Answer:14708
Step-by-step explanation:Exponential Functions:
y=abxy=ab^x this is not right not correct
y=ab
x
a=starting value = 13000a=\text{starting value = }13000
a=starting value = 13000
r=rate = 2.5%=0.025r=\text{rate = }2.5\% = 0.025
r=rate = 2.5%=0.025
Exponential Growth:\text{Exponential Growth:}
Exponential Growth:
b=1+r=1+0.025=1.025b=1+r=1+0.025=1.025
b=1+r=1+0.025=1.025
Write Exponential Function:
y=13000(1.025)xy=13000(1.025)^x
y=13000(1.025)
x
Put it all together
Plug in time for x:\text{Plug in time for x:}
Plug in time for x:
y=13000(1.025)5y=13000(1.025)^{5}
y=13000(1.025)
5
y=14708.30677y= 14708.30677
y=14708.30677
Evaluate
y≈14708y\approx 14708
y≈14708
Answer:
C. 
Step-by-step explanation:
The radius of the circle is 24 units, then the area of the whole circle is

The shaded circle is limited by 45° angle, so its area is
of the area of the whole circle.
The area of the shaded circle is
Average of 5 games = 13
Total of 5 games = 13 x 5 = 65
Average of 6 games = 17
Total of 6 games = 102
6th game = 102 - 65 = 37
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Answer: Her score for the 6th game as 37.
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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)