Answer:
(A) 0.125 probability
(B) 0.625 probability
(C) 660 miles
Step-by-step explanation:
The distance driven by a truck driver daily, falls between 300miles and 700miles and follows a uniform distribution.
(A) The probability that the truck driver goes more than 650 miles a day is:
[700 - 650] / [700 - 300] = 50/400 = 0.125
(B) The probability that the truck driver goes between 400 and 650 miles a day is:
[650 - 400] / [700 - 300] = 250/400 = 0.625
(C) The minimum number of miles the truck driver travels on the furthest 10% of days is given thus:
10% of 400 = 40
Subtract this from the farthest distance;
700miles - 40miles = 660miles
Y=(-x*x)-4x+3
y=-(x^2+4x-3)
4+y=-(x^2+4x+4)+3
y+4=-(x-2)^2.) +3
-4. -4
y=-((x-2)^2)-1
Answers:
Vertex: (2,-1)
AOS:x = 2
Domain: All Real Numbers
Range:[-1,Infinity)
Step-by-step explanation:
You are just replacing the 'n' with the number in the 'n box'.
So if this is the formula:

then:
B(-3)= 4-(-3/3)
B(-3)= 4-( -1)
B(-3)= 4+1
B(-3)= 5
B(0)= 4-(0/3)
B(0)= 4-0
B(0)= 4
B(3)= 4-(3/3)
B(3)= 4-1
B(3)= 3
a(i). Since you are given a velocity v. time graph, the distance will be represented by:

In this case, however, we can just use simple geometry to evaluate the area under the graph v(t). I split it up into 2 trapezoids, and 1 rectangle. So, the area will be as follows:





So, the particle traveled a total of
1275m assuming it never turned back (because it says to calculate distance).
a(iii). Deceleration is a word for negative acceleration. Acceleration is the first derivative of velocity, and so deceleration is too. So, we just need to find the slope of the line that passes through t = 30 because it has a linear slope (meaning the slope doesn't change). So, we can just use simple algebra instead of calculus to figure this out. Recall from algebra that slope (m):

So, let's just pick values. I'm going to pick (25, 30) and (35, 15). Let's plug and chug:

Since it's a negative value, this means that acceleration is negative but deceleration is positive (because deceleration is negative acceleration). So, your answer is:
The deceleration of the particle at t = 30s is 3/2 or 1.5.