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:
42 miles per gallon
Step-by-step explanation:
Answer:
2
Step-by-step explanation:
f(x)=2x^2+9x-5
When we are find how many times it intersects the x axis, we are finding the zero's. Set the equation equal to zero
0=2x^2+9x-5
Factor the equation
0 = (2x+1) (x-5)
2*1
1*-5 = -5
2*-5 +1*1 = -9
This checks for the first last and middle terms so we factored correctly
Then using the zero product property
2x+1 = 0 and x-5 =0
2x = -1 x=5
x = -1/2 and x=5
This function crosses the x axis 2 times
Answer:
a) P(X=2)= 0.29
b) P(X<2)= 0.59
c) P(X≤2)= 0.88
d) P(X>2)= 0.12
e) P(X=1 or X=4)= 0.24
f) P(1≤X≤4)= 0.59
Step-by-step explanation:
a) P(X=2)= 1 - P(X=0) - P(X=1) - P(X=3) - P(X=4)= 1-0.41-0.18-0.06-0.06= 0.29
b) P(X<2)= P(X=0) + P(X=1)= 0.41 + 0.18 = 0.59
c) P(X≤2)= P(X=0) + P(X=1) + P(X=2)=0.41+0.18+0.29= 0.88
d) P(X>2)=P(X=3) + P(X=4)=0.06+0.06= 0.12
e) P(X=1 or X=4)=P(X=1 ∪ X=4) = P(X=1) + P(X=4)=0.18+0.06= 0.24
f) P(1≤X≤4)=P(X=1) + P(X=2) + P(X=3) + P(X=4)=0.18+0.29+0.06+0.06= 0.59
