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Nana76 [90]
3 years ago
9

You invest 2,000,00 in stock plan and another in a saving account the stock plan decreases by 7% the first year and gains 10% th

e second year the saving account earns a 3.7% APR and compounds annually what is the difference in earning between the stock and saving account at the end of the second year
Mathematics
1 answer:
sineoko [7]3 years ago
3 0
After calculating the stock and savings account total I found out that the difference in earning between them is $104.738.
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I need help now, please. Please explain also. Thanks!
Slav-nsk [51]

Answer:

C)-1

Step-by-step explanation:

You can not spend a negative hour playing golf.

4 0
3 years ago
The price of gold is often reported per ounce. At the end of 2005, this price was $513. At the end of 2015, it was $1060. By wha
cricket20 [7]

Answer:

the answer is 106.63%.

8 0
3 years ago
What law can be used to determine that the conclusion is valid based on the given statements
ryzh [129]

Answer:

A. Law of detachment

Step-by-step explanation:

The Law of detachment implies that when one condition is fulfilled the other cannot be and vice versa, then it is made the conclusion.

This condition is made the conclusion.

The Acute and Obtuse are detached of each other.

The acute angle is one in which the value of the angle is less than 90 degrees and obtuse angle is one in which the angle is greater than 90 degrees but less than 180 degrees.

Thus angles less than 90 degrees are acute and greater than 90 degrees are obtuse.

The conclusion of the given statement  is valid based on the law of detachment as the condition has been made a conclusion.

8 0
3 years ago
Use the quadratic formula to solve the equation.
allochka39001 [22]
=1.3+0.1sqrt29, 1.3-0.1sqrt29
7 0
3 years ago
Use matrices and elementary row to solve the following system:
LiRa [457]

I assume the first equation is supposed to be

5x-3y+2z=13

and not

5x-3x+2x=4x=13

As an augmented matrix, this system is given by

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\4&-2&4&12\end{array}\right]

Multiply through row 3 by 1/2:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\2&-1&2&6\end{array}\right]

Add -1(row 2) to row 3:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&5&5\end{array}\right]

Multiply through row 3 by 1/5:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&1&1\end{array}\right]

Add -2(row 3) to row 1, and add 3(row 3) to row 2:

\left[\begin{array}{ccc|c}5&-3&0&11\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -3(row 2) to row 1:

\left[\begin{array}{ccc|c}-1&0&0&-1\\2&-1&0&4\\0&0&1&1\end{array}\right]

Multiply through row 1 by -1:

\left[\begin{array}{ccc|c}1&0&0&1\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -2(row 1) to row 2:

\left[\begin{array}{ccc|c}1&0&0&1\\0&-1&0&2\\0&0&1&1\end{array}\right]

Multipy through row 2 by -1:

\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&-2\\0&0&1&1\end{array}\right]

The solution to the system is then

\boxed{x=1,y=-2,z=1}

5 0
3 years ago
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