so change a value to a percentage format, we simply divide it by 100.
for example, 2% can be rewritten as 2/100 or 0.02 in decimal format.
now, let's first convert the mixed fraction to improper, and then do the percentage format and simplification.
![\bf \stackrel{mixed}{3\frac{1}{8}}\implies \cfrac{3\cdot 8+1}{8}\implies \stackrel{improper}{\cfrac{25}{8}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{~~\frac{25}{8}~~}{100}\implies \cfrac{~~\frac{25}{8}~~}{\frac{100}{1}}\implies \cfrac{25}{8}\cdot \cfrac{1}{100}\implies \cfrac{1}{8}\cdot \cfrac{25}{100}\implies \cfrac{1}{8}\cdot \cfrac{1}{4}\implies \cfrac{1}{32}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Cstackrel%7Bmixed%7D%7B3%5Cfrac%7B1%7D%7B8%7D%7D%5Cimplies%20%5Ccfrac%7B3%5Ccdot%208%2B1%7D%7B8%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B25%7D%7B8%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B~~%5Cfrac%7B25%7D%7B8%7D~~%7D%7B100%7D%5Cimplies%20%5Ccfrac%7B~~%5Cfrac%7B25%7D%7B8%7D~~%7D%7B%5Cfrac%7B100%7D%7B1%7D%7D%5Cimplies%20%5Ccfrac%7B25%7D%7B8%7D%5Ccdot%20%5Ccfrac%7B1%7D%7B100%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B8%7D%5Ccdot%20%5Ccfrac%7B25%7D%7B100%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B8%7D%5Ccdot%20%5Ccfrac%7B1%7D%7B4%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B32%7D%20)
Answer:
1/3
Step-by-step explanation:
If Mackenzie drinks 8 ounces of water per 24 hours, that could be written as a ratio and a fraction. 8:24, 8/24
8/24 in its simplest form is 1/3.
I would just choose a,d,and c
The minimum sample size required for a test with a confidence interval of
![100(1 - \alpha )%](https://tex.z-dn.net/?f=100%281%20-%20%5Calpha%20%29%25)
with a z-score of
![z_{ \alpha /2}](https://tex.z-dn.net/?f=z_%7B%20%5Calpha%20%2F2%7D)
and margin of error of E and a population proportion of p is given by:
![n= \frac{p(1-p)z_{ \alpha /2}^2 \alpha }{E^2}](https://tex.z-dn.net/?f=n%3D%20%5Cfrac%7Bp%281-p%29z_%7B%20%5Calpha%20%2F2%7D%5E2%20%5Calpha%20%7D%7BE%5E2%7D%20)
Given p = 34% = 0.34, E = 0.027,
![z_{ \alpha /2}=1.645](https://tex.z-dn.net/?f=z_%7B%20%5Calpha%20%2F2%7D%3D1.645)
Therefore,