The Slope of this line is -3
Answer:
y=1.003009+0.003453x
or
GPA=1.003009+0.003453(SAT Score)
Step-by-step explanation:
The least square regression equation can be written as
y=a+bx
In the given scenario y is the GPA and x is SAT score because GPA depends on SAT score.
SAT score (X) GPA (Y) X² XY
421 2.93 177241 1233.53
375 2.87 140625 1076.25
585 3.03 342225 1772.55
693 3.42 480249 2370.06
608 3.66 369664 2225.28
392 2.91 153664 1140.72
418 2.12 174724 886.16
484 2.5 234256 1210
725 3.24 525625 2349
506 1.97 256036 996.82
613 2.73 375769 1673.49
706 3.88 498436 2739.28
366 1.58 133956 578.28
sumx=6892
sumy=36.84
sumx²=3862470
sumxy=20251.42
n=13

b=9367.18/2712446
b=0.003453
a=ybar-b(xbar)
ybar=sum(y)/n
ybar=2.833846
xbar=sum(x)/n
xbar=530.1538
a=2.833846-0.003453*(530.1538)
a=1.003009
Thus, required regression equation is
y=1.003009+0.003453x.
The least-squares regression equation that shows the best relationship between GPA and the SAT score is
GPA=1.003009+0.003453(SAT Score)
9514 1404 393
Answer:
4 pounds
Step-by-step explanation:
Let b represent the pounds of bananas Mrs. Davis bought. The weight of strawberries would be (7-b) and their cost would be 4×0.50 = 2.00 per pound. Her total purchase was ...
0.50b + 2.00(7 -b) = 8.00
-1.50b +14 = 8 . . . . simplify
6 = 1.50b . . . . . . add 1.50b -8
4 = b . . . . . . . . divide by 1.50
Mrs. Davis bought 4 pounds of bananas.
Area of the rectangle is 18•3= 54
Area of the triangle is (10-3)•(18-6)/2= 7•12/2= 42
Area of the whole figure is 54+42=96
All our answers lie in the above statement.
Confidence Level:
The creator claims that 9 out 10 students will have the average score in the said range. Or in other words we can say that the creator is 90% confident about the result of the field test. So the confidence level is 90%.
Margin of Error:
The average score lies within 4% of 70%. This means the margin of error is 4% i.e. the average scores can deviate from 70% by 4% .
Confidence Interval:
Lower Limit = 70% - 4% = 66%
Upper Limit = 70% + 4% = 74%
Interpretation:
The exam creator is 90% confident that the average scores of seniors will be between 66% and 74%.