Answer:
1
Step-by-step explanation:
3(1)+5=8
3(0)+5=5
3(2)+5=11
3(3)+5=14
3(4)+5=17
.....
Need=51
S=12.75
G=6.5/hr•2hr=13
W=5.25/hr•5hr=26.25
S+g+w=12.75+13+26.25=52
Yes because total will be 52
......
5x+3≥23
5x≥20
x≥4
[4,8,12]
.......
9w
.....
38.88+1.87x
.....
9x-3
....
x=56
...
69=18+19+t
69=37+t
32=t
...
77-20.1-10.39=s
...
2
....
11 doesn't have an equation to solve.
...
12 doesn't have a solution to solve.
....
n≤7
...
32≤14+m
18≤m
....
120≤20+m
100≤m
...
d=50t
...
Y=5.5x
....
Y=4x
...
Dependent variable is distance.
Answer:
B
Step-by-step explanation:
Answer:
![P(35-2 < X 35+2) = P(33< X< 37)= P(X](https://tex.z-dn.net/?f=%20P%2835-2%20%3C%20X%2035%2B2%29%20%3D%20P%2833%3C%20X%3C%2037%29%3D%20P%28X%3C37%29%20-P%28X%3C32%29)
And using the cumulative distribution function we got:
![P(35-2 < X 35+2) = P(33< X< 37)= P(X](https://tex.z-dn.net/?f=%20P%2835-2%20%3C%20X%2035%2B2%29%20%3D%20P%2833%3C%20X%3C%2037%29%3D%20P%28X%3C37%29%20-P%28X%3C32%29%20%3D%20%5Cfrac%7B37-20%7D%7B50-20%7D%20-%5Cfrac%7B33-20%7D%7B50-20%7D%20%3D0.567-0.433%3D0.134%20)
The probability that preparation is within 2 minutes of the mean time is 0.134
Step-by-step explanation:
For this case we define the following random variable X= (minutes) for a lab assistant to prepare the equipment for a certain experiment , and the distribution for X is given by:
![X \sim Unif (a= 20, b =50)](https://tex.z-dn.net/?f=%20X%20%5Csim%20Unif%20%28a%3D%2020%2C%20b%20%3D50%29)
The cumulative distribution function is given by:
![F(x) = \frac{x-a}{b-a} , a \leq X \leq b](https://tex.z-dn.net/?f=%20F%28x%29%20%3D%20%5Cfrac%7Bx-a%7D%7Bb-a%7D%20%2C%20a%20%5Cleq%20X%20%5Cleq%20b)
The expected value is given by:
![E(X) = \frac{a+b}{2} = \frac{20+50}{2}=35](https://tex.z-dn.net/?f=%20E%28X%29%20%3D%20%5Cfrac%7Ba%2Bb%7D%7B2%7D%20%3D%20%5Cfrac%7B20%2B50%7D%7B2%7D%3D35)
And we want to find the following probability:
![P(35-2 < X 35+2) = P(33< X< 37)](https://tex.z-dn.net/?f=%20P%2835-2%20%3C%20X%2035%2B2%29%20%3D%20P%2833%3C%20X%3C%2037%29)
And we can find this probability on this way:
![P(35-2 < X 35+2) = P(33< X< 37)= P(X](https://tex.z-dn.net/?f=%20P%2835-2%20%3C%20X%2035%2B2%29%20%3D%20P%2833%3C%20X%3C%2037%29%3D%20P%28X%3C37%29%20-P%28X%3C32%29)
And using the cumulative distribution function we got:
![P(35-2 < X 35+2) = P(33< X< 37)= P(X](https://tex.z-dn.net/?f=%20P%2835-2%20%3C%20X%2035%2B2%29%20%3D%20P%2833%3C%20X%3C%2037%29%3D%20P%28X%3C37%29%20-P%28X%3C32%29%20%3D%20%5Cfrac%7B37-20%7D%7B50-20%7D%20-%5Cfrac%7B33-20%7D%7B50-20%7D%20%3D0.567-0.433%3D0.134%20)
The probability that preparation is within 2 minutes of the mean time is 0.134
F(-6) = 2(-6)+15(-6)-2 = -12-90-2 = -104
Answer:
Option B is correct.
Reason:It is an argument
Step-by-step explanation:
Given:
and ![BC \cong EF](https://tex.z-dn.net/?f=BC%20%5Ccong%20EF)
By transitive property: a = b and b = c then, a =c
⇒![AB \cong EF](https://tex.z-dn.net/?f=AB%20%5Ccong%20EF)
AB = EF [def of
segment]
By Segment Addition Postulate states that given two points A and C, and a third point B lies on the line segment AC if and only if the distances between the points satisfy the equation
AB + BC = AC.
Since, the line segment EG , F lies on the line segment;
then, by segment addition postulates we have;
EG = EF + FG
By substitution AB=EF
⇒ EG = AB + FG hence proved!
Since, in the fourth statement reason is: It is an argument