When it is perpendicular, the slope is the negative reciprocal so it would equal -3/2x. Plug in the points to the equation so you can find the intercept. -8=-3/2(4)+b
Combine: -8=-6+b
Add 6: -2=b
The equation would be f(x)=-3/2x-2
Hope this helped and sorry if I am wrong!
Answer: 4
Step-by-step explanation:
Given the following :
P = probability of success = 0.5
n = number of trials = 8
The expected value of a binomial distribution with probability of success P and number of trials n is defined by:
E(n, p) = n * p
Therefore, expected value when P = 0.5 and n = 8
E(8, 0.5) = 8 × 0.5
= 4
The expected value of the binomial distribution is 4
Answer:
Is that all to the problem?
Step-by-step explanation:
Answer:
Step-by-step explanation:
If the diagonals of the rectangle are congruent,
AC = BD
By using formula to calculate the distance between two points,
Distance between two points = 
Distance between A(0, -3) and C(2, 8),
AC = 
= 
= 
Similarly, distance between two points B(-4, 0) and D(6, 5),
BD = 
= 
= 
Therefore, both the diagonals are congruent.
Hence, given quadrilateral ABCD is a rectangle.
What two cars, I don’t see anything