68.0052/12 would be 5.6671. If you round that to the nearest hundredth it would be 5.67. Hope this helps!
Answer: 180
Step-by-step explanation:
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If you add the two purchases, you find 9 hotdogs and 9 hamburgers cost 24.75. One hotdog and one hamburger will cost 1/9 of that total, or 2.75.
... One hotdog and one hamburger costs $2.75.
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This is what your question asks. Usually, we want to know the individual costs of the items. If we multiply this total by 5, then we have the cost of 5 hotdogs and 5 hamburgers: 13.75. This $1.75 more than the cost of 5 hotdogs and 4 hamburgers, so the cost of a hamburger must be $1.75
... One hotdog costs $1.00
... One hamburger costs $1.75
keeping in mind that 4 months is not even a year, since there are 12 months in a year, 4 months is then 4/12 years.
let's assume is simple interest.
![\bf ~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$34300\\ r=rate\to 3.5\%\to \frac{3.5}{100}\dotfill &0.035\\ t=years\to \frac{4}{12}\dotfill &\frac{1}{3} \end{cases} \\\\\\ A=34300\left[ 1+(0.035)\left( \frac{1}{3} \right) \right]\implies A= 34300(1.011\overline{6})\implies A=34700.1\overline{6}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%20Amount%7D%20%5C%5C%5C%5C%20A%3DP%281%2Brt%29%5Cqquad%20%5Cbegin%7Bcases%7D%20A%3D%5Ctextit%7Baccumulated%20amount%7D%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cdotfill%20%26%20%5C%2434300%5C%5C%20r%3Drate%5Cto%203.5%5C%25%5Cto%20%5Cfrac%7B3.5%7D%7B100%7D%5Cdotfill%20%260.035%5C%5C%20t%3Dyears%5Cto%20%5Cfrac%7B4%7D%7B12%7D%5Cdotfill%20%26%5Cfrac%7B1%7D%7B3%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D34300%5Cleft%5B%201%2B%280.035%29%5Cleft%28%20%5Cfrac%7B1%7D%7B3%7D%20%5Cright%29%20%5Cright%5D%5Cimplies%20A%3D%2034300%281.011%5Coverline%7B6%7D%29%5Cimplies%20A%3D34700.1%5Coverline%7B6%7D)
