Use the given frequency distribution to find the class width, class midpoint, and class boundaries
1 answer:
Based on the table attached, the class boundaries are as follows: 43.5, 49.5, 53.5, 59.5, 63.5, 69.5, 73.5, 79.5.
<h3 /><h3>What is the frequency distribution about?</h3>Temperature (°F) Frequency
40-44 1
45-49 3
50-54 5
55-59 11
60-64 7
65-69 7
70-74 1
A. Viewing at the frequency table, we find that the lower class limits of two consecutive classes are 40 and 44, respectively. Thus the class width is:
Class width = 44 − 40= 4
B: One can find the class midpoints by putting the average of the lower and upper class limits. So the midpoint of the first class is seen below:
= 40 - 44 = = 42. In a similar way, we find the rest.
C. The Class boundaries is known to be that which separate the classes removing the class limits. The class boundaries are as follows:
43.5, 49.5, 53.5, 59.5, 63.5, 69.5, 73.5, 79.5
Therefore, Based on the table attached, the class boundaries are as follows: 43.5, 49.5, 53.5, 59.5, 63.5, 69.5, 73.5, 79.5.
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Answer:
a = - , b = -
Step-by-step explanation:
To obtain the required form use the method of completing the square
add/ subtract ( half the coefficient of the x- term)² to x² - 9x
y = x² + 2(- )x +
-
+ 14
= (x - )² -
+
= (x - )²-
← in the form (x + a)² + b
with a = - and b = -