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Rasek [7]
3 years ago
10

Peachtree Company borrows $30,000 from the local bank at 7% interest. The term of the note is five years, and the annual payment

s remain constant at $7,317. Determine the decrease in notes payable that Peachtree Company should record in the first year.
Mathematics
1 answer:
zubka84 [21]3 years ago
4 0

Answer:

$5,217

Step-by-step explanation:

Data provided;

Amount borrowed = $30,000

Interest rate = 7%

Term of note = 5 years

Annual payments = $7,317

Interest payment for the first year = Amount borrowed × Interest rate

= $30,000 × 7%

= $30,000 × 0.07

= $2,100

Hence,,

The Principal amount paid = Total amount paid - Interest

or

⇒ The Principal amount paid = $7,317 - $2,100

= $5,217

Hence,

the decrease in notes payable that Peachtree Company should record in the first year is $5,217

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2. When a large truckload of mangoes arrives at a packing plant, a random sample of 150 is selected and examined for
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a) The 90% confidence interval of the percentage of all mangoes on the truck that fail to meet the standards is: (7.55%, 12.45%).

b) The margin of error is: 2.45%.

c) The 90% confidence is the level of confidence that the true population percentage is in the interval.

d) The needed sample size is: 271.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions has the bounds given by the rule presented as follows:

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

The margin of error is given by:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

The variables are listed as follows:

  • \pi is the sample proportion, which is also the estimate of the parameter.
  • z is the critical value.
  • n is the sample size.

The confidence level is of 90%, hence the critical value z is the value of Z that has a p-value of \frac{1+0.90}{2} = 0.95, so the critical value is z = 1.645.

The sample size and the estimate are given as follows:

n = 150, \pi = \frac{15}{150} = 0.1

The margin of error is of:

M = z\sqrt{\frac{0.1(0.9)}{150}} = 0.0245 = 2.45\%

The interval is given by the estimate plus/minus the margin of error, hence:

  • The lower bound is: 10 - 2.45 = 7.55%.
  • The upper bound is: 10 + 2.45 = 12.45%.

For a margin of error of 3% = 0.03, the needed sample size is obtained as follows:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.645\sqrt{\frac{0.1(0.9)}{n}}

0.03\sqrt{n} = 1.645\sqrt{0.1(0.9)}

\sqrt{n} = \frac{1.645\sqrt{0.1(0.9)}}{0.03}

(\sqrt{n}})^2 = \left(\frac{1.645\sqrt{0.1(0.9)}}{0.03}\right)^2

n = 271 (rounded up).

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

3 0
2 years ago
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