Answer:
HELLOOOO
Step-by-step explanation:
Sorry I can't help but I hope you got it
False. That does not satify the equation
Answer:
Range = 13
Mean = 8.4
Variance= 21.24
Standard deviation= 4.61
Step-by-step explanation:
2, 10, 15, 3, 13, 9, 14, 7, 2, 9
For the range
Let the set of data be arranged inn ascending order
Range= higehest value- lowest value
Range = 15-2
Range= 13
For the mean
Mean = (2+2+3+7+9+9+10+13+14+15)/10
Mean = 84/10
Mean = 8.4
For variance
Variance=((2-8.4)²+(2-8.4)²+(3-8.4)²+(7-8.4)²+(9-8.4)²+(9-8.4)²+(10-8.4)²+(13-8.4)²+(14-8.4)²+(15-8.4)²)/10
Variance= (40.96+40.96+29.16+1.96+0.36+0.36+2.56+21.16+31.36+43.56)/10
Variance= 212.4/10
Variance= 21.24
Standard deviation= √variance
Standard deviation= √21.24
Standard deviation= 4.609
Approximately = 4.61
Answer:
Your answer would be:
Since 6 is in the tenths places, the value of the digit is 0.6
Hope this helps Buddy!
-Courtney
Answer:
a) ![f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000](https://tex.z-dn.net/?f=f%28t%29%3D0.001155%5B%5Cfrac%7B2%7D%7B3%7Dt%28t-1980%29%5E%7B3%2F2%7D-%5Cfrac%7B4%7D%7B15%7D%28t-1980%29%5E%7B5%2F2%7D%5D%2B264%2C034%2C000)
b) f(t=2015) = 264,034,317.7
Step-by-step explanation:
The rate of change in the number of hospital outpatient visits, in millions, is given by:
a) To find the function f(t) you integrate f(t):
![\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt](https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7Bdf%28t%29%7D%7Bdt%7Ddt%3Df%28t%29%3D%5Cint%20%5B0.001155t%28t-1980%29%5E%7B0.5%7D%5Ddt)
To solve the integral you use:
Next, you replace in the integral:
Then, the function f(t) is:
![f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'](https://tex.z-dn.net/?f=f%28t%29%3D0.001155%5B%5Cfrac%7B2%7D%7B3%7Dt%28t-1980%29%5E%7B3%2F2%7D-%5Cfrac%7B4%7D%7B15%7D%28t-1980%29%5E%7B5%2F2%7D%5D%2BC%27)
The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient visits.
Hence C' = 264,034,000
The function is:
b) For t = 2015 you have:
![f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7](https://tex.z-dn.net/?f=f%28t%3D2015%29%3D0.001155%5B%5Cfrac%7B2%7D%7B3%7D%282015%29%282015-1980%29%5E%7B1%2F2%7D-%5Cfrac%7B4%7D%7B15%7D%282015-1980%29%5E%7B5%2F2%7D%5D%2B264%2C034%2C000%5C%5C%5C%5Cf%28t%3D2015%29%3D264%2C034%2C317.7)
