precal

003. PC1 questions

Question: `q001 A straight line connects the points (3, 5) and (7, 17), while another straight line continues on from (7, 17) to the point (10, 29). Which line is steeper and on what basis to you claim your result?

Your solution:

Y2 – Y1 / X2 – X1

17 – 5 =12 / 7-3 = 4= 12/4 = slope of 3

29 – 17= 12 / 10 – 7= 3 = 12 / 3= slope of 4

Line 1 is steeper because it goes over less than line 2.

confidence assessment: 3

Given Solution:

`a The point (3,5) has x coordinate 3 and y coordinate 5. The point (7, 17) has x coordinate 7 and y coordinate 17. To move from (3,5) to (7, 17) we must therefore move 4 units in the x direction and 12 units in the y direction.

Thus between (3,5) and (7,17) the rise is 12 and the run is 4, so the rise/run ratio is 12/4 = 3.

Between (7,10) and (10,29) the rise is also 12 but the run is only 3--same rise for less run, therefore more slope. The rise/run ratio here is 12/3 = 4.

Self-critique (if necessary): OK

Self-critique rating:

Question: `q002. The expression (x-2) * (2x+5) is zero when x = 2 and when x = -2.5. Without using a calculator verify this, and explain why these two values of x, and only these two values of x, can make the expression zero.

Your solution:

(2 – 2=0) * (2(0) + 5)=5 * 0= 0

For the first value of 2, we find zero because the equation, from the start, is destined zero because we are subtracting 2 from 2, leaving zero then multiplying by zero.

(-2.5 – 2=-4.5) *(2(-2.5)= -5 + 5=0) -4.5 * 0 = 0

Anytime that you multiply anything by zero you always receive zero.

confidence assessment: 3

Given Solution:

`aIf x = 2 then x-2 = 2 - 2 = 0, which makes the product (x -2) * (2x + 5) zero.

If x = -2.5 then 2x + 5 = 2 (-2.5) + 5 = -5 + 5 = 0.which makes the product (x -2) * (2x + 5) zero.

The only way to product (x-2)(2x+5) can be zero is if either (x -2) or (2x + 5) is zero.

Note that (x-2)(2x+5) can be expanded using the Distributive Law to get

x(2x+5) - 2(2x+5). Then again using the distributive law we get

2x^2 + 5x - 4x - 10 which simplifies to

2x^2 + x - 10.

However this doesn't help us find the x values which make the expression zero. We are better off to look at the factored form.

STUDENT QUESTION

I think I have the basic understanding of how x=2 and x=-2.5 makes this equation 0

I was looking at the distributive law and I understand the basic distributive property as stated in algebra

a (b + c) = ab + ac and a (b-c) = ab – ac

but I don’t understand the way it is used here

(x-2)(2x+5)

x(2x+5) - 2(2x+5)

2x^2 + 5x - 4x - 10

2x^2 + x - 10.

Would you mind explaining the steps to me?

INSTRUCTOR RESPONSE

The distributive law of multiplication over addition states that

a (b + c) = ab + ac

and also that

(a + b) * c = a c + b c.

So the distributive law has two forms.

In terms of the second form it should be clear that, for example

(x - 2) * c = x * c - 2 * c.

Now if c = 2 x + 5 this reads

(x-2)(2x+5) = x * ( 2 x + 5) - 2 * (2 x + 5).

The rest should be obvious.

We could also have used the first form.

a ( b + c) = ab + ac so, letting a stand for (x - 2), we have

(x-2)(2x+5) = ( x - 2 ) * 2x + (x - 2) * 5.

This will ultimately give the same result as the previous. Either way we end up with 2 x^2 + x - 10.

Self-critique (if necessary):OK

Self-critique rating:

Question: `q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero?

Your solution:

(3x - 6) * (x + 4) * (x^2 - 4)

(3(2)= 6 – 6=0) * ((2) + 4=6) * (2^2=4 – 4=0)= 0 * 6 * 0= 0

if x equals two then the product will be zero

confidence assessment: 2

Given Solution:

`aIn order for the expression to be zero we must have 3x-6 = 0 or x+4=0 or x^2-4=0.

3x-6 = 0 is rearranged to 3x = 6 then to x = 6 / 3 = 2. So when x=2, 3x-6 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero.

x+4 = 0 gives us x = -4. So when x=-4, x+4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero.

x^2-4 = 0 is rearranged to x^2 = 4 which has solutions x = + - `sqrt(4) or + - 2. So when x=2 or when x = -2, x^2 - 4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero.

We therefore see that (3x - 6) * (x + 4) * (x^2 - 4) = 0 when x = 2, or -4, or -2. These are the only values of x which can yield zero.**

Self-critique (if necessary):

I didn’t continue working through the problem, once I found an answer I stopped. I understand why -2 and -4 would produce 0.

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Self-critique rating :

Question: `q004. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'.

Your solution:

Trapezoid A- 3, 7= 7 – 3= width =4

5, 9= 5 + 9= 14 / 2 height= 7

Trapezoid B- 10, 50, 50 – 10= width = 40

2, 4, 2+4=6/2= height of 3

Trapezoid A is four times the height of trapezoid B; however, Trapezoid B is 40 x the width of trapezoid A. Therefore Trapezoid B is much larger than A.

confidence assessment: 2

Given Solution:

`aYour sketch should show that while the first trapezoid averages a little more than double the altitude of the second, the second is clearly much more than twice as wide and hence has the greater area.

To justify this a little more precisely, the first trapezoid, which runs from x = 3 to x = 7, is 4 units wide while the second runs from x = 10 and to x = 50 and hence has a width of 40 units. The altitudes of the first trapezoid are 5 and 9,so the average altitude of the first is 7. The average altitude of the second is the average of the altitudes 2 and 4, or 3. So the first trapezoid is over twice as high, on the average, as the first. However the second is 10 times as wide, so the second trapezoid must have the greater area.

This is all the reasoning we need to answer the question. We could of course multiply average altitude by width for each trapezoid, obtaining area 7 * 4 = 28 for the first and 3 * 40 = 120 for the second. However if all we need to know is which trapezoid has a greater area, we need not bother with this step.

Question: `q005. Sketch graphs of y = x^2, y = 1/x and y = `sqrt(x) [note: `sqrt(x) means 'the square root of x'] for x > 0. We say that a graph increases if it gets higher as we move toward the right, and if a graph is increasing it has a positive slope. Explain which of the following descriptions is correct for each graph:

As we move from left to right the graph increases as its slope increases.

As we move from left to right the graph decreases as its slope increases.

As we move from left to right the graph increases as its slope decreases.

As we move from left to right the graph decreases as its slope decreases.

Your solution:

A. y = x^2

y = 1/x

y = `sqrt(x)

0, 1, 2, 3

A. 0,1, 4, 9

B. 0, 1, .5, .3

C. 0, 1, 1.4, 1.7

The slope increases as the graph decreases.

confidence assessment: 2

Given Solution:

`aFor x = 1, 2, 3, 4:

The function y = x^2 takes values 1, 4, 9 and 16, increasing more and more for each unit increase in x. This graph therefore increases, as you say, but at an increasing rate.

The function y = 1/x takes values 1, 1/2, 1/3 and 1/4, with decimal equivalents 1, .5, .33..., and .25. These values are decreasing, but less and less each time. The decreasing values ensure that the slopes are negative. However, the more gradual the decrease the closer the slope is to zero. The slopes are therefore negative numbers which approach zero.

Negative numbers which approach zero are increasing. So the slopes are increasing, and we say that the graph decreases as the slope increases.

We could also say that the graph decreases but by less and less each time. So the graph is decreasing at a decreasing rate.

For y = `sqrt(x) we get approximate values 1, 1.414, 1.732 and 2. This graph increases but at a decreasing rate.

Self-critique (if necessary): OK

Self-critique rating:

Question: `q006. If the population of the frogs in your frog pond increased by 10% each month, starting with an initial population of 20 frogs, then how many frogs would you have at the end of each of the first three months (you can count fractional frogs, even if it doesn't appear to you to make sense)? Can you think of a strategy that would allow you to calculate the number of frogs after 300 months (according to this model, which probably wouldn't be valid for that long) without having to do at least 300 calculations?

Your solution:

20 * .10%= 2 frogs. Plus original 20 = 22 frogs for first month. Second Month 22 *.10= 2.2 frogs. Plus original 20= 22 + 2.2 new frogs= 24.4 frogs. Third month 24.4 * .10=2.44 frogs. Plus original 24.4= 26.84 frogs.

Multiply .10% of the original 20 for 300 months would be 30, multiply this by 20(original frogs).

300 * .10= 30 *20= 600

confidence assessment: 1

Given Solution:

`aAt the end of the first month, the number of frogs in the pond would be (20 * .1) + 20 = 22 frogs. At the end of the second month there would be (22 * .1) + 22 = 24.2 frogs while at the end of the third month there would be (24.2 * .1) + 24.2 = 26.62 frogs.

The key to extending the strategy is to notice that multiplying a number by .1 and adding it to the number is really the same as simply multiplying the number by 1.1. 10 * 1.1 = 22; 22 * 1.1 = 24.2; etc.. So after 300 months you will have multiplied by 1.1 a total of 300 times. This would give you 20 * 1.1^300, whatever that equals (a calculator will easily do the arithmetic).

A common error is to say that 300 months at 10% per month gives 3,000 percent, so there would be 30 * 20 = 600 frogs after 30 months. That doesn't work because the 10% increase is applied to a greater number of frogs each time. 3000% would just be applied to the initial number, so it doesn't give a big enough answer.

Self-critique (if necessary):

For the calculation, I messed up in my addition. I understand what I did wrong. 22 + 2.2= 24.2 not 24.4.

Furthermore, I see I did exactly what you said would be a common error. I understand why it wouldn’t work, as well as I understand the increasing of 1.1%.

Self-critique rating :

Question: `q007. Calculate 1/x for x = 1, .1, .01 and .001. Describe the pattern you obtain. Why do we say that the values of x are approaching zero? What numbers might we use for x to continue approaching zero? What happens to the values of 1/x as we continue to approach zero? What do you think the graph of y = 1/x vs. x looks for x values between 0 and 1?

Your solution:

1/x

x = 1, .1, .01, .001

1, 10, 100, 1000

We are able to approach 0 by dividing 1 by any whole number greater than one. The values continue to grow smaller and smaller, always remaining positive. I imagine that the graph would be very close to 0 and 1, but never touching the zero.

confidence assessment: 1

Given Solution:

`aIf x = .1, for example, 1 / x = 1 / .1 = 10 (note that .1 goes into 1 ten times, since we can count to 1 by .1, getting.1, .2, .3, .4, ... .9, 10. This makes it clear that it takes ten .1's to make 1.

So if x = .01, 1/x = 100 Ithink again of counting to 1, this time by .01). If x = .001 then 1/x = 1000, etc..

Note also that we cannot find a number which is equal to 1 / 0. Deceive why this is true, try counting to 1 by 0's. You can count as long as you want and you'll ever get anywhere.

The values of 1/x don't just increase, they increase without bound. If we think of x approaching 0 through the values .1, .01, .001, .0001, ..., there is no limit to how big the reciprocals 10, 100, 1000, 10000 etc. can become.

The graph becomes steeper and steeper as it approaches the y axis, continuing to do so without bound but never touching the y axis.

This is what it means to say that the y axis is a vertical asymptote for the graph .

Self-critique (if necessary): OK

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Self-critique rating:

Question: `q008. At clock time t the velocity of a certain automobile is v = 3 t + 9. At velocity v its energy of motion is E = 800 v^2. What is the energy of the automobile at clock time t = 5?

Your solution:

v = 3 t + 9

800 v^2= 3(5)=15 +9 =24

800v^2=24

800/ v squared = 24 / 800

V squared = 33.333

v= 33.333 squared= 1111.111

???

confidence assessment: 2

.............................................

Given Solution:

`aFor t=5, v = 3 t + 9 = (3*5) + 9 = 24. Therefore E = 800 * 24^2 = 460800.

Ok, I see, we substitute 24 in for v since it is the velocity. I should have thought it out.

Question: `q009. Continuing the preceding problem, can you give an expression for E in terms of t?

Your solution:

E= 800 v squared = v = 3 t + 9

confidence assessment: 1

Given Solution:

`aSince v = 3 t + 9 the expression would be E = 800 v^2 = 800 ( 3t + 9) ^2. This is the only answer really required here.

For further reference, though, note that this expression could also be expanded by applying the Distributive Law:.

Since (3t + 9 ) ^ 2 = (3 t + 9 ) * ( 3 t + 9 ) = 3t ( 3t + 9 ) + 9 * (3 t + 9) = 9 t^2 + 27 t + 27 t + 81 = 9 t^2 + 54 t + 81, we get

E = 800 ( 9 t^2 + 54 t + 81) = 7200 t^2 + 43320 t + 64800 (check my multiplication because I did that in my head, which isn't always reliable).

Self-critique (if necessary): I’m clear on the answer.

Self-critique rating :

&#Good responses. Let me know if you have questions. &#

Task: `q003.

Check the rest of your disks.

Each disk should be accessed by browsing to the disk and running the file whose name most closely matches the name of the disk, or the HTML file in the root folder of the DVD. Insert each disk in turn into your drive, browse to the appropriate file, and run it. 

If all your disks work, indicate that they do. If you have trouble with any of them, or with these instructions, describe your problems in detail.

your response &&&&&&&&&&&&&&&&&&

(insert your response beginning in the next line; the next line is blank and doesn't include the #$... prompt):

All disk work

#$&* (your response should have gone on the line above this one)

*#&!

@& I'm glad everything works.*@

#(*!

&#Good responses. Let me know if you have questions. &#

&#Good responses. Let me know if you have questions. &#

Task: `q007. It is also not valid to justify a solution by copying a picture or a solution from a calculator (unless of course the problem specifies that the calculator is to be used in this manner).

• It is fine to use a calculator to do your arithmetic, but you must, for example, show the steps in solving an equation.

• If the solution of a problem includes a graph you must explain the behavior of that graph rather than just copying calculator output.

The key is that while a calculator can be very useful, operations like entering a function or an equation and copying output from a calculator is not a college-level skill.

If the process is part of the course, you have to show the steps of the process.

State this policy in your own words.

**** Your response (insert your response beginning in the next line; the next line is blank and doesn't include the #$... prompt):

You may use a calculator to do your work, you must not copy a picture directly from the calculator.

#$&* (your response should have gone on the line above this one)

*#&!

&#Your work looks good. Let me know if you have any questions. &#

#(*!

&#Your work looks good. Let me know if you have any questions. &#

• How can you use observations of position and clock time to determine whether the tape rolling along an incline is speeding up or slowing down?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

Based on the incline that the roll of tape was on, the tape was speeding up rather than slowing down. The incline gave the roll of tape increased velocity down the incline.

#$&*

@& We might assume that this would be the case, but the evidence of the clock and the tape would be more persuasive than an assumption. *@

• How can you use observations of position and clock time to determine whether the tape rolling along an incline is speeding up or slowing down?

answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):

Based on the incline that the roll of tape was on, the tape was speeding up rather than slowing down. The incline gave the roll of tape increased velocity down the incline.

#$&*

@& We might assume that this would be the case, but the evidence of the clock and the tape would be more persuasive than an assumption. *@

#(*!

&#This lab submittion looks good. See my notes. Let me know if you have any questions.

Revision isn't requested, but if you do choose to submit revisions, clarifications or questions, please insert them into a copy of this document, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

&#

#(*!

&#Good work on this lab exercise. Let me know if you have questions. &#

List the following in order, from least to greatest. Indicate 'ties': The object's initial speed, its final speed, its average speed, and the change in its speed as it rolled from one end of the book to the other.

your brief discussion/description/explanation:

The balls initail speed was 0 since it was at rest

the average speed was 14.05 inches per second

The final speed was the fastest of the three

The balls speed increased the further it traveled because it was rolling downhill.

#$&*

@& You have assumed that the final speed was the greatest, and of course this is consistent with your perception.

However measurements would be more persuasive. For example you could have timed the object down the first half of the ramp, then down the entire ramp, and concluded from your measurements that the average velocity on the entire ramp was greater than on the first half.

There would still be an assumption involved, that the trend of the ball's velocity is unchanging, but you would have established a trend based on hard evidence.*@

List the following in order, from least to greatest. Indicate 'ties': The object's initial speed, its final speed, its average speed, and the change in its speed as it rolled from one end of the book to the other.

your brief discussion/description/explanation:

The balls initail speed was 0 since it was at rest

the average speed was 14.05 inches per second

The final speed was the fastest of the three

The balls speed increased the further it traveled because it was rolling downhill.

#$&*

@& You have assumed that the final speed was the greatest, and of course this is consistent with your perception.

However measurements would be more persuasive. For example you could have timed the object down the first half of the ramp, then down the entire ramp, and concluded from your measurements that the average velocity on the entire ramp was greater than on the first half.

There would still be an assumption involved, that the trend of the ball's velocity is unchanging, but you would have established a trend based on hard evidence.*@

#(*!

&#This lab submittion looks good. See my notes. Let me know if you have any questions.

Revision isn't requested, but if you do choose to submit revisions, clarifications or questions, please insert them into a copy of this document, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

&#

You got at least 20 time intervals.  Based on your data what was the average of the first 20 time intervals?  Note that you could get this average by averaging the first 20 intervals.  My first few intervals were .15625, .15625, .1875, .171875, etc; I could just add up the first 20 intervals and divide by 20 to get the average.  However there is an easier and quicker way to get the result, so use the easier way if you can.

Give your result, number only, in the first line, and starting in the second line explain how you got it.

 .14065

I divided 2.813 from the last intervals clock time by 20

 

#$&*

@& That would include that .39 second interval at the beginning, which was not a quick-click interval. No need to redo the experiment, but you do need to be sure the data you include in a data report is relevant to the thing being measured.*@

You got at least 20 time intervals.  Based on your data what was the average of the first 20 time intervals?  Note that you could get this average by averaging the first 20 intervals.  My first few intervals were .15625, .15625, .1875, .171875, etc; I could just add up the first 20 intervals and divide by 20 to get the average.  However there is an easier and quicker way to get the result, so use the easier way if you can.

Give your result, number only, in the first line, and starting in the second line explain how you got it.

 .14065

I divided 2.813 from the last intervals clock time by 20

 

#$&*

@& That would include that .39 second interval at the beginning, which was not a quick-click interval. No need to redo the experiment, but you do need to be sure the data you include in a data report is relevant to the thing being measured.*@

#(*!

#$&*

When I did this activity the first few lines of my data were as follows:

event number clock time time interval

1 11.67188 11.67188

2 11.875 0.203125

3 12.0625 0.1875

4 12.20313 0.140625

5 12.375 0.171875

6 12.54688 0.171875

7 12.73438 0.1875

8 12.92188 0.1875

9 13.10938 0.1875

10 13.28125 0.171875

11 13.4375 0.15625

It looks like the same intervals keep popping up.  For example .1875 seconds occurs 5 times out of the first 10 intervals, .171875 seconds occurs three times, and .203125 seconds, .140625 seconds and .15625 seconds each occur once.

A frequency distribution for my time intervals would be as follows:

time interval frequency

,140625 1

.15625 1

.171875 3

.1875 5

.203125 1

What different time intervals did you observe in your first 20 intervals, and how many times did each occur?  List below the different time intervals you observed and the number of times each occurred.  List from the shortest to the longest interval, and use a comma between the time interval and its frequency.  For example my data above would be listed at

.140625, 1

.1565, 1

.171875, 3

.1875, 5

.203125, 1

Your list should be in exactly this format, with no other symbols or characters.

 .078, 1

.093, 1

.109, 2

.11, 3

.125, 4

.14, 2

.141, 2

.156, 2

.157, 1

.172, 1

.391, 1

 

#$&*

You may make any comments or ask any question about the process so far in the box below

 

 

#$&*

On the 10 intervals I've shown you, do you really think I managed to get a time of .1875 seconds, accurate to 4 significant figures, on half of the intervals?  If you do, I'm grateful for your confidence but I'm just not that good.  No human being has that much neurological and muscular control.

So why do you think the TIMER program reported that time so frequently?  Why weren't there times like .1830 seconds, or .1769 seconds?  Does this mean that the TIMER program is flawed?  Does that mean it's useless?

 I think it is because that it rounds off the intervals and it just so happened that the intervals where close enough for it to round them to the same interval numerous times. No the timer program is not flawed or useless, it still is accurate enough for what we are doing and more.

 

#$&*

Here are a few more lines of data, with an added column showing the difference between each time interval and the next.

clock time time interval difference from one time interval to next

9 13.10938 0.1875 -0.01563

10 13.28125 0.171875 -0.01563

11 13.4375 0.15625 0.03125

12 13.625 0.1875 -0.01563

13 13.79688 0.171875 0.015625

14 13.98438 0.1875 0.015625

15 14.1875 0.203125 -0.03125

16 14.35938 0.171875 -0.01563

17 14.51563 0.15625 0.03125

Take a good look at that last column and tell us what you see in those numbers, and what this tells you about the TIMER program

 Number that are just about the same. It looks to me like the intervals where taken too quick for the program to actually get a good accurate number.

 

#$&*

Now initialize the TIMER once more, and take a series of 10 relaxed breaths.  Every time you start to inhale, hit the TIMER button.

My results for the first 7 complete breaths are as follows:

series of relaxed breaths  

event number clock time time interval difference between time interval and next

1 1569.734 1569.734  

2 1582.75 13.01563 0.32812

3 1596.094 13.34375 3.90625

4 1613.344 17.25 2.70313

5 1633.297 19.95313 1.35937

6 1654.609 21.3125 4.23438

7 1680.156 25.54688 2.15625

8 1707.859 27.70313  

I didn't go on because the time between my breaths kept increasing, and I was afraid if I relaxed any more I might stop breathing altogether.  It's going to take either more statistical analysis to determine whether that's a real danger, or a little common sense.

Report your results by just entering your time intervals, one to each line, in the box below.  If I was entering my results I would enter

13.01563

13.34375

17.25

19.95313

21.3125

etc.

Enter your results in the same format:

 0.266

8.75

11.358

10.25

10.733

11.219

11.015

11.061

11.937

10.89

 

@& I doubt that you completed a breath in 0.266 seconds. Same caution as in my previous note.*@

#$&*

If you have any comments please insert them here

 I started the clock when I first inhaled

 

#$&*

Most likely you did not observe the same exact time interval twice, and if you did it did not happen nearly as often as when you did the fact clicks.

Why do you think this is exactly what we would expect?

 Because there are greater time intervals this time around

 

#$&*

Which of the following statements do you think is the most accurate?

a.  The TIMER program is capable of determining the time between two events accurately to within about .1 second.

b.  The TIMER program is capable of determining the time between two events accurately to within about .01 second.

c.  The TIMER program is capable of determining the time between two events accurately to within about .001 second.

d.  The TIMER program is capable of determining the time between two events accurately to within about .0001 second.

Enter your answer and your reasoning below:

 C, because from the time intervals in the previous sections it looks like the program has recorded different enough thousandth place numbers to say that it is accurate to the .001.

 

#$&*

Note that the TIMER.exe program can save your data in a format that can be read by a spreadsheet (the TIMER applet cannot do so).  This will be very handy in the future, so take a minute and do the following:

1.  Click on the button at the lower right of the TIMER form, entitled Click to File Data.

2.  A box will pop up allowing you to include an identifying message. You should generally choose to include such a message; for the data presently on your timer that might be 'series of regular breaths time at beginning of inhalation' or something similar.  Type in whatever you think would serve as a good identifier for this data and OK that box.

3.  A typical Save As window will appear.  Decide where to save your data and what to call it, and proceed to save it.  The program will save the file as a comma-delimited text file.

4.  Open your spreadsheet program (typically Excel) and choose File > Open.  Browse to the folder in which you just saved your data.  Below the File Name line will be a File Type line; set this either to Text Files or All Files so your file will appear.  Open it.

5.  A series of windows will typically appear.  In the first window make sure the file type chosen is Delimited, the proceed to the next window. 

6.  In the second window you will see a series of checkboxes; check the one entitled Comma, in order to select the comma-delimited file, then just click on Finish.

If you can't run the .exe program, you can't do Steps 1-3.  However all you need to do is copy the contents of the program to a text file, using copy-and-paste.  Save that text file, using any filename you wish.  Then proceed with steps 4-6 above.

Your data should appear in your spreadsheet, and can be manipulated as in any spreadsheet.

Copy a few lines of the data from your spreadsheet below:

 

 0.266

8.75

11.358

10.25

10.733

11.219

11.015

11.061

11.937

10.89

 

#$&*

Your instructor is trying to gauge the typical time spent by students on these experiments.  Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:  "

@& Good data, and good answers. However do see my notes about ensuring that reported data corresponds to the quantity being measured.*@