Answer:
Here is the graph I made, starting from (-3,1) to (-2,-1) it goes down 2 units and over 1 unit.
I realize that the equation is different, but do it in the same way, it'll help! It is given that the water taxi's path can be modeled by the equation y =0.5(x - 14)^2. Therefore, this is one of the equations in this system. Find a linear equation that will model the path of the water skier, which begins at the point (6,6) and ends at the point (8,-4). The slope is (-5). Use the slope and one point on the line to find the y-intercept of the line. The y-intercept of the line that passes through the points (6,6) and (8,-4) is (0,36). Thus, the equation is y=-5x+36. Now, to determine if it is possible for the water skier to collide with the taxi, we have to determine if there is a solution to the system of equations. To determine if there is a solution to the system of equations, solve the system using substitution. First, write the equation that models the water taxi's path in standard form. y=0.5(x - 14)^2-->0.5x^2-14x+98. Use substitution. Substitute for y in the equation and then solve for x. As the expression on the left side of the equation cannot easily be factored, use the Quadratic Formula to solve for x. Do x=-b(plusorminus)sqrrtb^2-4ac/2a. Identify a, b, and c. a=0.5, b=-9, and c=62. Substitute into the Quadratic Formula. If there is a negative number under the radical, there are NO solutions. Thus, the path of the water skier will never cross the path of the taxi.
In conclusion: It is not possible that the water skier could collide with the taxi as the two paths never cross.
Answer:
The expected number of appearance defects in a new car = 0.64
standard deviation = 0.93(approximately)
Step-by-step explanation:
Given -
7% had exactly three, 11% exactly two, and 21% only one defect in cars.
Let X be no of defects in a car
P(X=3) = .07 , P(X=2) = .11 , P(X=3) = .21
P(X=0 ) = 1 - .07 - .11 - .21 = 0.61
The expected number of appearance defects in a new car =
E(X) = 
= 
= 0.64
= variation of E(X) = 
= 
= 
= 0.8704
standard deviation = 
= 
= 0.93(approximately)
