methods like Crammer's rule, Gaussian Elimination etc. But if you notice eq. 3 and eq. 4 we have some zero entries in these equations and we might reduce the number of equations which will be easier and faster method to solve this system.

From eq. 3

0 + y -4z + 0 = 0

y = 4z

z = y/4

From eq. 4

x - 2y + 0 + 0 = -1

x = 2y - 1

Now put the values of x and z into eq. 1

x + 5y + 10z + 20w = 133

2y - 1 + 5y + 10(y/4) + 20w = 133

7y + (5/2)y + 20w = 133 + 1

9.5y + 20w = 134  eq. 5

Now put the values of x and z into eq. 2

x + y + z + w = 39

2y - 1 + y + y/4 + w = 39

3y + y/4 + w = 39 + 1

3.25y + w = 40  eq. 6

Now we have 2 equations ( eq. 5 and eq. 6) and 2 unknowns (y and w)

Solving these two equations simultaneously

9.5y + 20w = 134  eq. 5

3.25y + w = 40  eq. 6

Multiply eq. 6 by 20 and subtract it from eq. 5

9.5y + 20w = 134

65y + 20w = 800

-55.5y + 0 = -666

y = -666/-55.5 = 12

Put the value of y into eq. 5 to get the value of w

9.5(12) + 20w = 134

114 + 20w = 134

20w = 134 - 114

w = 20/20 = 1

Now we have got the value of y and we can substitute this value into eq. 3 and eq. 4 to get the values of x and z

From eq. 3

z = y/4 = 12/4 = 3

From eq. 4

x = 2y - 1 = 2(12) - 1 = 23

So we have solved the whole system of equations and got the following results

x = 23 $1 bills

y = 12 $5 bills

z = 3 $10 bills

w = 1 $20 bill

Verification:

Lets verify our results to check if we have got the correct numbers!

eq. 1  => x + 5y + 10z + 20w = 133

23 + 5(12) + 10(3) + 20(1) = 133

23 + 60 + 30 + 20 = 133

133 = 133 (hence satisfied)

eq. 2 => x + y + z + w = 39

23 + 12 + 3 + 1 = 39

39 = 39 (hence satisfied)

eq. 3 => 0 + y -4z + 0 = 0

0 + 12 -4(3) = 0

12 - 12 = 0

0 = 0 (hence satisfied)

eq. 4 => x - 2y + 0 + 0 = -1

23 - 2(12) + 0 + 0 = -1

23 - 24 = -1

-1 = -1 (hence satisfied)

c

step-by-step explanation:

it isn't d, because the interquartile range is from 27-57. and the interquartile range is basically half the data, it's pretty obvious.

it isn't b, because like i said earlier, the interquartile range is from 27-57. 57-27 is 30. the real range has to be longer

not a either. although it looks like it, 3/4 of the data is not greater than 27. but if c is wrong than this should be correct

the correct answer is c, because 13 is the point in which the plot starts