Answer: 0.75
Step-by-step explanation:
Given : Interval for uniform distribution : [0 minute, 5 minutes]
The probability density function will be :-

The probability that a given class period runs between 50.75 and 51.25 minutes is given by :-
![P(x>1.25)=\int^{5}_{1.25}f(x)\ dx\\\\=(0.2)[x]^{5}_{1.25}\\\\=(0.2)(5-1.25)=0.75](https://tex.z-dn.net/?f=P%28x%3E1.25%29%3D%5Cint%5E%7B5%7D_%7B1.25%7Df%28x%29%5C%20dx%5C%5C%5C%5C%3D%280.2%29%5Bx%5D%5E%7B5%7D_%7B1.25%7D%5C%5C%5C%5C%3D%280.2%29%285-1.25%29%3D0.75)
Hence, the probability that a randomly selected passenger has a waiting time greater than 1.25 minutes = 0.75
Answer: we can get the sequence of reasons with help of below explanation.
Step-by-step explanation:
Here It is given that, line segment BD is the perpendicular bisector of line segment AC. ( shown in below diagram)
By joining points B with A and B with C ( Construction)
We get two triangles ABD and CBD.
We have to prove that : Δ ABD ≅ Δ CBD
Statement Reason
1. AD ≅ DC 1. By the property of segment bisector
2. ∠BDA ≅ ∠ BDC 2. Right angles
3. BD ≅ BD 3. Reflexive
4. Δ ABD ≅ Δ CBD 4. By SAS postulate of congruence
Six nights.
1. Saturday night
2. Sunday night
3. Monday night
4. Tuesday night
5. Wednesday night
6. Thursday night
Answer:
vol. of triangular prism = lbh/2
vol. of rectangular prism = 18 of vol. of triangular prism
18x = 18 + 14x
4x = 18
x=4.5 ft
so vol. of rectangular prism = 6 × 4 . 5 × 3 = 81 ftcube
Answer:
x = 8 and y = 10
Step-by-step explanation:
Since the figures are congruent:
Angle PQR = EFG
6y + x = 68
Segment QR = FG
2x - 4 = 12
Solving that:
2x - 4 = 12
2x = 16
x = 8
Replacing x in 6y + x =68
6y + 8 = 68
6y = 60
y = 10