Using a graphing tool
Let's graph each of the cases to determine the solution of the problem
<u>case A)</u> 
see the attached figure N 
The range is the interval--------> (0,∞)
therefore
the function is not the solution
<u>case B)</u> 
see the attached figure N 
The range is the interval--------> (0,∞)
therefore
the function is not the solution
<u>case C)</u> 
see the attached figure N 
The range is the interval--------> (-∞,3)
therefore
the function is the solution
<u>case D)</u> 
see the attached figure N 
The range is the interval--------> (-3,∞)
therefore
the function is not the solution
<u>the answer is</u>
<span>2/15 if drawn without replacement.
1/9 if drawn with replacement.
Assuming that the chips are drawn without replacement, there are 6 * 5 different possibilities. And that's a low enough number to exhaustively enumerate them. So they are:
1,2 : 1,3 : 1,4 : 1,5 : 1,6
2,1 : 2,3 : 2,4 : 2,5 : 2,6
3,1 : 3,2 : 3.4 : 3,5 : 3,6
4,1 : 4,2 : 4.3 : 4,5 : 4,6
5,1 : 5,2 : 5.3 : 5,4 : 5,6
6,1 : 6,2 : 6.3 : 6,4 : 6,5
Of the above 30 possible draws, there are 4 that add up to 5. So the probability is 4/30 = 2/15
If the draw is done with replacement, then there are 36 possible draws. Once again, small enough to exhaustively list, they are:
1,1 : 1,2 : 1,3 : 1,4 : 1,5 : 1,6
2,1 : 2,2 : 2,3 : 2,4 : 2,5 : 2,6
3,1 : 3,2 : 3,3 : 3.4 : 3,5 : 3,6
4,1 : 4,2 : 4.3 : 4,4 : 4,5 : 4,6
5,1 : 5,2 : 5.3 : 5,4 : 5,5 : 5,6
6,1 : 6,2 : 6.3 : 6,4 : 6,5 : 6,6
And of the above 36 possibilities, exactly 4 add up to 5. So you have 4/36 = 1/9</span>
Answer:
Step-by-step explanation:
given is the Differential equation in I order linear as
Take Laplace on both sides
![L(y') +4L(y) = 48L(t)\\sY(s)-y(0) +4Y(s) = 48 *\frac{1}{s^2} \\Y(s) [s+4]=\frac{48}{s^2}+9\\Y(s) = \frac{1}{s^2(s+4)}+\frac{9}{s+4}](https://tex.z-dn.net/?f=L%28y%27%29%20%2B4L%28y%29%20%3D%2048L%28t%29%5C%5CsY%28s%29-y%280%29%20%2B4Y%28s%29%20%3D%2048%20%2A%5Cfrac%7B1%7D%7Bs%5E2%7D%20%5C%5CY%28s%29%20%5Bs%2B4%5D%3D%5Cfrac%7B48%7D%7Bs%5E2%7D%2B9%5C%5CY%28s%29%20%3D%20%5Cfrac%7B1%7D%7Bs%5E2%28s%2B4%29%7D%2B%5Cfrac%7B9%7D%7Bs%2B4%7D)
Now if we take inverse we get y(t) the solution
Thus the algebraic equation would be
Answer:
True, True, False, True, False, True
Step-by-step explanation:
<u>CDs have a higher mean than digital</u>
<u />
Let's check. CD mean is (1000 + 800 + 800 + 600 + 400 + 200)/6 = 633.33
Digital mean: (100 + 300 + 300 + 500 + 700 + 900)/6 = 466.67
CD's mean is higher. Since you are dividing byt he same number you also could have just added the amount and found which was a larger number. But yes, this is true.
<u>The range of digital is 800</u>
<u />
The range is the highest number inus the lowest number. For digital the highest is 900 and lowest is 100 so the ange is 900-100 = 800. So this is true.
<u>The median of CDs is 400.
</u>
<u />
The median is the middle value for odd numbers of values, or the average of the two middle values. 6 total values means you have to take the third and fourth and average them. in CDs the middle values are 800 and 600, the average is 700, so that is what the median is. this is false.
<u>Both have the same interquartile range.
</u>
<u />
Interquartile range is to find the middle number or numbers like in median, then take the two parts it is split into and find the "median" of those. Then subtract the larger one fromt he smaller one.
IQR of CD the first half is 200, 400, 600 so the middle here is 400, second half has a middle number of 800 so IQR here is 800-400=400
IQR of digital is 700-300 = 400 so yes both are the same.
<u>Both have the same median</u>
<u />
We know the medan of CDs is 700 then findign the median of digital is (300+500)/2 = 400. So no, the medians are not the same.
<u>Digital’s mean is around 467.
</u>
<u />
We founf the mean for digital to be 466.67 which rounds up to 467, So I would say it is true.
