Answer:
Usual, because the result is between the minimum and maximum usual values.
Step-by-step explanation:
To identify if the value is usual or unusual we're going to use the Range rule of thumbs which states that most values should lie within 2 standard deviations of the mean. If the value lies outside those limits, we can tell that it's an unusual value.
Therefore:
Maximum usual value: μ + 2σ
Minimum usual value: μ - 2σ
In this case:
μ = 153.1
σ = 18.2
Therefore:
Maximum usual value: 189.5
Minimum usual value: 116.7
Therefore, the value of 187 lies within the limits. Therefore, the correct option is D. Usual, because the result is between the minimum and maximum usual values.
Answer:
-2(5)^2
-2(25)
-50
Step-by-step explanation:
Here's a way to do it.
Let 4e +2 = 5n +1 . . . . . . for some integer n
Then e = (5n -1)/4 = n + (n -1)/4
We want (n-1)/4 to be an integer, so let it be integer m.
... m = (n -1)/4
... 4m = n -1
... 4m +1 = n
Substituting this into our expression for e gives
... e = (5(4m+1) -1)/4 = (20m +4)/4 = 5m +1
e = 5m+1 for any integer m
Negative 3143 over 1000 divide -3.143 by 1000 and there’s your answer
Answer:
7
4
Step-by-step explanation:
The <u>actual values</u> are shown on the given graph as <u>blue points</u>.
The <u>line of regression</u> is shown on the given graph as the <u>red line</u>.
From inspection of the graph, in the year 2000 the actual rainfall was 43 cm, shown by point (2000, 43). It appears that the regression line is at y = 50 when x is the year 2000.
⇒ Difference = 50 - 43 = 7 cm
<u>In 2000, the actual rainfall was </u><u>7</u><u> centimeters below what the model predicts</u>.
From inspection of the graph, in the year 2003 the actual rainfall was 44 cm, shown by point (2003, 40). It appears that the regression line is at y = 40 when x is the year 2003.
⇒ Difference = 44 - 40 = 4 cm
<u>In 2003, the actual rainfall was </u><u>4</u><u> centimeters above what the model predicts.</u>