Answer:
choice 1) -6 and choice 2) -5
Step-by-step explanation:
choice 1)
-(1/2)-6)+6 > 8
9 > 8
choice 2)
-(1/2)(-5)+6 > 8
8 1/2 > 8
Answer
∴ The true statement is Answer:
The true statement is BD ≅ CE ⇒ 3rd answer
Step-by-step explanation:
- There is a line contained points B , C , D , E
- All points are equal distance from each other
- That means the distance of BC equal the distance of CD and equal
the distance of DE
∴ BC = CD = DE
- That means the line id divided into 3 equal parts, each part is one
third the line
∴ BC = 1/3 BE
∴ CD = 1/3 BE
∴ DE = 1/3 BE
∵ BC = CD
∴ C is the mid-point of BD
∴ BC = 1/2 BD
∵ CD = DE
∴ D is the mid-point of DE
∴ CD = 1/2 CE
- Lets check the answers
* BD = one half BC is not true because BC = one half BD
* BC = one half BE is not true because BC = one third BE
* BD ≅ CE is true because
BD = BC + CD
CE = CD + DE
BC ≅ DE and CD is common
then BD ≅ CE
* BC ≅ BD is not true because BC is one half BD
∴ The true statement is BD ≅ CE
plz mark me brainliest
Answer:
Step-by-step explanation:
If the (X) in this means times then
1/15
56% can be represented with 56/100. That fraction can be reduced (or simplified) to 14/25.
Answer:
51/4
Step-by-step explanation:
To begin with you have to understand what is the distribution of the random variable. If X represents the point where the bus breaks down. That is correct.
X~ Uniform(0,100)
Then the probability mass function is given as follows.
Now, imagine that the D represents the distance from the break down point to the nearest station. Think about this, the first service station is 20 meters away from city A, and the second station is located 70 meters away from city A then the mid point between 20 and 70 is (70+20)/2 = 45 then we can represent D as follows
Now, as we said before X represents the random variable where the bus breaks down, then we form a new random variable 
, 
is a random variable as well, remember that there is a theorem that says that
![E[Y] = E[D(X)] = \int\limits_{-\infty}^{\infty} D(x) f(x) \,\, dx](https://tex.z-dn.net/?f=E%5BY%5D%20%3D%20E%5BD%28X%29%5D%20%3D%20%5Cint%5Climits_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%20D%28x%29%20f%28x%29%20%5C%2C%5C%2C%20dx)
Where 
is the probability mass function of X. Using the information of our problem
![E[Y] = \int\limits_{-\infty}^{\infty} D(x)f(x) dx \\= \frac{1}{100} \bigg[ \int\limits_{0}^{20} x dx +\int\limits_{20}^{45} (x-20) dx +\int\limits_{45}^{70} (70-x) dx +\int\limits_{70}^{100} (x-70) dx \bigg]\\= \frac{51}{4} = 12.75](https://tex.z-dn.net/?f=E%5BY%5D%20%3D%20%5Cint%5Climits_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%20%20D%28x%29f%28x%29%20dx%20%5C%5C%3D%20%5Cfrac%7B1%7D%7B100%7D%20%5Cbigg%5B%20%5Cint%5Climits_%7B0%7D%5E%7B20%7D%20x%20dx%20%2B%5Cint%5Climits_%7B20%7D%5E%7B45%7D%20%28x-20%29%20dx%20%2B%5Cint%5Climits_%7B45%7D%5E%7B70%7D%20%2870-x%29%20dx%20%2B%5Cint%5Climits_%7B70%7D%5E%7B100%7D%20%28x-70%29%20dx%20%20%5Cbigg%5D%5C%5C%3D%20%5Cfrac%7B51%7D%7B4%7D%20%3D%2012.75)
