Answer:
See below
Step-by-step explanation:
9 and 8.4 are less than 1 apart, so 9-8.4 must be less than 1. The actual difference is 9-8.4=0.6 which verifies that our answers are true since 0.6<1.
16 would be the only answer the others are there to confuse you
a)
well, she put 4000, and she earned in interest 960, so her accumulated amount is just their sum, 4960.
b)
now, it doesn't say, so we're assuming is <u>simple interest</u>, as opposed to compound interest.

c)
let's make the rate 1% greater then, and check
![\bf ~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill\\ P=\textit{original amount deposited}\dotfill & \$4000\\ r=rate\to \stackrel{8+1}{9\%}\to \frac{9}{100}\dotfill&0.09\\ t=years\dotfill &3 \end{cases} \\\\\\ I=(4000)(0.09)(3)\implies I=1080 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{at 9\%}}{1080}-\stackrel{\textit{at 8\%}}{960}\implies \stackrel{\textit{that much more}}{120}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%7D%20%5C%5C%5C%5C%20I%20%3D%20Prt%5Cqquad%20%5Cbegin%7Bcases%7D%20I%3D%5Ctextit%7Binterest%20earned%7D%5Cdotfill%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cdotfill%20%26%20%5C%244000%5C%5C%20r%3Drate%5Cto%20%5Cstackrel%7B8%2B1%7D%7B9%5C%25%7D%5Cto%20%5Cfrac%7B9%7D%7B100%7D%5Cdotfill%260.09%5C%5C%20t%3Dyears%5Cdotfill%20%263%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20I%3D%284000%29%280.09%29%283%29%5Cimplies%20I%3D1080%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bat%209%5C%25%7D%7D%7B1080%7D-%5Cstackrel%7B%5Ctextit%7Bat%208%5C%25%7D%7D%7B960%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bthat%20much%20more%7D%7D%7B120%7D)
$69.48/6=11.58
3 feet in a yard