Suraj took a slice of pizza from the freezer and put it in the oven. The pizza's temperature (in degrees Celsius) as a function
of time (in minutes) is graphed.
How long did it take the pizza to reach 0 degrees Celsius?
How long did it take the pizza to reach 0 degrees Celsius?
2 answers:
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Question not correct, so i have attached the correct question.
Answer:
SE = 0.59
Step-by-step explanation:
The mean of the students height is;
x' = (53 + 52.5 + 54 + 51 + 50.5 + 49.5 + 48 + 53 + 52 + 50)/10
x' = 51.35
Now, deviation from the mean for each height;
53 - 51.35 = 1.65
52.5 - 51.35 = 1.15
54 - 51.35 = 2.65
51 - 51.35 = -0.35
50.5 - 51.35 = -0.85
49.5 - 51.35 = -1.85
48 - 51.35 = -3.35
53 - 51.35 = 1.65
52 - 51.35 = 0.65
50 - 51.35 = -1.35
Now, square of the deviations above;
1.65² = 2.7225
1.15² = 1.3225
2.65² = 7.0225
-0.35² = 0.1225
-0.85² = 0.7225
-1.85² = 3.4225
-3.35² = 11.2225
1.65² = 2.7225
0.65² = 0.4225
-1.35² = 1.8225
Sum of the squared deviations;
2.7225 + 1.3225 + 7.0225 + 0.1225 + 0.7225 + 3.4225 + 11.2225 + 2.7225 + 0.4225 + 1.8225 = 31.525
Let's divide the sum by the DF of n - 1 i.e 10 - 1 = 9.
Thus;
31.525/9 = 3.50278
Taking the square root of that gives us the standard deviation.
Thus;
s = √3.50278
s = 1.8716
Formula for standard error is;
SE = s/√n
SE = 1.8716/√10
SE = 0.59
