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alex41 [277]
3 years ago
10

A bank manager wants to encourage new customers to open accounts with principals of at least ​$2 comma 500. He decides to make a

poster advertising a simple interest rate of 3​%. What must the principal be if the bank manager also wants to advertise that one can earn ​$10 the first​ month? Can the poster correctly​ say, "Open an account of ​$2 comma 500 and earn at least ​$10 interest in 1​ month!"?
Mathematics
1 answer:
prisoha [69]3 years ago
4 0

Answer:

To earn $10 in 1st month, the principal must be $4,000.

No, the poster cannot claim that "Open an account of ​$2 comma 500 and earn at least ​$10 interest in 1​ month!".

Step-by-step explanation:

We have been given that a bank manager wants to encourage new customers to open accounts with principals of at least ​$2,500. He decides to make a poster advertising a simple interest rate of 3​%. We are asked to find the principal to advertise that one can earn ​$10 the first​ month.

We will use simple interest formula to solve our given problem.

I=Prt, where,

I = Amount of interest,

P = Principal amount,

r = Annual interest rate in decimal form,

t = Time in years.

r=3\%=\frac{3}{100}=0.03

1 month will be equal to \frac{1}{12}.

\$10=P\cdot0.03\cdot \frac{1}{12}

\$10=0.0025P

\frac{\$10}{0.0025}=\frac{0.0025P}{0.0025}

P=\$4000

Therefore, to earn $10 in 1st month, the principal must be $4,000.

Now, we will check if an account of ​$2,500 can earn at least ​$10 interest in 1​ month.

I=\$2500\cdot 0.03\cdot \frac{1}{12}

I=\$6.25

Since the account earns an amount of $6.25 in one month, therefore, the poster is not true.

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Solve for XX. Assume XX is a 2×22×2 matrix and II denotes the 2×22×2 identity matrix. Do not use decimal numbers in your answer.
sveticcg [70]

The question is incomplete. The complete question is as follows:

Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.

\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]· X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right] =<em>I</em>.

First, we have to identify the matrix <em>I. </em>As it was said, the matrix is the identiy matrix, which means

<em>I</em> = \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

So, \left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]· X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right] =  \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

Isolating the X, we have

X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]= \left[\begin{array}{cc}2&8\\-6&-9\end{array}\right] -  \left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

Resolving:

X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]= \left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]

X·\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]=\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]

Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,

X=\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]⁻¹·\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]

Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.

So,

\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]·\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]=\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]

9a - 3b = 1

7a - 6b = 0

9c - 3d = 0

7c - 6d = 1

Resolving these equations, we have a=\frac{2}{11}; b=\frac{7}{33}; c=\frac{-1}{11} and d=\frac{-3}{11}. Substituting:

X= \left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11}  \end{array}\right]·\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]

Multiplying the matrices, we have

X=\left[\begin{array}{ccc}\frac{8}{11} &\frac{26}{11} \\\frac{39}{11}&\frac{198}{11}  \end{array}\right]

6 0
3 years ago
In 1987​, 522 comma 000 people worked in the air transportation industry. In 2003​, the number was 480 comma 000. ​(a) Find a li
umka2103 [35]

Answer:

the linear equation is :

y= -2625x+5737875

and 440000 employees will be in year 2271

Step-by-step explanation:

in order to find the linerar equation let us first convert the given data into (x,y ) co-ordinates i.e

x1= 1987 y1= 522,000  

and

 

x2= 2003, y2 = 480000

according to two point form:

y-y1 = (y2-y1) \frac{x-x1}{x2-x1}

In this example we have x1=1987 , y1=522000 , x2=2003 and y2=480000. So,

y - 522000 = \frac{480000-522000}{2003-1987}  (x-1987)\\y- 522000 = -2625(x-1987)\\y- 522000= -2625x+5215875\\\\y= -2625x+5737875

now to find the year when there will be 440000 employs working , we simply put y= 440000 in the above derived equation

440000= -2625x+ 5737875\\\\2526x = 5737875 -440000\\2526x =  5736555\\x= \frac{5736555}{2526} \\x= 2271

7 0
3 years ago
Is this problem convergent or divergent?
Svetllana [295]

Answer:

Its divergent

Step-by-step explanation:

<em><u>HOPE</u></em><em><u> </u></em><em><u>YOU</u></em><em><u> </u></em><em><u>LIKED</u></em><em><u> </u></em><em><u>IT</u></em>

6 0
2 years ago
HELP ME PLEASEEEEEEEEEEEEEEEE
dsp73
The answer should be 18.66
7 0
2 years ago
Scrieți multipli lui 7 mai mari decât 10 și mai mici decât 20​
beks73 [17]
<h2>Helloooo!!! Marie Here!!!</h2><h2 />

The answer is here:

  • 7 . 2 = 14, 10 >14 > 50

  • 7 . 3 = 21, 10 > 21 > 50

  • 7 . 4 = 28, 10 > 28 > 50

  • 7 . 5 = 35, 10 > 35 > 50

  • 7 . 6 = 42, 10 > 42 > 50

<h3>Hope This Helps! Have A GREAT Day!</h3>

3 0
3 years ago
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