All categories

3 years ago

Anna has a 30-year, 5.75% mortgage on her $250,000 home. She has been

Mathematics

2 answers:

poizon [28]3 years ago

4 0

Answer:

Her new monthly payment is now $1,378.91¢

Step-by-step explanation:

For us to calculate the new monthly mortgage payment that Anna will start paying from now on, we need to input the formula for calculating monthly mortgage payments.

The formula is:-

$M = P [\frac{r(1+r)^{n} }{(1+r)^{n}-1}]$

Where M is the monthly mortgage payment.

P is the principal

r is the monthly interest rate calculated by dividing your annual interest rate by 12

n is the number of payments(the number of months you will be paying the loan).

In this case, the new principal that Anna must pay back is $231,905.47¢. The annual interest rate has been reduced to 5.17% from 5.75% so the new monthly interest rate will be obtained by dividing the new annual interest rate by 12

= 5.17%/2

= 0.431%

This is the new monthly interest rate.

Since she has been paying her mortgage loan diligently for 5 complete years. It means she now has just 25 years to complete the payment. If 12 months make up one year, then there are - 12 × 25 = 300 more months to go.

300 is therefore "n" that is required for the calculation.

All the terms needed for the calculation of her new monthly mortgage is now complete.

P = $231,905.47¢

r = 0.431%

n = 300

$M = 231,905.47[\frac{0.00431(1+0.00431)^{300} }{(1+0.00431)^{300} -1}]$

$= 231,905.47[\frac{0.00431(3.634)}{2.634}]$

= 231,905.47 × 0.005946

M = $1,378.91¢

Therefore her new monthly mortgage payment will become $1,378.91¢

Mademuasel [1]3 years ago

4 0

Alternative aaaapppeexx answer if it’s lowered to 5.25% is $1389.69!

You might be interested in

Last week, De swam five miles farther than three times the distance his father swam. De swam 23 miles last week. Choose the equa

melomori [17]

Answer:

3m+5 equals 23

Step-by-step explanation:

If De swam 5 miles longer than three times what his dad seam then you would make his father's distance m and plug in the numbers.

4 0

2 years ago

In analyzing the city and highway fuel efficiencies of many cars and trucks, the mean difference in fuel efficiencies for the 60

Olegator [25]

Answer:

$7.38-1.96\frac{2.51}{\sqrt{601}}=7.18$

$7.38+1.96\frac{2.51}{\sqrt{601}}=7.58$

For this case the confidence interval is given by :

$7.18 \leq \mu \leq 7.58$

Step-by-step explanation:

Data provided

$\bar X=7.38$ represent the sample mean for the fuel efficiencies

$\mu$ population mean

s=2.51 represent the sample standard deviation

n=601 represent the sample size

Confidence interval

The formula for the confidence interval of the true mean is given by:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom for this case is given by:

$df=n-1=601-1=600$

The Confidence level for this case is 0.95 or 95%, and the significance level $\alpha=0.05$ and $\alpha/2 =0.025$ and the critical value is given by $t_{\alpha/2}=1.96$

Replcing into the formula for the confidence interval is given by:

$7.38-1.96\frac{2.51}{\sqrt{601}}=7.18$

$7.38+1.96\frac{2.51}{\sqrt{601}}=7.58$

For this case the confidence interval is given by :

$7.18 \leq \mu \leq 7.58$

8 0

3 years ago

The graph shows the relationship between the distance a truck can travel and the amount of gasoline used.

sp2606 [1]

Answer:

7 mpg

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Stream Black Cult

riadik2000 [5.3K]

Sorry, but I cannot understand the first part of this question

Nor the second part it's not finished.

<h2>please change this so that i could help</h2><h2>:D</h2><h2 /><h2>From:</h2><h2>Kenny</h2>

4 0

2 years ago

According to a recent survey, the population distribution of number of years of education for self-employed individuals in a c

creativ13 [48]

Answer:

a) X: number of years of education

b) Sample mean = 13.5, Sample standard deviation = 0.4

c) Sample mean = 13.5, Sample standard deviation = 0.2

d) Decrease the sample standard deviation

Step-by-step explanation:

We are given the following in the question:

Mean, μ = 13.5 years

Standard deviation,σ = 2.8 years

a) random variable X

X: number of years of education

Central limit theorem:

If large random samples are drawn from population with mean $\mu$ and standard deviation $\sigma$ , then the distribution of sample mean will be normally distributed with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

b) mean and the standard for a random sample of size 49

$\mu_{\bar{x}} = \mu = 13.5\\\\\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{2.8}{\sqrt{49}} = 0.4$

c) mean and the standard for a random sample of size 196

$\mu_{\bar{x}} = \mu = 13.5\\\\\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{2.8}{\sqrt{196}} = 0.2$

d) Effect of increasing n

As the sample size increases, the standard error that is the sample standard deviation decreases. Thus, quadrupling sample size will half the standard deviation.

7 0

3 years ago