The remainder of the exam is to be answered on the exam in the spaces provided (

Question

The remainder of the exam is to be answered on the exam in the spaces provided (no scantron answers). Please write ledgeably the spaces provided Ca 27. Draw a linear graph of the following data on the graph paper provided. Draw a proper graph, with appropriate labels, and using a proper scale. Turn your graph in with your exam. (graph 5 points) These data were obtained by measuring the pressure in a sealed container as a function of temperature. Ul Temperature Pressure (kPa) 300100 345 395 445 500 120 140 152 163 187 201 542 580 a. (3 pts) What is the slope of the line? (show how you determined the slope) b. (3 pts) What are the units of the slope? c. (3 pts) What is the equation of the line? d. (8 pts) At what temperature is the pressure 175 kPa? obtained this value.) (mark point on your graph where you

Explanation / Answer

I will answer, for now, the question 28. Question 27 post it in another question thread ok?.

First the median lenght it's the middle value of all the data values there. As you put in there, the middle values of the 10 values are 10.10 and 10.15, so the mean would be 10.125 cm

The mean it's just the average of all those data, and it's in fact 10.145. This is because, indeed there are 2 numbers that repeat twice, but still they need to be taken account in the average calculations so:

mean = 9.9+9.95+10+10.05+10.10+10.15+10.15+10.2+10.2+10.75 / 10 = 10.145 cm

the average deviation would be the sum of the mean value with each of the value, elevated to 2, and the standard deviation would be the result of this, between the number of values - 1, and taking the root square. so:

SD = [(10.145-9,9)2+(10.145-9,95)2+(10.145-10.05)2+(10.145-10.1)2+2(10.145-10.15)2+2(10.145-10,2)2+(10.145-10.75)2/ 9 ]1/2

SD = (0.4812/9)1/2 = 0.23 cm

The average deviaton is 0.4812

Hope this helps

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