find the slope of the line on the graph write your answer as a fraction or a whole number not a mixed number or decimal
1 answer:
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Answer:
2 and a half cup
Step-by-step explanation:
A detailed look at the image attached will provide further explanation.
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
Answer:
The correct option is;
∠A ≅ ∠X
Step-by-step explanation:
The given coordinates of the points of triangle ACB are;
A(-4, 4), C(-1, 3), B(-4, 0)
The given coordinates of the points of triangle XYZ are;
X(0, 8), Y(8, 8), Z(6, 2), therefore, we have
The length. l. of segment is given by the following formula;
For the length of the segment AC; (x₁, y₁) = (-4, 4), (x₂, y₂) = (-1, 3), l = √(10)
For the length of the segment AB; (x₁, y₁) = (-4, 4), (x₂, y₂) = (-4, 0), l = 4
For the length of the segment BC; (x₁, y₁) = (-4, 0), (x₂, y₂) = (-1, 3), l = 3·√2
For the length of the segment XY; (x₁, y₁) = (0, 8), (x₂, y₂) = (8, 8), l = 8
For the length of the segment XZ; (x₁, y₁) = (0, 8), (x₂, y₂) = (6, 2), l = 6·√2
For the length of the segment ZY; (x₁, y₁) = (6, 2), (x₂, y₂) = (8, 8), l = 2·√(10
Therefore;
XY ~ AB, XZ ~ BC, ZY ~ AC
Which gives;
∠A ≅ ∠X, ∠B ≅ ∠Y, ∠C ≅ ∠Z
