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daser333 [38]
3 years ago
13

Tamara has $500 she is looking to save for a class trip. She wants to earn the most possible interest and will not need access t

o her money for a full year. Which type of savings account will be best for Tamara?
Computers and Technology
2 answers:
Lapatulllka [165]3 years ago
7 0

Answer:

Certificate of deposit

Explanation:

A certificate of deposit is a type of savings account that has a fixed date in which the money can be withdrawn and it offers a higher interest rate than a regular savings account in which you have access to your money at any time. According to this, the answer is that the type of savings account that will be best for Tamara is a certificate of deposit because it pays a higher interest rate than standard savings accounts and she doesn't need the money for a year.

Novay_Z [31]3 years ago
6 0
Certificate Of Deposit- It will be unaccesible and will help her the best.
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What would you have to know about the pivot columns in an augmented matrix in order to know that the linear system is consistent
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The Rouché-Capelli Theorem. This theorem establishes a connection between how a linear system behaves and the ranks of its coefficient matrix (A) and its counterpart the augmented matrix.

rank(A)=rank\left ( \left [ A|B \right ] \right )\:and\:n=rank(A)

Then satisfying this theorem the system is consistent and has one single solution.

Explanation:

1) To answer that, you should have to know The Rouché-Capelli Theorem. This theorem establishes a connection between how a linear system behaves and the ranks of its coefficient matrix (A) and its counterpart the augmented matrix.

rank(A)=rank\left ( \left [ A|B \right ] \right )\:and\:n=rank(A)

rank(A)

Then the system is consistent and has a unique solution.

<em>E.g.</em>

\left\{\begin{matrix}x-3y-2z=6 \\ 2x-4y-3z=8 \\ -3x+6y+8z=-5  \end{matrix}\right.

2) Writing it as Linear system

A=\begin{pmatrix}1 & -3 &-2 \\  2& -4 &-3 \\ -3 &6  &8 \end{pmatrix} B=\begin{pmatrix}6\\ 8\\ 5\end{pmatrix}

rank(A) =\left(\begin{matrix}7 & 0 & 0 \\0 & 7 & 0 \\0 & 0 & 7\end{matrix}\right)=3

3) The Rank (A) is 3 found through Gauss elimination

(A|B)=\begin{pmatrix}1 & -3 &-2  &6 \\  2& -4 &-3  &8 \\  -3&6  &8  &-5 \end{pmatrix}

rank(A|B)=\left(\begin{matrix}1 & -3 & -2 \\0 & 2 & 1 \\0 & 0 & \frac{7}{2}\end{matrix}\right)=3

4) The rank of (A|B) is also equal to 3, found through Gauss elimination:

So this linear system is consistent and has a unique solution.

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