Find class boundaries, midpoint, and width for the class. $$ 12.8-15.0 $$

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UpStudy AI · Accepted Answer

To find the class boundaries, midpoint, and width for the class interval $$ 12.8 - 15.0 $$, we can follow these steps:

1. **Class Boundaries**: The class boundaries are found by taking the lower limit of the class and subtracting 0.05 to find the lower boundary, and taking the upper limit and adding 0.05 to find the upper boundary.

- Lower boundary: $$ 12.8 - 0.05 = 12.75 $$
   - Upper boundary: $$ 15.0 + 0.05 = 15.05 $$

Therefore, the class boundaries are $$ 12.75 - 15.05 $$.

2. **Midpoint**: The midpoint of the class interval is calculated by averaging the lower and upper limits of the class.

$$
   \text{Midpoint} = \frac{\text{Lower limit} + \text{Upper limit}}{2} = \frac{12.8 + 15.0}{2}
   $$

3. **Width**: The width of the class interval is calculated by subtracting the lower limit from the upper limit.

$$
   \text{Width} = \text{Upper limit} - \text{Lower limit} = 15.0 - 12.8
   $$

Now, let's calculate the midpoint and width.

I'll perform the calculations for the midpoint and width.
Calculate the value by following steps:
- step0: Calculate:
  $$15-12.8$$
- step1: Subtract the numbers:
  $$2.2$$
Calculate or simplify the expression $$ (12.8 + 15.0) / 2 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\left(12.8+15\right)}{2}$$
- step1: Remove the parentheses:
  $$\frac{12.8+15}{2}$$
- step2: Add the numbers:
  $$\frac{27.8}{2}$$
- step3: Convert the expressions:
  $$\frac{\frac{139}{5}}{2}$$
- step4: Multiply by the reciprocal:
  $$\frac{139}{5}\times \frac{1}{2}$$
- step5: Multiply the fractions:
  $$\frac{139}{5\times 2}$$
- step6: Multiply:
  $$\frac{139}{10}$$
Now that we have performed the calculations, here are the results:

1. **Class Boundaries**: 
   - Lower boundary: $$ 12.75 $$
   - Upper boundary: $$ 15.05 $$
   - Therefore, the class boundaries are $$ 12.75 - 15.05 $$.

2. **Midpoint**: 
   - The midpoint is $$ 13.9 $$.

3. **Width**: 
   - The width of the class interval is $$ 2.2 $$.

### Summary:
- **Class Boundaries**: $$ 12.75 - 15.05 $$
- **Midpoint**: $$ 13.9 $$
- **Width**: $$ 2.2 $$
Class boundaries: 12.75 and 15.05 Midpoint: 13.9 Width: 2.2

Find class boundaries, midpoint, and width for the class. \[ 12.8-15.0 \]

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