Answer:
Kindly check explanation
Step-by-step explanation:
Given the following :
Equation of regression line :
Yˆ = −114.05+2.17X
X = Temperature in degrees Fahrenheit (°F)
Y = Number of bags of ice sold
On one of the observed days, the temperature was 82 °F and 66 bags of ice were sold.
X = 82°F ; Y = 66 bags of ice sold
1. Determine the number of bags of ice predicted to be sold by the LSR line, Yˆ, when the temperature is 82 °F.
X = 82°F
Yˆ = −114.05+2.17(82)
Y = - 114.05 + 177.94
Y = 63.89
Y = 64 bags
2. Compute the residual at this temperature.
Residual = Actual value - predicted value
Residual = 66 - 64 = 2 bags of ice
Answer:
-0.394875
Step-by-step explanation:
(9/16)*(-3/4)
9/16=.5265
-3/4=-.75
.5265*(-.75)=-0.394875
Through trial and error:
(7x-6)(3x-2)
Usually it works best to test out a number of possibilities :)
Answer:
8a^2+a-7
Step-by-step explanation:
Answer: (a) 0.006
(b) 0.027
Step-by-step explanation:
Given : P(AA) = 0.3 and P(AAA) = 0.70
Let event that a bulb is defective be denoted by D and not defective be D';
Conditional probabilities given are :
P(D/AA) = 0.02 and P(D/AAA) = 0.03
Thus P(D'/AA) = 1 - 0.02 = 0.98
and P(D'/AAA) = 1 - 0.03 = 0.97
(a) P(bulb from AA and defective) = P ( AA and D)
= P(AA) x P(D/AA)
= 0.3 x 0.02 = 0.006
(b) P(Defective) = P(from AA and defective) + P( from AAA and defective)
= P(AA) x P(D/AA) + P(AAA) x P(D/AAA)
= 0.3(0.02) + 0.70(0.03)
= 0.027
