If you were to reverse it it would be 7/6, but if you need it in another form, tell me.
The given equation is the best line that approximates the linear
relationship between the midterm score and the score in the final exam.
- AJ's residual is 0.3, which is not among the given options, therefore, the correct option is. <u>E. None of these</u>.
Reasons:
The given linear regression line equation is; 
= 25.5 + 0.82·
Where;
= Final exam score;
= The midterm score;
AJ score in the first test, = 90
AJ's actual score in the exam = 99
Required:
The value of AJ's residual
Solution:
By using the regression line equation, we have;
The predicted exam score, = 25.5 + 0.82 × 90 = 99.3
- The residual score = Predicted score - Actual score
∴ AJ's residual = 99.3 - 99 = 0.3
AJ's residual = 0.3
Therefore, the correct option is option E;
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y = 7 + 3x
y = 3x + 7
y = mx + b, therefore the y-intercept is 7.
3y = 6x + 12
y = 2x + 4
y = mx + b, therefore the y-intercept is 4.
Answer:
24.5 unit²
Step-by-step explanation:
Area of ∆
= ½ | x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) |
= ½ | (-1)(3 -(-4)) + 6(-4 -3) + (-1)(3 - 3) |
= ½ | -7 - 42 |
= ½ | - 49 |
= ½ (49)
= 24.5 unit²
<u>Method 2:</u>
Let the vertices are A, B and C. Using distance formula:
AB = √(-1-6)² + (3-3)² = 7
BC = √(-6-1)² + (-4-3)² = 7√2
AC = √(-1-(-1))² + (4-(-3))² = 7
Semi-perimeter = (7+7+7√2)/2
= (14+7√2)/2
Using herons formula:
Area = √s(s - a)(s - b)(s - c)
here,
s = semi-perimeter = (14 + 7√2)/2
s - a = S - AB = (14+7√2)/2 - 7 = (7 + √2)/2
s - b = (14+7√2)/2 - 7√2 = (14 - 7√2)/2
s - c = (14+7√2)/2 - 7 = (7 + √2)/2
Hence, on solving for area using herons formula, area = 49/2 = 24.5 unit²
Answer:
B. 12 pieces
Step-by-step explanation:
75 / 6 1/4 or 75 / 6.25 = 12
