solution;
a) From the given information,
N = 53,min=42 and max = 129
K = 1+3.322log₁₀53
= 1+3.322(1.724)
= 1+5.728
= 6.728
Number of classes is 7.
b) find the width of the class interval first
width = max – min /classes
= 129-42/7
= 87/7 =12.42
Take the next largest number 13 as class width. Since the minimum value is 42
The lower limit iof the first class could be 40.
So,
The upper limit of the last class is,
40+(13 x 7) = 131
As the upper limit of the last class covers the entire range of values,
It is better to take the lower limit of the first class as 40
The lower limit of the first calss could be 40.
The answer is 12 1/4. how I got my answer.
Their are 4 quarters to a whole, so that means all I have to do is multiply 4×3=12, then add 1/4+12= 12 1/4.
I hope this helps!
Sure. So the hundreds digit is after the tens digit which is 3, so therefore it's 4. The millions digit is 7 places to the left, so it's 7. Ten thousands digits are 5 places to the left, so it's 9. The hundred thousands is 6 to the left, therefore 8. Finally, the thousands is 4 places to the left, so it's 0. Hope this helped!
Oops for Jerry. Don't let this happen to you!
An overdrawn account by $15 is a negative $15. When you add the $10 service charge, you now have negative $25. When you add your $60, you will only have the difference between $25 and $60 left in the account.
It looks like this:
-$15 + $60 = ?
60 - 15 = 45 dollars in the account.
Good news for Jerry. My bank charges $35 service charge to be overdrawn. (Don't ask me how I know this...)