Answer:
112.569 ( D )
Step-by-step explanation:
Applying the estimated Regression Equation
y = b1X1 + b2X2 + a
b1 = ((SPX1Y)*(SSX2)-(SPX1X2)*(SPX2Y)) / ((SSX1)*(SSX2)-(SPX1X2)*(SPX1X2)) = 596494.5/635355.88 = 0.93884
b2 = ((SPX2Y)*(SSX1)-(SPX1X2)*(SPX1Y)) / ((SSX1)*(SSX2)-(SPX1X2)*(SPX1X2)) = 196481.5/635355.88 = 0.30925
a = MY - b1MX1 - b2MX2 = 149.25 - (0.94*61.31) - (0.31*193.88) = 31.73252
y = 0.939X1 + 0.309X2 + 31.733
For x1 ( age ) =39, and x2(weight) =143
y = (0.93884*39) + (0.30925*143) + 31.73252= 112.569
where
Sum of X1 = 981
Sum of X2 = 3102
Sum of Y = 2388
Mean X1 = 61.3125
Mean X2 = 193.875
Mean Y = 149.25
attached is the Tabular calculation of the required values needed for estimated regression equation
Answer:
5 m
Step-by-step explanation:
You know the area of a parallelogram is the product of its base length and height:
area = base × height
Fill in the given values, and solve for height:
60 m² = (12 m) × height
(60 m²)/(12 m) = height = (60/12) m
height = 5 m
The height is 5 meters.
Answer:
p^5 + q^5
Step-by-step explanation:
(p + q)^5
p^5 + q^5 Distribute the fifth power to the letters in the paranthesis
Answer:
1 and 3 are both perpendicular to segment NY
Step-by-step explanation:
1. Find the slope of line NY
slope of NY = 5-(-7)/-11-5 = - 3/4
Any line that is perpendicular to NY should have a slope of the inverse of negative slope of NY.
2.Find the slope of perpendicular lines
the inverse of negative slope of NY = - (-4/3) = 4/3
Answer:
The correct option is;
c. Her score was better than those of 60% of all test takers
Step-by-step explanation:
Percentile Score is the score of an exam or test candidate with regard to the relative performance of the other persons that took part in the exam or test.
The percentile score is calculated by converting candidates scores into a scale consisting of the scores of all candidates and ranging from 0 to 100
A 60th percentile score means that the student performed better than 60% of all the candidates of the test and 40% of the test takers scored the same as or better than the student.
