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Bas_tet [7]
1 year ago
11

Vanessa can afford a $1405-per-month house loan payment. If she is being offered a 20-year house loan with an APR of 4.8%, compo

unded monthly, which of these expressions represents the value of the most money she can borrow?
Mathematics
2 answers:
DochEvi [55]1 year ago
3 0
To solve this we are going to use the present value formula: PV=P[ \frac{1-(1+ \frac{r}{n})^{-kt} }{ \frac{r}{n} } ]
where 
PV is the present value
P is the periodic payment
r is the interest rate in decimal form
n is the number of times the interest is compounded per year 
k is the number of payments per year
t is the number of years

We know for our problem that she can afford a $1405-per-month house loan payment, so P=1405 and k=12. We also know that the loan is a 20-year house loan, so t=20. To convert the interest rate to decimal form, we are going to divide the rate by 100% r= \frac{4.8}{100}=0.048. Since the interest rate is compounded monthly, it is compounded 12 times per year; therefore, n=12. Lets replace those values in our formula:

PV=P[ \frac{1-(1+ \frac{r}{n})^{-kt} }{ \frac{r}{n} } ]
PV=1405[ \frac{1-(1+ \frac{0.048}{12})^{-(12)(20)} }{ \frac{0.048}{12} } ]
PV=216501.09

We can conclude that the expression that represents the value of the most money she can borrow is: PV=1405[ \frac{1-(1+ \frac{0.048}{12})^{-(12)(20)} }{ \frac{0.048}{12} } ]. Evaluating the expression we can prove that she can borrow $216,501.09
Paraphin [41]1 year ago
3 0

\frac{(1405)((1+0.004)^2^4^0-1)}{(0.004)(1+0.004)^2^4^0} Apex

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Step-by-step explanation:

The random variable <em>X</em> is defined as the amount of sodium consumed.

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If X ~ N (µ, σ²), then Z=\frac{X-\mu}{\sigma}, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z ~ N (0, 1).

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P(x_{1}

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Compute the value of <em>x</em>₁ and <em>x</em>₂ as follows:

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Thus, the middle 30% of American men consume between 9.5 grams to 10.1 grams of sodium.

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