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Maru [420]
3 years ago
15

When the new principal on a loan is $79,946.13 and the interest rate is 8 percent, the monthly payment is $587.20. Find the inte

rest, the payment to principal, and the new balance. STEP 1: Find the interest.
Mathematics
1 answer:
lesya [120]3 years ago
7 0

9514 1404 393

Answer:

  • to interest: $532.97
  • to principal: $54.23
  • new balance: $79,891.90

Step-by-step explanation:

The interest is found by multiplying the monthly rate by the balance on the loan. For the first month, the balance is the loan amount.

  $79,946.13 × 0.08 ×(1/12) . . . . . one month = 1/12 year

  = $532.97

The interest amount in the first payment is $532.97.

__

The amount of the first payment that goes to principal is what is left after the interest is paid:

  $587.20 -532.97 = $54.23 . . . amount to principal

__

The new balance is the previous balance less the amount to principal:

  $79,946.13 -54.23 = $79,891.90 . . . new balance

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Answer:

Step-by-step explanation:

Part A

Mean = (2.2 + 2.4 + 2.5 + 2.5 + 2.6 + 2.7)/6 = 2.48

Standard deviation = √(summation(x - mean)²/n

n = 6

Summation(x - mean)² = (2.2 - 2.48)^2 + (2.4 - 2.48)^2 + (2.5 - 2.48)^2 + (2.5 - 2.48)^2 + (2.6 - 2.48)^2 + (2.7 - 2.48)^2 = 0.1484

Standard deviation = √(0.1484/6

s = 0.16

Standard error = s/√n = 0.16/√6 = 0.065

Part B

Confidence interval is written as sample mean ± margin of error

Margin of error = z × s/√n

Since sample size is small and population standard deviation is unknown, z for 98% confidence level would be the t score from the student t distribution table. Degree of freedom = n - 1 = 6 - 1 = 5

Therefore, z = 3.365

Margin of error = 3.365 × 0.16/√6 = 0.22

Confidence interval is 2.48 ± 0.22

Part C

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 2.3

For the alternative hypothesis,

H1: µ > 2.3

This is a right tailed test

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 6

Degrees of freedom, df = n - 1 = 6 - 1 = 5

t = (x - µ)/(s/√n)

Where

x = sample mean = 2.48

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s = samples standard deviation = 0.16

t = (2.48 - 2.3)/(0.16/√6) = 2.76

We would determine the p value using the t test calculator. It becomes

p = 0.02

Assuming significance level, alpha = 0.05.

Since alpha, 0.05 > than the p value, 0.02, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the mean absolute refractory period for all mice when subjected to the same treatment increased.

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