Answer:
The minimum sample needed to provide a margin of error of 3 or less is 52.
Step-by-step explanation:
The confidence interval for population mean (<em>μ</em>) is:
The margin of error is:
<u>Given:</u>
MOE = 3
<em>σ </em>= 11
The critical value for 95% confidence interval is: 
**Use the <em>z</em>-table for critical values.
Compute the sample size (<em>n</em>) as follows:
Thus, the minimum sample needed to provide a margin of error of 3 or less is 52.
Answer:
cool
Step-by-step explanation:
Answer:
20 - 6c
Step-by-step explanation:
because that shows that 6c is the cost of them and you'd take the cost of them away from how much he payed i guess
To get the solution, we are looking for, we need to point out what we know.
1. We assume, that the number 112 is 100% - because it's the output value of the task.
2. We assume, that x is the value we are looking for.
3. If 112 is 100%, so we can write it down as 112=100%.
4. We know, that x is 200% of the output value, so we can write it down as x=200%.
5. Now we have two simple equations:
1) 112=100%
2) x=200%
where left sides of both of them have the same units, and both right sides have the same units, so we can do something like that:
112/x=100%/200%
6. Now we just have to solve the simple equation, and we will get the solution we are looking for.
7. Solution for what is 200% of 112
112/x=100/200
(112/x)*x=(100/200)*x - we multiply both sides of the equation by x
112=0.5*x - we divide both sides of the equation by (0.5) to get x
112/0.5=x
224=x
x=224
now we have:
200% of 112=224
●I tried my best I was never good with percentages my least favorite.... Please let me know if you got it wrong. If you do I'm sorry.
After thorough researching, if a jet in calm air conditions travels with velocity vector (389, 389) and the wind velocity (in mph) at the plane’s cruising altitude is given by (0, 50), then the plane’s true speed is 600 mph. The correct answer to the following given statement above is with the letter C.
