To solve this problem you must apply the proccedure shown below:
1. You have the following logarithm:
<span>log(2)n=4
2. Therefore, you con rewrite it as below:
loga(b)=logb/loa
</span>
3. Therefore, you have:
log(2)n=4⇒log(n)/log(2)=4
4. Then, you obtain:
log(n)=4log(2)
5. Therefore, as you can see, the answer for the exercise shown above is the last option, which is:
log(n)=4log(2)
Answer: The critical value for a two-tailed t-test = 2.056
The critical value for a one-tailed t-test = 1.706
Step-by-step explanation:
Given : Degree of freedom : df= 26
Significance level : 
Using student's t distribution table , the critical value for a two-tailed t-test will be :-
The critical value for a two-tailed t-test = 2.056
Again, Using student's t distribution table , the critical value for a one-tailed t-test will be :-
The critical value for a one-tailed t-test = 1.706
Answer:
40
Step-by-step explanation:
If we go by 20's (20% of 100) 100% is 5. Since were dealing with 120% in our case we'll be dividing by 6 <u>to figure out what every 20% is</u>. All we need to do it's divide 48 by 6 and we get the answer 8. So in our case every 20% is equal to 8. now to solve for 100% all we need to do is subtract 8 (20%) from 120% (48) and we find 100% is 40. We could also do 8 x 5. (<em>remember every 1 is 20% in our case</em>) which is also 40. Therefore, your answer is 40.
If you don't quite understand what I'm talking about please let me know and I'll elaborate.. thanks! have a nice day and good luck with your quiz!
Look at deonomators
assuming that the deonomenators are 5x+15y and 2x+6y
find their LCM
factor
5x+15y=5(x+3y)
2x+6y=2(x+3y)
LCM=10(x+3y)=10x+30y
multiply 2/(5x+15y) by 2/2=4/(10x+30y)
multiply 1/(2x+6y) by 5/5=5/(10x+30y)
if we add them
9/(10x+30y)
