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3 years ago

Solve for the given time.

Mathematics

1 answer:

AlekseyPX3 years ago

3 0

9514 1404 393

Answer:

$319,525.64

Step-by-step explanation:

It appears you want the interest to accumulate for 15 years before you pay anything on the loan. The value of $180,000 appreciating at 3.9% annually will be ...

$180,000·(1 +0.039)^15 = $319,525.64

If your lender allows you to wait 15 years before making any payment, your payment will be $319,525.64.

<em>Additional comment</em>

Home loans are usually paid with monthly payments computed from the formula ...

A = P(r/12)/(1 -(1 +r/12)^(-12t))

where A is the monthly payment on principal P, r is the annual interest rate, t is the number of years.

For your $180,000 loan, the monthly payment is ...

A = $180,000·(0.039/12)/(1 -(1 +0.039/12)^(-12·15)) ≈ $1322.44

The total of those payments is ...

180·$1322.44 = $238,039.20

If you pay $1322.44 monthly on the loan, your total payment is $238,039.20.

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4 years ago

Read 2 more answers

The number of chocolate chips in an 18-ounce bag of Chips Ahoy! chocolate chip cookies is approximately normally distributed wit

sineoko [7]

Answer:

a) 0.013

b) 0.7

c) 0.866

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 1262

Standard Deviation, σ = 118

We are given that the distribution of number of chocolate chips is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(fewer than 1000 chocolate chips)

P(x < 1000)

$P( x < 1000) = P( z < \displaystyle\frac{1000 - 1262}{118}) = P(z < -2.220)$

Calculation the value from standard normal z table, we have,

$P(x < 1000) = 0.013 = 1.3\%$

b) P(contains more than 1200 chocolate chips)

$P(x > 1200) = P(z > \displaystyle\frac{1200-1262}{118}) = P(z > -0.5254)\\\\P( z > -0.5254) = 1 - P(z \leq -0.5254)$

Calculating the value from the standard normal table we have,

$1 - 0.300 = 0.7 = 70\%\\P( x > 1200) = 0.7$

c)P(contains between 1000 and 1400 chocolate chips)

$P(1000 \leq x \leq 1400) = P(\displaystyle\frac{1000 - 1262}{118} \leq z \leq \displaystyle\frac{1400-1262}{118}) = P(-2.22 \leq z \leq 1.169)\\\\= P(z \leq 1.169) - P(z < -2.22)\\= 0.879 - 0.013 = 0.866 = 86.6\%$

$P(1000 \leq x \leq 1400) = 86.6\%$

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3 years ago