pre-laboratory assignment
1. state the number of significant figures in each of the following numbers and explain your answer.
(a) 37.60
(b) 0.0130
(c) 13000
(d) 1.3400
2. perform the indicated operations to the correct number of significant figures using the rules for significant figures.
(a)
37.60
× 1.23
(b) 6.7 \overline{) 8.975}
(c)
3.765
+ 1.2
+37.21
3–6. three students named abe, barb, and cal make measurements (in m) of the length of a table using a meter stick. each student’s measurements are tabulated in the table below along with the mean, the standard deviation from the mean, and the standard error of the measurements.

| student | $l_1$ | $l_2$ | $l_3$ | $l_4$ | $\bar{l}$ | $\sigma_{n-1}$ | $\alpha$ |
|---------|-------|-------|-------|-------|-----------|----------------|-----------|
| abe | 1.4717| 1.4711| 1.4722| 1.4715| 1.4716 | 0.00046 | 0.0002 |
| barb | 1.4753| 1.4759| 1.4756| 1.4749| 1.4754 | 0.00043 | 0.0002 |
| cal | 1.4719| 1.4723| 1.4727| 1.4705| 1.4719 | 0.00096 | 0.0005 |

note that in each case only one significant figure is kept in the standard error $\alpha$, and this determines the number of significant figures in the mean. the actual length of the table is determined by very sophisticated laser measurement techniques to be 1.4715 m.

Question

pre-laboratory assignment
1. state the number of significant figures in each of the following numbers and explain your answer.
(a) 37.60
(b) 0.0130
(c) 13000
(d) 1.3400
2. perform the indicated operations to the correct number of significant figures using the rules for significant figures.
(a)
  37.60
× 1.23
(b) 6.7 \overline{) 8.975}
(c)
  3.765
+ 1.2
+37.21
3–6. three students named abe, barb, and cal make measurements (in m) of the length of a table using a meter stick. each student’s measurements are tabulated in the table below along with the mean, the standard deviation from the mean, and the standard error of the measurements.

| student | $l_1$ | $l_2$ | $l_3$ | $l_4$ | $\bar{l}$ | $\sigma_{n-1}$ | $\alpha$ |
|---------|-------|-------|-------|-------|-----------|----------------|-----------|
| abe     | 1.4717| 1.4711| 1.4722| 1.4715| 1.4716    | 0.00046        | 0.0002    |
| barb    | 1.4753| 1.4759| 1.4756| 1.4749| 1.4754    | 0.00043        | 0.0002    |
| cal     | 1.4719| 1.4723| 1.4727| 1.4705| 1.4719    | 0.00096        | 0.0005    |

note that in each case only one significant figure is kept in the standard error $\alpha$, and this determines the number of significant figures in the mean. the actual length of the table is determined by very sophisticated laser measurement techniques to be 1.4715 m.

Accepted Answer

### Problem 1: Significant Figures Count

#### Part (a): 37.60
# Explanation:
## Step1: Recall sig fig rules
Non - zero digits and trailing zeros after decimal are significant.
37.60 has 3, 7, 6 (non - zero) and 0 (trailing after decimal).
## Step2: Count significant figures
Count: 3, 7, 6, 0 → 4 significant figures.
# Answer: 4

#### Part (b): 0.0130
# Explanation:
## Step1: Recall sig fig rules
Leading zeros are not significant, non - zero digits and trailing zero after non - zero are significant.
0.0130: leading zeros (0.0) not significant, 1, 3 (non - zero), and trailing 0.
## Step2: Count significant figures
Count: 1, 3, 0 → 3 significant figures.
# Answer: 3

#### Part (c): 13000
# Explanation:
## Step1: Recall sig fig rules
Trailing zeros in a whole number without decimal are ambiguous, but here we assume they are placeholders. Non - zero digits 1 and 3 are significant.
## Step2: Count significant figures
Count: 1, 3 → 2 significant figures (assuming the zeros are not significant).
# Answer: 2

#### Part (d): 1.3400
# Explanation:
## Step1: Recall sig fig rules
Non - zero digits and trailing zeros after decimal are significant.
1.3400 has 1, 3, 4 (non - zero) and two trailing zeros after decimal.
## Step2: Count significant figures
Count: 1, 3, 4, 0, 0 → 5 significant figures.
# Answer: 5

### Problem 2: Operations with Significant Figures

#### Part (a): $37.60\times1.23$
# Explanation:
## Step1: Recall multiplication rule
For multiplication, the result has the same number of significant figures as the least precise measurement. 37.60 has 4 sig figs, 1.23 has 3 sig figs. So result should have 3 sig figs.
## Step2: Perform multiplication
$37.60\times1.23 = 46.248$
## Step3: Round to 3 sig figs
$46.248\approx46.2$ (wait, no: 37.60 is 4 sig figs, 1.23 is 3 sig figs. The rule is that the number of sig figs in the result is equal to the number of sig figs in the least precise measurement. So 3 sig figs. $37.60\times1.23 = 46.248$, rounded to 3 sig figs is 46.2? Wait, no, 37.60 has 4, 1.23 has 3. So the product should have 3 sig figs. $37.60\times1.23 = 46.248$, so 46.2 (if we take 3 sig figs: 46.2). Wait, actually, 37.60 is 4, 1.23 is 3. So the limiting factor is 3 sig figs. So $37.60\times1.23=46.248\approx46.2$ (no, 46.248 rounded to 3 significant figures: the first three significant figures are 4, 6, 2, the next digit is 4 which is less than 5, so it's 46.2? Wait, no, 46.248: 4 (1st), 6 (2nd), 2 (3rd), 4 (4th). So we keep 46.2? Wait, no, 37.60 × 1.23: 37.60 has 4, 1.23 has 3. So the result should have 3 sig figs. So 46.2 (since 46.248 rounded to 3 sig figs is 46.2? Wait, no, 46.248: the third significant figure is 2, the next digit is 4, so we don't round up. So 46.2. Wait, but let's calculate again: 37.60 * 1.23. 37 * 1.23 = 45.51, 0.60*1.23 = 0.738, total is 45.51+0.738 = 46.248. Yes. So with 3 sig figs, it's 46.2.
# Answer: 46.2

#### Part (b): $8.975\div6.7$
# Explanation:
## Step1: Recall division rule
For division, the result has the same number of significant figures as the least precise measurement. 8.975 has 4 sig figs, 6.7 has 2 sig figs. So result should have 2 sig figs.
## Step2: Perform division
$8.975\div6.7\approx1.33955$
## Step3: Round to 2 sig figs
$1.33955\approx1.3$ (wait, no: 6.7 has 2 sig figs, 8.975 has 4. So the result should have 2 sig figs. 1.33955 rounded to 2 sig figs: the first two are 1 and 3, the next digit is 3 which is less than 5, so 1.3? Wait, no, 8.975 ÷ 6.7: 6.7 is two sig figs. Let's do the division more accurately. 6.7 × 1.3 = 8.71, 6.7 × 1.4 = 9.38. 8.975 - 8.71 = 0.265. 0.265/6.7≈0.0396. So total is 1.3 + 0.0396≈1.3396. So rounded to two sig figs is 1.3? Wait, no, 1.3396: first two sig figs 1 and 3, the third digit is 3, which is less than 5, so we keep it 1.3? Wait, but 6.7 has two sig figs, so the answer should have two sig figs. So 1.3.
# Answer: 1.3

#### Part (c): $3.765 + 1.2+37.21$
# Explanation:
## Step1: Recall addition rule
For addition, the result has the same number of decimal places as the least precise measurement. 3.765 has 3 decimal places, 1.2 has 1 decimal place, 37.21 has 2 decimal places. The least is 1 decimal place.
## Step2: Perform addition
$3.765+1.2 + 37.21=3.765+38.41 = 42.175$
## Step3: Round to 1 decimal place
$42.175\approx42.2$ (since the first decimal place is 1, the next digit is 7 which is more than 5, so we round up the first decimal place: 42.2)
# Answer: 42.2

### Problem 3 - 6: (Assuming we need to analyze the data, but the question is not fully clear. If we assume we need to check which student's mean is closest to the actual length 1.4715 m)

Abe's mean: 1.4716 m, Barb's mean: 1.4754 m, Cal's mean: 1.4719 m.

The difference for Abe: $|1.4716 - 1.4715|=0.0001$ m

The difference for Barb: $|1.4754 - 1.4715| = 0.0039$ m

The difference for Cal: $|1.4719 - 1.4715|=0.0004$ m

So Abe's mean is closest to the actual length.

# Answer: Abe's measurements have a mean (1.4716 m) closest to the actual length (1.4715 m)

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QUESTION IMAGE

Question

Explanation:

Problem 1: Significant Figures Count

Part (a): 37.60

Step1: Recall sig fig rules

Step2: Count significant figures

Step1: Recall sig fig rules

Step2: Count significant figures

Step1: Recall sig fig rules

Step2: Count significant figures

Answer:

Part (b): 0.0130

student	$l_1$	$l_2$	$l_3$	$l_4$	$\bar{l}$	$\sigma_{n-1}$	$\alpha$
barb	1.4753	1.4759	1.4756	1.4749	1.4754	0.00043	0.0002
cal	1.4719	1.4723	1.4727	1.4705	1.4719	0.00096	0.0005