For (-7,3) and (5,3) the answer is 12. You add the different points, -7 and 5 because one is negative and one is positive. You do not add 3 and 3 because they are the same.
(3,-6) and (3,-10) = 4 because -6 and -10 are negatives. The answer is always positive.
(8,0) and (8,-8) = 8.
Answer:
b. more money
Step-by-step explanation:
Randy's original position lets him make:
50*500 = $25,000
The new position will let him make:
$27,500
Therefore,
original position < new position = $25,000 < $27,500
-10 + a = 6a - 7a
Rearrange the left hand side
a - 10 = 6a - 7a
Simplify 6a - 7a
a - 10 = -a
Add a on both sides
2a - 10 = 0
Add 10 on both sides
2a = 10
Divide by 2 on both sides
a = 5
Answer:
15.866%
Step-by-step explanation:
We solve using z score formula
z = (x-μ)/σ, where
x is the raw score = 3,282 gallons of kerosene
μ is the population mean = 3,087 gallons of kerosene
σ is the population standard deviation = 195 gallons
For x > 3282 gallons
z = 3282 - 3087/195
z = 1
Probability value from Z-Table:
P(x<3282) = 0.84134
P(x>3282) = 1 - P(x<3282) = 0.15866
Converting to percentage
0.15866 × 100 = 15.866%
The percent of flights on these particular routes that will burn more than 3,282 gallons of kerosene is 15.866%
Answer:
Step-by-step explanation:
The formula to calculate the forecast could be determine by using the exponential smoothing method :
![Ft = F(t-1) + \alpha [A(t-1) - F(t-1)]](https://tex.z-dn.net/?f=Ft%20%3D%20F%28t-1%29%20%2B%20%20%5Calpha%20%5BA%28t-1%29%20-%20F%28t-1%29%5D)
Where ,Ft is the Forecast for period t
F(t-1) is the Forecast for the period previous to t
A(t-1) is the Actual demand for the period previous to t
= Smoothing constant
To get the forecast for may and june the above formula with 
and april forecast of 500 will be used
For march
For April
For May
So forecast for May = 536.25
