Answer:
2
Step-by-step explanation:
4x+3=x+9
4x-x=9-3
3x=6
x=6/3
x=2
Answer:
Step-by-step explanation:
Given that 33.4% of people have sleepwalked.
Sample size n =1459
Sample favourable persons = 526
Sample proportion p = 
Sample proportion p is normal for large samples with mean = 0.334 and
std error = 
a) P(526 or more of the 1459 adults have sleepwalked.)
b) Yes, because hardly 1.4% is the probability
c) 33.4 is very less compared to the average. Either sample should be improved representing the population or population mean should be increased accordingly.
Well i have always gone by pemdas which is() exponits mulyiply or divide which ever comes first add subtract which ever comes first also
So take 13, 8 and 4 and add those together to get the budget she needs for the month.
13+8+4=25
so she needs $25 in all, to find how much more than 16 she needs you take 25 and minus 16 from it.
25-16=19
so she needs $19 more to have a balanced budget.
Wow !
OK. The line-up on the bench has two "zones" ...
-- One zone, consisting of exactly two people, the teacher and the difficult student.
Their identities don't change, and their arrangement doesn't change.
-- The other zone, consisting of the other 9 students.
They can line up in any possible way.
How many ways can you line up 9 students ?
The first one can be any one of 9. For each of these . . .
The second one can be any one of the remaining 8. For each of these . . .
The third one can be any one of the remaining 7. For each of these . . .
The fourth one can be any one of the remaining 6. For each of these . . .
The fifth one can be any one of the remaining 5. For each of these . . .
The sixth one can be any one of the remaining 4. For each of these . . .
The seventh one can be any one of the remaining 3. For each of these . . .
The eighth one can be either of the remaining 2. For each of these . . .
The ninth one must be the only one remaining student.
The total number of possible line-ups is
(9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) = 9! = 362,880 .
But wait ! We're not done yet !
For each possible line-up, the teacher and the difficult student can sit
-- On the left end,
-- Between the 1st and 2nd students in the lineup,
-- Between the 2nd and 3rd students in the lineup,
-- Between the 3rd and 4th students in the lineup,
-- Between the 4th and 5th students in the lineup,
-- Between the 5th and 6th students in the lineup,
-- Between the 6th and 7th students in the lineup,
-- Between the 7th and 8th students in the lineup,
-- Between the 8th and 9th students in the lineup,
-- On the right end.
That's 10 different places to put the teacher and the difficult student,
in EACH possible line-up of the other 9 .
So the total total number of ways to do this is
(362,880) x (10) = 3,628,800 ways.
If they sit a different way at every game, the class can see a bunch of games
without duplicating their seating arrangement !
