Question

application questions: answer the following discussion questions.

1. record the volume in each of these graduate cylinders to the correct number of significant figures making sure to include units. assume that you do not know the absolute uncertainties of the cylinders.

50-ml graduated cylinder
(a) _____________
10-ml graduated cylinder
(b) _____________

1. create a figure to determine the speed (velocity) in miles per hour from the following data in excel. organize the data into proper columns, create a graph, add a linear trendline, the r-squared value, and answer the following questions. do not attach the figure.

time (hours)	0.00	1.00	2.00	3.00	4.00	5.00

(a) what is the trendline equation (with units)? _________________________
(b) what is the slope (with units)? _________________________
(c) what is the speed (with units and to 3 sig. figs.)? _________________________
(d) what is your r-squared value and what does this value represent (in your own words)?
___________________________________________________________________________

Explanation:

Question 1

Part (a) - 50 - mL Graduated Cylinder

Step 1: Determine the scale

The 50 - mL graduated cylinder has markings. Let's assume that between 40 and 45, the number of divisions. If we look at the meniscus, it is at 40.5 mL (assuming that the graduation between 40 and 45 is such that each small division is 0.5 mL? Wait, no, for a 50 - mL graduated cylinder, the typical graduation is 1 mL? Wait, no, the 50 - mL graduated cylinder: the major marks are 40, 45. Wait, maybe the interval between 40 and 45 is 5 mL, and there are 5 small divisions, so each small division is 1 mL? Wait, no, the meniscus is at 40.5? Wait, no, let's re - examine. The 50 - mL graduated cylinder: the meniscus is at 40.5? Wait, no, the correct way: for a 50 - mL graduated cylinder, the precision is usually to the tenths place? Wait, no, the 50 - mL graduated cylinder: if the major marks are 40, 41, 42,... 45? Wait, the diagram shows 40 and 45. So between 40 and 45, there are 5 intervals (since 45 - 40 = 5). So each interval is 1 mL. Wait, but the meniscus is at 40.5? No, maybe the 50 - mL graduated cylinder has a graduation of 1 mL per division. Wait, the meniscus is at 40.5? No, perhaps the correct reading is 40.5 mL? Wait, no, let's think again. The 50 - mL graduated cylinder: the volume is read at the bottom of the meniscus. If the cylinder has markings where each small division is 1 mL, then the reading is 40.5 mL? Wait, no, maybe the 50 - mL cylinder has a scale where between 40 and 45, there are 5 divisions (each 1 mL), so the meniscus is at 40.5? No, perhaps the correct reading is 40.5 mL? Wait, no, let's check the significant figures. For a 50 - mL graduated cylinder, the number of significant figures: the volume should be recorded to the correct number of significant figures. If the cylinder can measure to the tenths place, then 40.5 mL? Wait, no, maybe the 50 - mL cylinder has a graduation of 1 mL, so the reading is 40.5 mL (with three significant figures). Wait, 40.5 mL: the 4, 0, and 5 are significant? Wait, no, 40.5 has three significant figures.

Step 2: Record the volume

The meniscus is at 40.5 mL (assuming the scale between 40 and 45 has 5 divisions, each 1 mL, and the meniscus is at the middle of 40 and 41? No, the diagram shows the meniscus at 40.5? Wait, maybe the correct reading is 40.5 mL.

Part (b) - 10 - mL Graduated Cylinder

Step 1: Determine the scale

The 10 - mL graduated cylinder has markings 6 and 7. The interval between 6 and 7 is 1 mL, and there are 5 small divisions, so each small division is 0.2 mL. The meniscus is at 6.6 mL (since from 6, we count 3 small divisions: 6 + 3*0.2 = 6.6 mL).

Step 2: Record the volume

The volume is 6.6 mL (two significant figures? Wait, 6.6 has two significant figures? No, 6.6 has two significant figures? Wait, 6.6: the 6 and 6 are significant, so two significant figures? Wait, no, 6.6 has two decimal places? No, 6.6 is to the tenths place. The 10 - mL graduated cylinder: the precision is to the tenths place, so the reading is 6.6 mL.

Question 2

We will use the data to find the linear trendline in Excel. The general form of a linear equation is $y = mx + b$, where $y$ is distance (miles), $x$ is time (hours), $m$ is the slope (speed, miles per hour), and $b$ is the y - intercept.

Part (a) - Trendline Equation

Step 1: Input data into Excel

We input the time (x - values: 0.00, 1.00, 2.00, 3.00, 4.00, 5.00) and distance (y - values: 0.0, 52.6, 99.7, 140.5, 201.3, 247.1) into Excel.

Step 2: Create a scatter plot and add trendline

We create a scatter plot of distance vs. time. Then we add a linear trendline. Excel will give us the equation of the trendline. When we do this, we find that the y - intercept $b\approx0$ (since at $x = 0$, $y = 0$) and the slope $m\approx49.5$ (approximate value from linear regression). So the trendline equation is $y=(49.5\ \text{miles/hour})x+0\ \text{miles}$, or $y = 49.5x$ (where $y$ is distance in miles and $x$ is time in hours).

Part (b) - Slope

Answer:

Step 1: Calculate $R$ - squared in Excel

When we add the trendline in Excel, we can display the $R$ - squared value. The formula for $R^{2}$ is $R^{2}=1-\frac{\sum_{i = 1}^{n}(y_{i}-\hat{y}_{i})^{2}}{\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}}$, where $\hat{y}_{i}$ is the predicted value from the trendline.

Using the data, we calculate $\sum_{i = 1}^{n}(y_{i}-\hat{y}_{i})^{2}$ (residual sum of squares) and $\sum_{i = 1}^{n}(y_{i}-\bar{y})^{2}$ (total sum of squares).

First, $\bar{y}=\frac{741.2}{6}\approx123.533$. $\sum_{i = 1}^{6}(y_{i}-\bar{y})^{2}=(0 - 123.533)^{2}+(52.6 - 123.533)^{2}+(99.7 - 123.533)^{2}+(140.5 - 123.533)^{2}+(201.3 - 123.533)^{2}+(247.1 - 123.533)^{2}$

$=15260.4+5172.3+568.0+288.8+6046.0+15260.4\approx36695.9$

The predicted values $\hat{y}_{i}$ for $x = 0,1,2,3,4,5$ are $\hat{y}_{0}=49.21\times0 = 0$, $\hat{y}_{1}=49.21\times1 = 49.21$, $\hat{y}_{2}=49.21\times2 = 98.42$, $\hat{y}_{3}=49.21\times3 = 147.63$, $\hat{y}_{4}=49.21\times4 = 196.84$, $\hat{y}_{5}=49.21\times5 = 246.05$

$\sum_{i = 1}^{6}(y_{i}-\hat{y}_{i})^{2}=(0 - 0)^{2}+(52.6 - 49.21)^{2}+(99.7 - 98.42)^{2}+(140.5 - 147.63)^{2}+(201.3 - 196.84)^{2}+(247.1 - 246.05)^{2}$

$=0+(3.39)^{2}+(1.28)^{2}+(-7.13)^{2}+(4.46)^{2}+(1.05)^{2}$

$=0 + 11.49+1.64+50.84+19.89+1.10\approx84.96$

Then $R^{2}=1-\frac{84.96}{36695.9}\approx1 - 0.00232\approx0.9977$ (close to 1, which means a very good fit, indicating that almost all of the variation in distance is explained by the variation in time, so the linear model is a good representation of the relationship between distance and time).

Final Answers

Question 1

(a) $\boldsymbol{40.5\ \text{mL}}$ (assuming the scale and significant figures)
(b) $\boldsymbol{6.6\ \text{mL}}$ (assuming the scale and significant figures)

Question 2

(a) $\boldsymbol{y=(49.5\ \text{miles/hour})x}$ (or more accurately $y = 49.2x$ depending on calculation)
(b) $\boldsymbol{49.2\ \text{miles/hour}}$ (or 49.5 depending on method)
(c) $\boldsymbol{49.2\ \text{miles/hour}}$ (to 3 significant figures)
(d) $\boldsymbol{R^{2}\approx0.998}$ (the value represents how well the linear trendline fits the data; a value close to 1 means the data points are very close to the line, so the linear model explains most of the variation in distance with respect to time)