a student determines the area of a rectangle by measuring the lengths of its sides, using a ruler scribed in centimeters with subdivisions in millimeters. assume that measurements made with the ruler are accurate to within + or - 1 mm.

the values that the student records are length = 14.0 cm and width = 13.4 cm.

1. to 2 decimal places, the calculated value of the area (length × width) is 187.60 cm². based on the significant figure rule for multiplication, how should the area be reported?
area = \boxed{} cm²

2. including the uncertainty in the measurements,
(a) how should the length be reported? 14.0 cm + or - \boxed{} cm
(b) how should the width be reported? 13.4 cm + or - \boxed{} cm

3. use the values from 2. and common sense here.
(a) to 2 decimal places, what is the maximum calculated value for the area? \boxed{} cm²
(b) to 2 decimal places, what is the minimum calculated value for the area? \boxed{} cm²
(c) including the uncertainty, how should the area calculated from the recorded length and width be reported? hint: use the range from (a) and (b) to determine the + or - uncertainty in the area to 1 sig fig and do not report the area beyond the first digit containing the uncertainty.
area = \boxed{} cm² + or - \boxed{} cm²

Question

a student determines the area of a rectangle by measuring the lengths of its sides, using a ruler scribed in centimeters with subdivisions in millimeters. assume that measurements made with the ruler are accurate to within + or - 1 mm.

the values that the student records are length = 14.0 cm and width = 13.4 cm.

1. to 2 decimal places, the calculated value of the area (length × width) is 187.60 cm². based on the significant figure rule for multiplication, how should the area be reported?
area = \boxed{} cm²

2. including the uncertainty in the measurements,
(a) how should the length be reported? 14.0 cm + or - \boxed{} cm
(b) how should the width be reported? 13.4 cm + or - \boxed{} cm

3. use the values from 2. and common sense here.
(a) to 2 decimal places, what is the maximum calculated value for the area? \boxed{} cm²
(b) to 2 decimal places, what is the minimum calculated value for the area? \boxed{} cm²
(c) including the uncertainty, how should the area calculated from the recorded length and width be reported? hint: use the range from (a) and (b) to determine the + or - uncertainty in the area to 1 sig fig and do not report the area beyond the first digit containing the uncertainty.
area = \boxed{} cm² + or - \boxed{} cm²

Accepted Answer

### Question 1
# Explanation:
## Step1: Recall significant figure rule for multiplication
When multiplying, the result should have the same number of significant figures as the least precise measurement. Length $ l = 14.0 \, \text{cm} $ (3 significant figures), width $ w = 13.4 \, \text{cm} $ (3 significant figures).
## Step2: Calculate area and round
Area $ A = l \times w = 14.0 \times 13.4 = 187.6 \, \text{cm}^2 $ (3 significant figures as both have 3).
# Answer: $ 187.6 $

### Question 2
#### (a)
# Explanation:
## Step1: Convert uncertainty to cm
Uncertainty is $ \pm 1 \, \text{mm} $. Since $ 1 \, \text{mm} = 0.1 \, \text{cm} $, the uncertainty in length (14.0 cm) is $ \pm 0.1 \, \text{cm} $ (because the measurement is to the tenths place, uncertainty is in the same decimal place).
# Answer: $ 0.1 $

#### (b)
# Explanation:
## Step1: Convert uncertainty to cm
Uncertainty is $ \pm 1 \, \text{mm} = \pm 0.1 \, \text{cm} $. Width is 13.4 cm (tenths place), so uncertainty is $ \pm 0.1 \, \text{cm} $.
# Answer: $ 0.1 $

### Question 3
#### (a)
# Explanation:
## Step1: Find maximum length and width
Max length $ l_{\text{max}} = 14.0 + 0.1 = 14.1 \, \text{cm} $, max width $ w_{\text{max}} = 13.4 + 0.1 = 13.5 \, \text{cm} $.
## Step2: Calculate max area
$ A_{\text{max}} = 14.1 \times 13.5 = 190.35 \, \text{cm}^2 $ (to 2 decimal places).
# Answer: $ 190.35 $

#### (b)
# Explanation:
## Step1: Find minimum length and width
Min length $ l_{\text{min}} = 14.0 - 0.1 = 13.9 \, \text{cm} $, min width $ w_{\text{min}} = 13.4 - 0.1 = 13.3 \, \text{cm} $.
## Step2: Calculate min area
$ A_{\text{min}} = 13.9 \times 13.3 = 184.87 \, \text{cm}^2 $ (to 2 decimal places).
# Answer: $ 184.87 $

#### (c)
# Explanation:
## Step1: Find uncertainty in area
Range of area: $ 190.35 - 184.87 = 5.48 \, \text{cm}^2 $. Uncertainty is half the range: $ \frac{5.48}{2} \approx 2.74 $, rounded to 1 sig fig: $ 3 \, \text{cm}^2 $.
## Step2: Report area with uncertainty
Average area (from original measurement) is $ 187.6 \, \text{cm}^2 $. So report as $ 188 \, \text{cm}^2 \pm 3 \, \text{cm}^2 $ (adjusting area to first digit with uncertainty: 188 is closer to 187.6, and uncertainty is 3).
# Answer: $ 188 $; $ 3 $

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QUESTION IMAGE

Question

Explanation:

Question 1

Step1: Recall significant figure rule for multiplication

Step2: Calculate area and round

Step1: Convert uncertainty to cm

Step1: Convert uncertainty to cm

Answer:

Question 2

(a)