C. If a toy company uses a 1:7 ratio for their action figures to the real people they are
based on. How tall is Dwayne "The Rock" Johnson is
his action figure is 11 inches
tall?
1
112
would be
7



Answer :

To determine how tall Dwayne "The Rock" Johnson is based on the 1:7 ratio and the height of his action figure, you can follow these steps:

### Step-by-Step Solution

1. Understand the Ratio:
- The toy company uses a ratio of 1:7, which means 1 inch on the action figure corresponds to 7 inches in real life.

2. Identify Given Value:
- The action figure is 11 inches tall.

3. Set Up the Proportion:
- Based on the 1:7 ratio, we can set up the following proportion:
[tex]\[ \frac{\text{Height of the toy}}{\text{Height of the real person}} = \frac{1}{7} \][/tex]
- Let [tex]\( H \)[/tex] represent the height of the real person.

4. Plug in the Given Values:
- Since we know the height of the toy (11 inches), we substitute this value into the proportion:
[tex]\[ \frac{11 \text{ inches}}{H} = \frac{1}{7} \][/tex]

5. Solve for [tex]\( H \)[/tex]:
- To isolate [tex]\( H \)[/tex], you need to cross-multiply the equation:
[tex]\[ 11 \times 7 = H \times 1 \][/tex]
[tex]\[ 77 = H \][/tex]

6. Conclusion:
- The height of Dwayne "The Rock" Johnson would be 77 inches.

### Final Height
- To convert inches to feet, note that there are 12 inches in a foot:
[tex]\[ 77 \text{ inches} = \frac{77}{12} \text{ feet} \][/tex]
[tex]\[ 77 \text{ inches} \approx 6 \text{ feet and } 5 \text{ inches} \][/tex]

So, Dwayne "The Rock" Johnson would be approximately 6 feet and 5 inches tall if his action figure, which is based on a 1:7 ratio, is 11 inches tall.

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