Sorry I know this late, but the answer is 40 MPH, I don't know if this helps! I HOPE SO THO!! <3
Answer is in this link: cctvsd/hvvdj.com
Hey there!!
What is slope-intercept form :
... y = mx + b
( a ) Given :
... ( 2 , -2 ) and slope=1.4
... y = mx + b
... -2 = 2×1.4 + b
... -2 = 2.8 + b
... -4.8 = b
The slope-intercept form :
... y = 1.4x - 4.8
( b ) Given :
... ( -1 , 4 ) and slope = -3.
y = mx + b
... 4 = -3×-1 + b
... 4 = 3 + b
... 1 = b
The slope-intercept form :
... y = -3x + 1
Note :
( m = slope and b = y-intercept )
Hope my answer helps!!
Observe the given data distribution table carefully.
The 5th class interval is given as,
The upper limit (UL) and lower limit (LL) of this interval are,
Thus, the upper-class limit of this 5th class is 17.4.
Percentage of depreciation of the car that Marie bought = 15%
Present value of the car = 13000
Let us assume that the value of the car at the time of buying = x
So
The percentage valuation of the car now = (100 - 15) percent
= 85 percent
Then
(85/100) * x = 13000
85x = 13000 * 100
85x = 1300000
x = 1300000/85
= 15294.12
So the actual cost of the car is 15294.12. I hope the procedure is clear enough for you to understand.
