Answer:
554 executives should be surveyed.
Step-by-step explanation:
We have that to find our 
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Z-table as such z has a p-value of 
.
That is z with a pvalue of 
, so Z = 1.96.
Now, find the margin of error M as such
In which 
is the standard deviation of the population and n is the size of the sample.
Standard deviation of 3 hours.
This means that 
The 95% level of confidence is to be used. How many executives should be surveyed?
n executives should be surveyed, and n is found with 
. So
Rounding up:
554 executives should be surveyed.
Answer:
option (3) is correct.
Step-by-step explanation:
Given 
We have to solve for e.
Consider the given statement,
Cross multiply, we get,
Taking square root both sides , we get,
We know square root of 9 is 3.
Thus, option (3) is correct.
Answer:
43
Step-by-step explanation:
204-75=129
129/3=43
Answer:
5,400
Step-by-step explanation:
5% = 270
270 ÷ 5 = 54
54 × 100 = 5400
Answer:
Step-by-step explanation:
2005 AMC 8 Problems/Problem 20
Problem
Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise. In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 24$
Solution
Alice moves $5k$ steps and Bob moves $9k$ steps, where $k$ is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, $14k$, is a multiple of $12$. Since this number must be a multiple of $12$, as stated in the previous sentence, $14$ has a factor $2$, $k$ must have a factor of $6$. The smallest number of turns that is a multiple of $6$ is $\boxed{\textbf{(A)}\ 6}$.
See Also
2005 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 19 Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
