Answer:
It's form. Linear is a normal line, exponential is a curvy line, and quadratic is a u shape with the vertice on the y-axis line, making half of the u on the left quadrant and the other half on the right quadrant
Step-by-step explanation:
first is linear, second is exponential and third is quadratic
Answer:
A continuous probability distribution having a rectangular shape, where the probability is evenly distributed over an interval of numbers is a(n) __uniform__________ distribution
Step-by-step explanation:
Given that there is a continuous probability distribution having a rectangular shape, where the probability is evenly distributed over an interval of numbers
Since the pdf is rectangular in shape and total probability is one we can say all values in the interval would be equally likely
Say if the interval is (a,b) P(X) = p the same for all places
Since total probability is 1,
we get integral of P(X)=p(b-a) =1
Or p=
this is nothing but a uniform distribution continuous defined in the interval
A continuous probability distribution having a rectangular shape, where the probability is evenly distributed over an interval of numbers is a(n) __uniform__________ distribution
By angle addition postulate,
m(∠ABD) + m(∠DBC) = m(∠ABC)
Now substitute the values of each angle,
(5x + 5)° + (3x + 12)° = 153°
Combine like terms of the expression,
(5x + 3x) + (5 + 12) = 153
8x + 17 = 153
8x = 153 - 17
x =
x = 17
Substitute the value of 'x' to get the measure of each angle,
m(∠ABD) = (5x + 5)
= 5(17) + 5
= 90°
m(∠DBC) = (3x + 12)
= 3(17) + 12
= 63°
Learn more,
Answer:
y = -4x² + 32x - 48
Step-by-step explanation:
The standard form of a quadratic equation is
y = ax² + bx + c
We must find the equation that passes through the points:
(2, 0), (6,0), and (3, 12)
We can substitute these values and get three equations in three unknowns.
0 = a(2²) + b(2) + c
0 = a(6²) + b(6) + c
12 = a(3²) + b(3) + c
We can simplify these to get the system of equations:
(1) 0 = 4a + 2b + c
(2) 0 = 36a + 6b + c
(3) 12 = 9a + 3b + c
Eliminate c from equations (1) and (2). Subtract (1) from (2).
(4) 0 = 32a + 4b
Eliminate c from equations (2) and (3). Subtract (3) from (2).
(5) -12 = 27a - 3b
Simplify equations (4) and (5).
(6) 0 = 8a + b
(7) -4 = 9a - b
Eliminate b by adding equations (6) and (7).
(8) a = -4
Substitute (4) into (6).
0 = -32 + b
(9) b = 32
Substitute a and b into (1)
0 = 4(-4) + 2(32) + c
0 = -16 + 64 + c
0 = 48 + c
c = -48
The coefficients are
a= -4, b = 32, c = -48
The quadratic equation is
y = -4x² + 32x - 48
The diagram below shows the graph of your quadratic equation and the three points through which it passes.
