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lys-0071 [83]
3 years ago
14

Item 20 You are riding a roller coaster when a shoe falls off your foot. The function y=200−16t2 represents the height y (in fee

t) of the shoe t seconds after it falls off your foot. The shoe lands on the top of a 31-foot-tall building. After how many seconds does the shoe hit the building?
Mathematics
2 answers:
AleksandrR [38]3 years ago
7 0
Assuming you mean y = 200 - 16t^2, we have all the required information needed to solve this problem. The y-value is the height of the building and the t-value represents the number of seconds after the shoe fell off.

Since we are trying to solve <em>for t</em>, we will be using our y-value. The problem states that we landed on a building with a height of 31 feet. We can plug this into the y-value, since that is what y is defined as (the height of the building).

Now we have:
31 = 200 - 16t^2

We can solve this to find t:
-169 = -16t^2
169 = 16t^2
10.5625 = t^2
3.25, -3.25 = t

We have found two answers for t. However, the negative value is not a solution because we can not have a negative number of seconds. Thus, 3.25 is the only value of t which works for this problem.

Since t is the value we are trying to find, we have our answer. The shoe hits the building after 3.25 seconds.
Gekata [30.6K]3 years ago
5 0

So what is the correct answer ?

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Answer:

The mortgage chosen is option A;

15-year mortgage term with a 3% interest rate because it has the lowest total amount paid over the loan term of $270,470

Step-by-step explanation:

The details of the home purchase are;

The price of the home = $275,000

The mode of purchase of the home = Mortgage

The percentage of the loan amount payed as down payment = 20%

The amount used as down payment for the loan = $55,000

The principal of the mortgage borrowed, P = The price of the house - The down payment

∴ P = $275,000 - 20/100 × $275,000 = $275,000 - $55,000 = $220,000

The principal of the mortgage, P = $220,000

The formula for the total amount paid which is the cost of the loan is given as follows;

Outstanding \ Loan \ Balance = \dfrac{P \cdot \left[\left(1+\dfrac{r}{12} \right)^n -  \left(1+\dfrac{r}{12} \right)^m \right] }{1 - \left(1+\dfrac{r}{12} \right)^n }

The formula for monthly payment on a mortgage, 'M', is given as follows;

M = \dfrac{P \cdot \left(\dfrac{r}{12} \right) \cdot \left(1+\dfrac{r}{12} \right)^n }{\left(1+\dfrac{r}{12} \right)^n - 1}

A. When the mortgage term, t = 15-years,

The interest rate, r = 3%

The number of months over which the loan is payed, n = 12·t

∴ n = 12 months/year × 15 years = 180 months

n = 180 months

The monthly payment, 'M', is given as follows;

M =

The total amount paid over the loan term = Cost of the mortgage

Therefore, we have;

220,000*0.05/12*((1 + 0.05/12)^360/( (1 + 0.05/12)^(360) - 1)

M = \dfrac{220,000 \cdot \left(\dfrac{0.03}{12} \right) \cdot \left(1+\dfrac{0.03}{12} \right)^{180} }{\left(1+\dfrac{0.03}{12} \right)^{180} - 1}  \approx 1,519.28

The minimum monthly payment for the loan, M ≈ $1,519.28

The total amount paid over loan term, A = n × M

∴ A ≈ 180 × $1,519.28 = $273,470

The total amount paid over loan term, A ≈ $270,470

B. When t = 20 year and r = 6%, we have;

n = 12 × 20 = 240

\therefore M = \dfrac{220,000 \cdot \left(\dfrac{0.06}{12} \right) \cdot \left(1+\dfrac{0.06}{12} \right)^{240} }{\left(1+\dfrac{0.06}{12} \right)^{240} - 1}  \approx 1,576.15

The total amount paid over loan term, A = 240 × $1,576.15 ≈ $378.276

The monthly payment, M = $1,576.15

C. When t = 30 year and r = 5%, we have;

n = 12 × 30 = 360

\therefore M = \dfrac{220,000 \cdot \left(\dfrac{0.05}{12} \right) \cdot \left(1+\dfrac{0.05}{12} \right)^{360} }{\left(1+\dfrac{0.05}{12} \right)^{360} - 1}  \approx 1,181.01

The total amount paid over loan term, A = 360 × $1,181.01 ≈ $425,163

The monthly payment, M ≈ $1,181.01

The mortgage to be chosen is the mortgage with the least total amount paid over the loan term so as to reduce the liability

Therefore;

The mortgage chosen is option A which is a 15-year mortgage term with a 3% interest rate;

The total amount paid over the loan term = $270,470

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See explanation below.

Step-by-step explanation:

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Answer:

D. The center remains constant, and the area in the tails of the distribution decreases.

Step-by-step explanation:

Hello!

Be it two independent random variables, X~N(μ;σ²) and U~Xₙ², the variable t is determined by the quotient between a random variable N(0;1) and the square root of a Chi-Square variable divided by its degrees of freedom:

t= \frac{(X-Mu)/Sigma}{\sqrt{U/n} }

As a consequence of this, the structure of the distribution depends on the parameter n (degrees of freedom), it is centered in zero and has a bell-shape similar to the normal distribution.

It has a mean and variance:

E(t)= 0 for n > 1

V(t)= \frac{n}{n - 2} n > 2

As you can see the variance of the distribution is directly affected by its degrees of freedom, which means that when the degrees of freedom change, the variance of the distribution change and so does its shape.

When ↑n ⇒ ↑V(t) ⇒ The area under the tails increases.

When ↓n ⇒ ↓V(t) ⇒ The area under the tails decreases.

In this example, the degrees of freedom of the distribution decreased from 40 to 20, then the variance of the distribution decreases and it "flattens", i.e. the area under the tails gets lowered.

The E(t) isn't affected by the modification of n, so the distribution remains centered in zero.

I hope this helps!

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