Answer:
Step-by-step explanation:
Given a triangle has points:
(-5,1),(2,1), (2,-1)
Let us label the points:
A(2,1),
B(-5,1) and
C(2,-1)
To find:
Distance between (−5, 1) and (2, −1) i.e. BC.
Horizontal leg AB and
Vertical leg, AC.
Solution:
Please refer to the attached diagram for the labeling of the points on xy coordinate plane.
We can simply use Distance formula here, to find the distance between two coordinates.
Distance formula
:
For BC:
Horizontal leg, AC:
Vertical Leg, AB:
Answer:
V(x)=(x+9) * (x-4) * (x+6)
Step-by-step explanation:
A fish tank is normally in the shape of a rectangular prism. The volume of a rectangular prism can be calculated using the following formula
V = w * h * l
where w represents the width, h represents the heigh, l represents the length, and V represents the volume of the rectangular prism/fish tank. Therefore, we can use the function provided in the question and simply add 3 units to the length and 2 units to the width in order for it to work for our new fish tank.
V(x)=(x+9) * (x-4) * (x+6)
We have to determine which value is equivalent to | f ( i ) | if the function is: f ( x ) = 1 - x. We know that for the complex number: z = a + b i , the absolute value is: | z | = sqrt( a^2 + b^2 ). In this case: | f ( i )| = | 1 - i |. So: a = 1, b = - 1. | f ( i ) | = sqrt ( 1^2 + ( - 1 )^2) = sqrt ( 1 + 1 ) = sqrt ( 2 ). Answer: <span>C. sqrt( 2 )</span>
Answer:
$26.91
Step-by-step explanation:
11.96 divided 4 = 2.99
2.99 x 9 = $26.91
(a) 0.059582148 probability of exactly 3 defective out of 20
(b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.
(a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So
0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)
= 0.05^3 * 0.95^17 * 20! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)
= 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)
= 0.05^3 * 0.95^17 * 20*19*3
= 0.000125* 0.418120335 * 1140
= 0.059582148
(b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.
The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)
= 0.05*0.95^3*24/(1!3!)
= 0.05*0.857375*24/6
= 0.171475
So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.
