We just need to <u>plug in 7 for B! </u>
Let's do it !
2 + (10 - 7)
<em>We always start with the parenthesis first ! It's just a rule in math :) </em>
10 - 7 = ?
10 - 7 = 3
Now we have ...
2 + 3
2 + 3 = 5
Your final answer is 5!
I hope I helped :) Just comment if you have any questions and remember to vote me "brainliest" if you like my answer !!!
To answer your question, this could be the possible answer and i hope you understand and interpret it correctly:
<span>[Integrate [0, 1/2] xcos(pi*x
let u=x so that du=dx
and v=intgral cos (xpi)dx
v=(1/pi)sin(pi*x)
integration by parts
uv-itgral[0,1/2]vdu just plug ins
(1/pi)sinpi*x]-(1/pi)itgrlsin(pi*x)dx from 0 to 1/2
(1/pi)x sinpi*x - (1/pi)[-(1/pi) cos pi*x] from 0 to 1/2
=(1/2pi)+(1/pi^2)[cos pi*x/2-cos 0]
=1/2pi - 1/2pi^2
=(pi-2)/2pi^2 ans</span>
There is no hard and fast rule to select the class width. It largely depends on our application.However, one thing that should be kept in mind is that the number of classes should neither to be too low nor too high. So keeping this thing in mind, the class width is select.
The range of the data is = Maximum- Minimum = 96 - 11 = 85
10 classes will be most suited for this data.
The class width for each data can be calculated as:
Class Width = Range / Number of Classes = 85/10 = 8.5
Class width is always rounded to nearest next integer. So the class width will be 9 in this case.
So, the best value of class width or interval width for the given data will be 9.
