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Juli2301 [7.4K]

2 years ago

6

Thomas is taking out a business loan of $20,000 over 5 years. He is given an APR of 6.12% he would like his monthly payments to

not exceed $350 a month. Looking at the first few rows of the amortization table below, will his monthly payment be below $350 ? How much over/ under will hi actual payments be from his goal of $350? How much over/ under will his actual payments be from his goal of $350 per month?

1 answer:

Mila [183]2 years ago

8 0

1. Thomas' monthly payment for the loan exceeds $350.

2. An excess monthly payment of $37.77 exists.

3. Thus, the actual monthly payment is $387.77.

<h3>What is interest?</h3>

The interest is the finance charge for the loan, representing the cost of the time value of money.

The interest and the monthly payments can be calculated using an online finance calculator as follows:

<h3>Data and Calculations:</h3>

Business Loan = $20,000

Down Payment = 0%

Loan Term = 5 years or 60 months (5 x 12)

Interest Rate = 6.12%

Monthly Mortgage Payment = $387.77

Total of 60 Mortgage Payments = $23,266.38 ($387.77 x 60)

Total Interest = $3,266.38

Thus, Thomas' actual monthly payment is $387.77.

Learn more about calculating interests at brainly.com/question/25545513

#SPJ1

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<u>Step-by-step explanation:</u>

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Then combining their charges together to get:

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4 years ago

In a recent poll of 2,000 household, 29% had pets.how many households did not have pets?

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3 years ago

Determine which of the sets of vectors is linearly independent. A: The set where p1(t) = 1, p2(t) = t2, p3(t) = 3 + 3t B: The se

defon

Answer:

The set of vectors A and C are linearly independent.

Step-by-step explanation:

A set of vector is linearly independent if and only if the linear combination of these vector can only be equalised to zero only if all coefficients are zeroes. Let is evaluate each set algraically:

$p_{1}(t) = 1$ , $p_{2}(t)= t^{2}$ and $p_{3}(t) = 3 + 3\cdot t$ :

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (3 +3\cdot t) = 0$

$(\alpha_{1}+3\cdot \alpha_{3})\cdot 1 + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot t = 0$

The following system of linear equations is obtained:

$\alpha_{1} + 3\cdot \alpha_{3} = 0$

$\alpha_{2} = 0$

$\alpha_{3} = 0$

Whose solution is $\alpha_{1} = \alpha_{2} = \alpha_{3} = 0$ , which means that the set of vectors is linearly independent.

$p_{1}(t) = t$ , $p_{2}(t) = t^{2}$ and $p_{3}(t) = 2\cdot t + 3\cdot t^{2}$

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot t + \alpha_{2}\cdot t^{2} + \alpha_{3}\cdot (2\cdot t + 3\cdot t^{2})=0$

$(\alpha_{1}+2\cdot \alpha_{3})\cdot t + (\alpha_{2}+3\cdot \alpha_{3})\cdot t^{2} = 0$

The following system of linear equations is obtained:

$\alpha_{1}+2\cdot \alpha_{3} = 0$

$\alpha_{2}+3\cdot \alpha_{3} = 0$

Since the number of variables is greater than the number of equations, let suppose that $\alpha_{3} = k$ , where $k\in\mathbb{R}$ . Then, the following relationships are consequently found:

$\alpha_{1} = -2\cdot \alpha_{3}$

$\alpha_{1} = -2\cdot k$

$\alpha_{2}= -2\cdot \alpha_{3}$

$\alpha_{2} = -3\cdot k$

It is evident that $\alpha_{1}$ and $\alpha_{2}$ are multiples of $\alpha_{3}$ , which means that the set of vector are linearly dependent.

$p_{1}(t) = 1$ , $p_{2}(t)=t^{2}$ and $p_{3}(t) = 3+3\cdot t +t^{2}$

$\alpha_{1}\cdot p_{1}(t) + \alpha_{2}\cdot p_{2}(t) + \alpha_{3}\cdot p_{3}(t) = 0$

$\alpha_{1}\cdot 1 + \alpha_{2}\cdot t^{2}+ \alpha_{3}\cdot (3+3\cdot t+t^{2}) = 0$

$(\alpha_{1}+3\cdot \alpha_{3})\cdot 1+(\alpha_{2}+\alpha_{3})\cdot t^{2}+3\cdot \alpha_{3}\cdot t = 0$

The following system of linear equations is obtained:

$\alpha_{1}+3\cdot \alpha_{3} = 0$

$\alpha_{2} + \alpha_{3} = 0$

$3\cdot \alpha_{3} = 0$

Whose solution is $\alpha_{1} = \alpha_{2} = \alpha_{3} = 0$ , which means that the set of vectors is linearly independent.

The set of vectors A and C are linearly independent.

4 0

3 years ago