Answer:
Multiples [10] - 10, 20, 30, 40, 50
Multiples [11] - 11, 22, 33, 44, and 55
<u><em>Find the Even numbers of these multiples</em></u>
Multiples [10]
- - 10 = Even
- - 20 = Even
- - 30 = Not Even
- - 40 = Even
- - 50 = Not Even
- <u><em>10 × 1 = 10</em></u>
- <u><em>10 × 2 = 20</em></u>
- <u><em>10 × 3 = 30</em></u>
- <u><em>10 × 4 = 40</em></u>
- <u><em>10 × 5 = 50</em></u>
Multiples [11]
- - 11 = Not Even
- - 22 = Even
- - 33 = Not Even
- - 44 = Even
- - 55 = Not Even
- <u><em> 11 × 1 = 11</em></u>
- <u><em> 11 × 2 = 22</em></u>
- <u><em> 11 × 3 = 33</em></u>
- <u><em> 11 × 4 = 44</em></u>
- <u><em> 11 × 5 = 55</em></u>
- <u><em> 11 × 6 = 66</em></u>
- <u><em> 11 × 7 = 77</em></u>
- <u><em> 11 × 8 = 88</em></u>
- <u><em> 11 × 9 = 99</em></u>
Step-by-step explanation:
.... I messed up, sorry. I thought that 20 said 11
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:
![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) 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)
Answer:
Type 1 error:
D)Reject the null hypothesis that the percentage of high school students who graduate is equal to 55 % when that percentage is actually equal to 55 %.
Type 2 error:
B,)Fail to reject the null hypothesis that the percentage of high school students who graduate is equal to 55 % when that percentage is actually less than 55 %.
Step-by-step explanation:
When something that is true, is been rejected, then it's reffered to as Type I, on the other hand when
Whensomething that is false is been failed to be rejected then it's reffered to as that is Type II
Type I ;
This is to reject Hypothesis when Hypothesis is true, i.e rejecting of the null when it's true.
For instance from the question,
The percentage of high school students who graduate is equal to 55%. Then to get the Type1 , One would say the percentage of high students who graduated is not 55% when it is in actual sense
Type II ;
This is happen when we accept a false null hypothesis which means it takes place when We fail to reject Hypothesis when it is False.
For instance, from the question which says The percentage of high school students who graduate is equal to 55%. Then for type II to occur One would say it is 55% when it is really not 55%.
When x=1, 7x+3 = 10.
When x=2, 7x+3 = 17.
When the value of 'x' changes from '1' to '2', the value of 7x+3 changes
from '10' to '17' . . . a distance of '7' .
That's what the ' 7x ' in the equation means, and that's what we mean
when we say that the 'slope' of the equation is '7'. 'y' changes '7' for
every '1' that 'x' changes.