To solve this problem, we need to set up an inequality that represents the condition under which the machine dispenses a candy bar.
First, let's interpret what nickels and dimes represent in terms of their values:
- A nickel (n) is worth \[tex]$0.05.
- A dime (d) is worth \$[/tex]0.10.
The machine dispenses candy based on the value of the coins inserted. We are given that candy costs up to \[tex]$0.50, and we need to find the inequality that represents the possible combinations of nickels and dimes that meet or exceed this value.
Here's how we can set it up using our knowledge of the values of the coins:
1. Each nickel has a value of \$[/tex]0.05, so we can write the total value of [tex]$n$[/tex] nickels as:
[tex]\[
0.05n
\][/tex]
2. Each dime has a value of \[tex]$0.10, so we can write the total value of $[/tex]d[tex]$ dimes as:
\[
0.10d
\]
3. The total value (in dollars) of $[/tex]n[tex]$ nickels and $[/tex]d[tex]$ dimes together is:
\[
0.05n + 0.10d
\]
4. Candy costs up to \$[/tex]0.50, so the value of the coins must be greater than or equal to \$0.50:
[tex]\[
0.05n + 0.10d \geq 0.50
\][/tex]
Now, let's find the inequality that matches this expression:
First, we can multiply every term of the inequality by 20 to clear the decimals and make the coefficients integers:
[tex]\[
20(0.05n + 0.10d) \geq 20(0.50)
\][/tex]
[tex]\[
1n + 2d \geq 10
\][/tex]
[tex]\[
n + 2d \geq 10
\][/tex]
After comparing with the given options, the equivalent inequality is:
[tex]\[
\frac{n}{20} + \frac{d}{10} \geq \frac{1}{2}
\][/tex]
So, the correct answer is:
[tex]\[ B. \frac{n}{20} + \frac{d}{10} \geq \frac{1}{2} \][/tex]
