There are twenty chicken strips in the Family Chicken Strip Bucket, which are the same as in five orders; each order costs $3.99 separately. Each medium order of fries costs $2.79 separately, and each medium soda costs $1.69 separately. The items separately cost a total of 

\(\displaystyle \$ 3.99 \times 5 + \$2.79 \times 4 + \$1.69 \times 4\)

\(\displaystyle = \$ 19.95 + \$ 11.16+ \$6.76\)

\(\displaystyle = \$ 37.87\)

Subtract $30; the savings amount to 

\(\displaystyle \$ 37.87 - \$30.00 = \$ 7.87\)

Multiply the terms as is, ignoring the decimal.

\(\displaystyle 24\times 3 = 72\)

Since there is one decimal from the original problem, the decimal place is one unit from the right.  Therefore, the final answer must have a decimal one unit to from the right.

The answer is \(\displaystyle 7.2\).

To find the sum, you need to add. Line up all numbers vertically. Since one number has a decimal to the tenths place, one has a decimal to the hundredths place, and one has no decimal, be extra careful to line up the digits correctly! To help you, add a zero to the end of 2.4, and add two zeroes to the end of 55 so that all the digits line up correctly.

\(\displaystyle \begin{matrix} &\ 2.40 \\ & 55.00\\ + & \ 6.21 \end{matrix}\)

Now, add down each column, starting on the right side and moving to the left.

\(\displaystyle 2.40 + 55.00+ 6.21 = 63.61\)

The sum is 63.61.

To find the answer, you must add up all three items. Be careful to add $0.49 TWO times because Samantha bought 2 bouncy balls:

 \(\displaystyle 0.49+0.49+4.75=5.73\)

Samantha spent $5.73 at the toy store.

Add the "whole" parts of the numbers:  \(\displaystyle 33 + 22 = 55\)

Add the "decimal" parts of the numbers: \(\displaystyle 0.4 + 0.7 = 1.1\)

Add both parts together: \(\displaystyle 55 + 1.1 = 56.1\)

\(\displaystyle 33.4 + 22.7 = 56.1\)

You add all three values and you get \(\displaystyle \$147.80\).

The items Ken purchases are as follows: 

Hamburger: $4.89

Medium fries: $2.79

Medium soda: $1.69

Hot dog: $4.39

Small fries: $2.29

Small soda: $1.09

The sum of the prices of these items:

  \(\displaystyle \$ 4.89\)

    \(\displaystyle 2.79\)

    \(\displaystyle 1.69\)

    \(\displaystyle 4.39\)

    \(\displaystyle 2.29\)

    \(\displaystyle \underline{1.09}\)

\(\displaystyle \$ 17.14\)

Subtract from $20:

\(\displaystyle \$20.00 - \$ 17.14 = \$ 2.86\)

Rounding this up - Ken needs to give the clerk three more $1 bills.

When we add with multi-digit numbers with decimals, we start with the numbers furthest to the right and work our way towards the left.

Add the numbers in the hundredths place:

\(\displaystyle 4 + 3 = 7\)

Your work should look something like this:

\(\displaystyle \begin{array}{r}\overset{\ ~}{ 4 } \overset{\ ~}{ 6 .}\overset{\ ~ }{ 3 } 4 \\ + \overset{\ ~ }{ 4 }\overset{\ ~ }{ 6 .}\overset{ }{ 3 } \overset{~\ }{ 3 } \\ \hline \space{} \, \space{\, } \space{\, } 7 \end{array}\)

Add the numbers in the tenths place:

\(\displaystyle 3 + 3 = 6\)

Your work should look something like this:

*Notice that the decimal moves directly down when we are adding decimal numbers. 

\(\displaystyle \begin{array}{r}\overset{\ ~}{ 4 } \overset{\ ~ }{ 6 .}\overset{\ ~ }{ 3 } 4 \\ + \overset{\ ~ }{ 4 }\overset{\ ~ }{ 6 .}\overset{ }{ 3 } \overset{~\ }{ 3 } \\ \hline \space{} \, \space{\,. } 6 \space{\, } 7 \end{array}\)

Add the numbers in the ones place:

\(\displaystyle 6 + 6 = 12\)

Because this sum is greater than 9, we write the 2 from the ones place and carry the 1 from the tens place over to the left. Your work should look something like this:

\(\displaystyle \begin{array}{r}\overset{\ 1 }{ 4 } \overset{\ ~ }{ 6 .}\overset{\ ~ }{ 3 } 4 \\ + \overset{\ ~ }{ 4 }\overset{\ ~ }{ 6 .}\overset{ }{ 3 } \overset{~\ }{ 3 } \\ \hline \space{} \, 2 \space{\,. } 6 \space{\, } 7 \end{array}\)

Add the numbers in the tens place including the one carried over:

\(\displaystyle 4 + 4 + 1 = 9\)

Your final answer should be: \(\displaystyle 92.67\)

\(\displaystyle \begin{array}{r}\overset{\ 1 }{ 4 } \overset{\ ~ }{ 6 .}\overset{\ ~ }{ 3 } 4 \\ + \overset{\ ~ }{ 4 }\overset{\ ~ }{ 6 .}\overset{ }{ 3 } \overset{~\ }{ 3 } \\ \hline 9 \space{} \, 2 \space{\,. } 6 \space{\, } 7 \end{array}\)

When we add with multi-digit numbers with decimals, we start with the numbers furthest to the right and work our way towards the left.

Add the numbers in the hundredths place:

\(\displaystyle 5 + 3 = 8\) 

Your work should look something like this:

\(\displaystyle \begin{array}{r}\overset{\ ~}{ 3 } \overset{\ ~}{ 3 .}\overset{\ ~ }{ 8 } 5 \\ + \overset{\ ~ }{ 4 }\overset{\ ~ }{ 3 .}\overset{ }{ 8 } \overset{~\ }{ 3 } \\ \hline \space{} \, \space{\, } \space{\, } 8 \end{array}\)

Add the numbers in the tenths place:

\(\displaystyle 8 + 8 = 16\) 

Because this sum is greater than 9, we write the 6 from the ones place and carry the 1 from the tens place over to the left. Your work should look something like this:

Your work should look something like this:

*Notice that the decimal moves directly down when we are adding decimal numbers. 

\(\displaystyle \begin{array}{r}\overset{\ ~}{ 3 } \overset{\ 1 }{ 3 .}\overset{\ ~ }{ 8 } 5 \\ + \overset{\ ~ }{ 4 }\overset{\ ~ }{ 3 .}\overset{ }{ 8 } \overset{~\ }{ 3 } \\ \hline \space{} \, \space{\,. } 6 \space{\, } 8 \end{array}\) 

Add the numbers in the ones place including the one carried over:

\(\displaystyle 3 + 3 + 1 = 7\)

Your work should look something like this:

\(\displaystyle \begin{array}{r}\overset{\ ~ }{ 3 } \overset{\ 1 }{ 3 .}\overset{\ ~ }{ 8 } 5 \\ + \overset{\ ~ }{ 4 }\overset{\ ~ }{ 3 .}\overset{ }{ 8 } \overset{~\ }{ 3 } \\ \hline \space{} \, 7 \space{\,. } 6 \space{\, } 8 \end{array}\) 

Add the numbers in the tens place:

\(\displaystyle 3 + 4 = 7\)

Your final answer should be: \(\displaystyle 77.68\)

\(\displaystyle \begin{array}{r}\overset{\ ~ }{ 3 } \overset{\ 1 }{ 3 .}\overset{\ ~ }{ 8 } 5 \\ + \overset{\ ~ }{ 4 }\overset{\ ~ }{ 3 .}\overset{ }{ 8 } \overset{~\ }{ 3 } \\ \hline 7 \space{} \, 7 \space{\,. } 6 \space{\, } 8 \end{array}\) 

When we add with multi-digit numbers with decimals, we start with the numbers furthest to the right and work our way towards the left.

Add the numbers in the hundredths place:

\(\displaystyle 7 + 5 = 12\) 

Because this sum is greater than 9, we write the 2 from the ones place and carry the 1 from the tens place over to the left. Your work should look something like this:

\(\displaystyle \begin{array}{r}\overset{\ ~}{ 3 } \overset{\ ~}{ 3 .}\overset{\ 1 }{ 7 } 7 \\ + \overset{\ ~ }{ 1 }\overset{\ ~ }{ 5 .}\overset{ }{ 7 } \overset{~\ }{ 5 } \\ \hline \space{} \, \space{\, } \space{\, } 2 \end{array}\)

Add the numbers in the tenths place including the one carried over:

\(\displaystyle 7 + 7 + 1 = 15\)

Because this sum is greater than 9, we write the 5 from the ones place and carry the 1 from the tens place over to the left. Your work should look something like this:

*Notice that the decimal moves directly down when we are adding decimal numbers. 

\(\displaystyle \begin{array}{r}\overset{\ ~}{ 3 } \overset{\ 1 }{ 3 .}\overset{\ 1 }{ 7 } 7 \\ + \overset{\ ~ }{ 1 }\overset{\ ~ }{ 5 .}\overset{ }{ 7 } \overset{~\ }{ 5 } \\ \hline \space{} \, \space{\,. } 5 \space{\, } 2 \end{array}\)

Add the numbers in the ones place including the one carried over:

\(\displaystyle 3 + 5 + 1 = 9\)

 Your work should look something like this: 

\(\displaystyle \begin{array}{r}\overset{\ ~ }{ 3 } \overset{\ 1 }{ 3 .}\overset{\ 1 }{ 7 } 7 \\ + \overset{\ ~ }{ 1 }\overset{\ ~ }{ 5 .}\overset{ }{ 7 } \overset{~\ }{ 5 } \\ \hline \space{} \, 9 \space{\,. } 5 \space{\, } 2 \end{array}\)

Add the numbers in the tens place:

\(\displaystyle 3 + 1 = 4\)

Your final answer should be: \(\displaystyle 49.52\)

\(\displaystyle \begin{array}{r}\overset{\ ~ }{ 3 } \overset{\ 1 }{ 3 .}\overset{\ 1 }{ 7 } 7 \\ + \overset{\ ~ }{ 1 }\overset{\ ~ }{ 5 .}\overset{ }{ 7 } \overset{~\ }{ 5 } \\ \hline 4 \space{} \, 9 \space{\,. } 5 \space{\, } 2 \end{array}\)