Pretty sure it's 1 since 8 counts as one but 0 doesn't
Answer:
Step-by-step explanation:
Let x represent the amount on deposit at 12%. Then 5900-x is the amount deposited at 6%, and the total interest earned is ...
12%·x +6%·(5900 -x) = 540
0.06x + 354 = 540 . . . . . . . . eliminate parentheses
0.06x = 186 . . . . . . . . . . . . . . subtract 354
x = 3100 . . . . . . . . . . . . . . . . . divide by 0.06
Scott invested $3100 at 12% and $2800 at 6%.
_____
<em>Quick sanity check</em>
The average interest earned is $540/$5900 ≈ 0.0915 ≈ 9.15%. This is more than the average of 12% and 6%, which is (12+6)/2 = 9 percent. Hence more than half of the money must be deposited at 12%.
TO solve this question on "how large of a sample must Beth have to get a margin of error less than 0.04", we will use the margin of error formula:
![E=Z\sqrt{\frac{\hat p(1-\hat p)}{n}}](https://tex.z-dn.net/?f=%20E%3DZ%5Csqrt%7B%5Cfrac%7B%5Chat%20p%281-%5Chat%20p%29%7D%7Bn%7D%7D%20)
Here, for a 80% confidence level, ![Z=1.28](https://tex.z-dn.net/?f=%20Z%3D1.28%20)
(given)
Thus, margin of error, E=0.04
rearranging we get:
![n=(\frac{Z}{E})^2\hat p(1-\hat p)](https://tex.z-dn.net/?f=%20n%3D%28%5Cfrac%7BZ%7D%7BE%7D%29%5E2%5Chat%20p%281-%5Chat%20p%29%20)
Plugging in gives us:
![n=(\frac{1.28}{0.04})^2\times 0.5(1-0.5)=1024\times0.25=256](https://tex.z-dn.net/?f=%20n%3D%28%5Cfrac%7B1.28%7D%7B0.04%7D%29%5E2%5Ctimes%200.5%281-0.5%29%3D1024%5Ctimes0.25%3D256%20)
Thus, Beth's sample for the true proportion of high school students in the area who attend their home basketball games must have 256 students.