well, keeping in mind that a year has 12 months, that means that 8 months is 8/12 of a year, when Mrs Rojas pull her money out.
![~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$6000\\ r=rate\to 4\%\to \frac{4}{100}\dotfill &0.04\\ t=years\to \frac{8}{12}\dotfill &\frac{2}{3} \end{cases} \\\\\\ A=6000[1+(0.04)(\frac{2}{3})]\implies A=6000\left( \frac{77}{75} \right)\implies A=6160](https://tex.z-dn.net/?f=~~~~~~%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%246000%5C%5C%20r%3Drate%5Cto%204%5C%25%5Cto%20%5Cfrac%7B4%7D%7B100%7D%5Cdotfill%20%260.04%5C%5C%20t%3Dyears%5Cto%20%5Cfrac%7B8%7D%7B12%7D%5Cdotfill%20%26%5Cfrac%7B2%7D%7B3%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D6000%5B1%2B%280.04%29%28%5Cfrac%7B2%7D%7B3%7D%29%5D%5Cimplies%20A%3D6000%5Cleft%28%20%5Cfrac%7B77%7D%7B75%7D%20%5Cright%29%5Cimplies%20A%3D6160)
well, she put in 6000 bucks, got back 160 extra, that's the interest earned in the 8 months.
what if she had left her money for 1 whole year, then
![~~~~~~ \textit{Simple Interest Earned Amount} \\\\ A=P(1+rt)\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill & \$6000\\ r=rate\to 4\%\to \frac{4}{100}\dotfill &0.04\\ t=years\dotfill &1 \end{cases} \\\\\\ A=6000[1+(0.04)(1)]\implies A=6240](https://tex.z-dn.net/?f=~~~~~~%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%246000%5C%5C%20r%3Drate%5Cto%204%5C%25%5Cto%20%5Cfrac%7B4%7D%7B100%7D%5Cdotfill%20%260.04%5C%5C%20t%3Dyears%5Cdotfill%20%261%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%3D6000%5B1%2B%280.04%29%281%29%5D%5Cimplies%20A%3D6240)
so had she left it in for a year, she'd have gotten 6240, namely 240 in interest, well, what fraction of a year's interest was earned? or worded differently, what fraction is 160(8 months) of 240(1 year)?
Answer:
B) 105.20
Step-by-step explanation:
491.60 + 157.68 - 68.44 = 581.20
• The ball is at the same height as the building between 8 and 10 seconds after it is thrown. TRUE - the height is zero somewhere in that interval, hence the ball is the same height from which it was thrown, the height of the roof of the building.
• The height of the ball decreases and then increases. FALSE - at t=2, the height is greater than at t=0.
• The ball reaches its maximum height about 4 seconds after it is thrown. TRUE - the largest number in the table corresponds to t=4.
• The ball hits the ground between 8 and 10 seconds after it is thrown. FALSE - see statement 1.
• The height of the building is 81.6 meters. FALSE - the maximum height above the building is 81.6 meters. Since the ball continues its travel to a distance 225.6 meters below the roof of the building, the building is at least that high.
1. TRUE
2. False
3. TRUE
4. False
5. False
Hello from MrBillDoesMath!
Answer:
88
Discussion:
Angle Q is an inscribed angle so
measure angle Q = 44 =
(1/2) measure of intercepted arc.
In other words,
44 = 1/2 mST => multiply both sides by 2
m ST = 44*2 = 88
which is the second choice
Thank you,
MrB
