Minimum required sample size for a desired margin of error and confidence level when it is a proportion problem: n = (z2÷margin of error2)*p-hat*q-hat
The maximum value of p-hat*q-hat occurs where p-hat = .5 (found by taking the derivative of (p-hat)*(1-p-hat) and setting it equal to 0 to find the maximum. n = ( 2.5762( for 99% confidence interval)÷.0482 )*.5*.5 = 720.028 or 721
24.5= <span><span>−<span>2.5c</span></span>−<span>4<span>c
</span></span></span><span>24.5=<span><span><span>−<span>2.5c </span></span>+ </span>−<span>4c</span></span></span><span>24.5 = <span>(<span><span>−<span>2.5c</span></span>+<span>−<span>4c</span></span></span>) </span></span><span>24.5=<span>−<span>6.5c </span></span></span><span>24.5=<span>−<span>6.5<span>c
</span></span></span></span><span>−<span>6.5c</span></span>=<span>24.5
</span><span><span><span>−<span>6.5c </span></span><span>− 6.5 </span></span>=<span>24.5 <span>− 6.5 </span></span></span><span>c=<span>−<span>3.769231
These are all the steps that you do to solve this equation. I worked it out fully and the answer is above these words.</span></span></span>
Answer:
about 14.28% for 1/7th
Step-by-step explanation:
Midpoint = ((x1+x2)/2 , (y1+y2)/2) = (2+14)/2), (1+(-8)/2) = (8,-3.5)
Answer:
40 tickets
Step-by-step explanation:
5/8 is .63
64 x .63 = 40.32 rounded is equal to 40 tickets