Answer:
What do you need help on?
Step-by-step explanation:
The polygons are similar.
This is because dividing the corresponding sides forms the same ratio, as shown by the three equations below
35/28 = 1.25
25/20 = 1.25
(15.5)/(12.4) = 1.25
So the larger figure on the right has side lengths that are 1.25 times larger compared to the corresponding sides of the figure on the left.
You'll need to flip the figure on the left so that the side labeled "20" is along the top, and the "28" is along the bottom.
After this flip happens, also note that the angle arc markings match up. The bottom pairs of angles of each figure are shown with a single arc, while the top angles are shown as double arcs. This helps visually show which angles pair up and are congruent to one another.
Because we have similar proportions as discussed earlier, and congruent pairs of angles like this, this shows the two figures are similar quadrilaterals. The one on the right is simply an enlarged scaled up copy of the figure on the left.
Answer:
Step-by-step explanation:
From the given information:
the mean ![(\mu) = 115 \times 20](https://tex.z-dn.net/?f=%28%5Cmu%29%20%3D%20115%20%5Ctimes%2020)
= 2300
Standard deviation = ![20 \times \sqrt{115}](https://tex.z-dn.net/?f=20%20%5Ctimes%20%5Csqrt%7B115%7D)
Standard deviation (SD) = 214.4761
TO find:
a) ![P(x > 3500)= P(Z > \dfrac{3500-\mu}{214.4761})](https://tex.z-dn.net/?f=P%28x%20%3E%203500%29%3D%20P%28Z%20%3E%20%5Cdfrac%7B3500-%5Cmu%7D%7B214.4761%7D%29)
![P(x > 3500)= P(Z > \dfrac{3500-2300}{214.4761})](https://tex.z-dn.net/?f=P%28x%20%3E%203500%29%3D%20P%28Z%20%3E%20%5Cdfrac%7B3500-2300%7D%7B214.4761%7D%29)
![P(x > 3500)= P(Z > \dfrac{1200}{214.4761})](https://tex.z-dn.net/?f=P%28x%20%3E%203500%29%3D%20P%28Z%20%3E%20%5Cdfrac%7B1200%7D%7B214.4761%7D%29)
![P(x > 3500)= P(Z >5.595)](https://tex.z-dn.net/?f=P%28x%20%3E%203500%29%3D%20P%28Z%20%3E5.595%29)
From the Z-table, since 5.595 is > 3.999
![P(x > 3500)=1-0.9999](https://tex.z-dn.net/?f=P%28x%20%3E%203500%29%3D1-0.9999)
P(x > 3500) = 0.0001
b)
Here, the replacement time for the mean ![(\mu) = \dfrac{0+0.5}{2}](https://tex.z-dn.net/?f=%28%5Cmu%29%20%3D%20%5Cdfrac%7B0%2B0.5%7D%7B2%7D)
= 0.25
Replacement time for the Standard deviation ![\sigma = \dfrac{0.5-0}{\sqrt{12}}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cdfrac%7B0.5-0%7D%7B%5Csqrt%7B12%7D%7D)
![\sigma = 0.1443](https://tex.z-dn.net/?f=%5Csigma%20%3D%200.1443)
For 115 component, the mean time = (115 × 20)+(114×0.25)
= 2300 + 28.5
= 2328.5
Standard deviation = ![\sqrt{(115\times 20^2) +(114\times (0.1443)^2)}](https://tex.z-dn.net/?f=%5Csqrt%7B%28115%5Ctimes%2020%5E2%29%20%2B%28114%5Ctimes%20%280.1443%29%5E2%29%7D)
= ![\sqrt{(115\times 400) +(114\times 0.02082249}](https://tex.z-dn.net/?f=%5Csqrt%7B%28115%5Ctimes%20400%29%20%2B%28114%5Ctimes%200.02082249%7D)
= ![\sqrt{(46000) +2.37376386}](https://tex.z-dn.net/?f=%5Csqrt%7B%2846000%29%20%2B2.37376386%7D)
= ![\sqrt{(46000) +(2.37376386)}](https://tex.z-dn.net/?f=%5Csqrt%7B%2846000%29%20%2B%282.37376386%29%7D)
= ![\sqrt{46002.374}](https://tex.z-dn.net/?f=%5Csqrt%7B46002.374%7D)
= 214.482
Now; the required probability:
![P(x > 4125) = P(Z > \dfrac{4125- 2328.5}{214.482})](https://tex.z-dn.net/?f=P%28x%20%3E%204125%29%20%3D%20P%28Z%20%3E%20%5Cdfrac%7B4125-%202328.5%7D%7B214.482%7D%29)
![P(x > 4125) = P(Z > \dfrac{1796.5}{214.482})](https://tex.z-dn.net/?f=P%28x%20%3E%204125%29%20%3D%20P%28Z%20%3E%20%5Cdfrac%7B1796.5%7D%7B214.482%7D%29)
![P(x > 4125) = P(Z >8.376)](https://tex.z-dn.net/?f=P%28x%20%3E%204125%29%20%3D%20P%28Z%20%3E8.376%29)
![P(x > 4125) =1- P(Z](https://tex.z-dn.net/?f=P%28x%20%3E%204125%29%20%3D1-%20%20P%28Z%20%3C8.376%29)
From the Z-table, since 8.376 is > 3.999
P(x > 4125) = 1 - 0.9999
P(x > 4125) = 0.0001
Mean is when you take all the numbers and add them together and divide by the amount of numbers there
Example: Find the mean of :3,4,6,7
So you a 3+4+6+7=20 divide that by 4(because there you added 4 numbers) and that equals 5