Solution:- Given numbers to compare are 512 and 521 .
As they are 3 digit numbers
So, we have to compare hundreds place.
But hundreds are equal in both the numbers with digit 5.
Next we have to compare tens place.
Case 1 :- 1 ten is smaller in 512 than 2 tens in 521 .
So we get the result that,
512 is smaller than 521
or <em> 512 < 521</em>
Case 2:-2 tens is greater in 521 than 1 ten in 512 .
So we get the result that,
521 is greater than 512
or <em> 521 > 512</em>
10 • 4 = _ • 5
10•4=40
40/5=8
Answer:
1. 
2. 
3. 
Step-by-step explanation:
Given information:


(1)
We need to find the value of P(s₁|I).





Therefore the value of P(s₁|I) is
.
(2)
We need to find the value of P(s₂|I).





Therefore the value of P(s₂|I) is
.
(3)
We need to find the value of P(s₃|I).





Therefore the value of P(s₃|I) is
.
Answer:
I need more information to answer this.
To best emphasize the number of defects. Manager should use graph 3 (refer the image shown):
If we talk about graph 1, it can also be used but usually we put the time line on the horizontal axis, for the convenience and the quantity to be measured on the y-axis. In the graph 1, the time is placed on the vertical axis (x-axis) so it would not be a good pick for the manager.
Same is the case with graph 2 again we have time on the vertical axis. So it is not a good idea to with graph 2.
Graph 3 could be the best to emphasize the number of defects because first of all time is placed on the horizontal axis and the quantity to be shown is on the vertical axis. Secondly, the range of the vertical axis is less so it is easy to observe the data set on the graph quite distinctively. Therefore, graph 3 is the best pick.
Graph 4 is placed correctly in terms of vertical and horizontal axes but the range of vertical axis is quite high due to which the dispersion or the display of the data is quite compressed and it gets hard to visualize.