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Law Incorporation [45]
3 years ago
15

Google has just purchased your website that reviews GPS systems. You made a profit of $65,000. You would like to invest the $65,

000 into an account that pays you an annual compounded interest rate of 5.5%. You want to make annual withdrawals over the next 10 years such that by the end of this 10 year period, the amount remaining in the account will be zero dollars. Determine, from the given information, the amount of the annual withdrawals. Round your answer to the nearest cent.
a.$6,500
b.$7,023.45
c.$7,251.31
d.$8,623.40
Mathematics
2 answers:
katen-ka-za [31]3 years ago
8 0

Answer: ITs D


Step-by-step explanation:


d1i1m1o1n [39]3 years ago
6 0
Hi there
The formula of the present value of annuity ordinary is
Pv=pmt [(1-(1+r)^(-n))÷r]
So we need to solve for pmt (the amount of the annual withdrawals)
PMT=pv÷ [(1-(1+r)^(-n))÷r]
Pv present value 65000
R interest rate 0.055
N time 10 years
PMT=65,000÷((1−(1+0.055)^(
−10))÷(0.055))
=8,623.40....answer

Hope it helps
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Find a set of vectors {u⃗ ,v⃗ }{u→,v→} in r4r4 that spans the solution set of the equations {w−x+2y+4z5w+2x−y+3z==0,0. {w−x+2y+4
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Given

w-x+2y+4z=0 \\ 5w+2x-y+3z=0

We can rewrite it in matrix form as:

\left[\begin{array}{cccc}1&-1&2&4\\5&2&-1&3\\0&0&0&0\\0&0&0&0\end{array}\right]   \left[\begin{array}{c}w\\x\\y\\z\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right] \\  \\ \left[\begin{array}{cccc}1&-1&2&4\\5&2&-1&3\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \ -5R_1+R_2\rightarrow R_2
\left[\begin{array}{cccc}1&-1&2&4\\0&7&-11&-17\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \  \frac{1}{7} R_2\rightarrow R_2 \\  \\ \left[\begin{array}{cccc}1&-1&2&4\\0&1&-\frac{11}{7}&-\frac{17}{7}\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \ R_1+R_2\rightarrow R_1
\\  \\ \left[\begin{array}{cccc}1&0&\frac{3}{7}&\frac{11}{7}\\0&1&-\frac{11}{7}&-\frac{17}{7}\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right] \\  \\ \Rightarrow w= -\frac{3}{7} y- \frac{11}{7} z \\ x=\frac{11}{7} y+ \frac{17}{7} z \\ y=free \\ z=free \\  \\ =y\left\ \textless \ -\frac{3}{7},\frac{11}{7}1,0\right\ \textgreater \ +z\left\ \textless \ - \frac{11}{7},\frac{17}{7},0,1\right\ \textgreater \

Thus, the solution set is a span of \{\left\ \textless \ -\frac{3}{7},\frac{11}{7}1,0\right\ \textgreater \ ,\left\ \textless \ - \frac{11}{7},\frac{17}{7},0,1\right\ \textgreater \ \}
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