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Juliette [100K]
3 years ago
14

Use the distance formula to write an equation of the parabola with focus F(0, -13) and directrix y=13

Mathematics
1 answer:
babunello [35]3 years ago
3 0

Answer:

-1/52 x^2

Step-by-step explanation:

4p(y-k)=(x-h)^2

4(-13)y=x^2

-52y=x^2

y=-1/52 x^2

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Which graph shows the same end behavior as the graph of f(x) = 2x6 - 2x2 - 5?
aleksandrvk [35]

Step-by-step explanation:

End behavior of a polynomial function is the behavior of the graph of f(x) as x tends towards infinity in the positive or negative sense.

  Given function:

            f(x)  = 2x⁶   -    2x²   -  5

To find the end behavior of a function:

  • Find the degree of the function. it is the highest power of the variable.

  Here the highest power is 6

  • Find the value of the leading coefficient. It is the number before the variable with the highest power.

   Here it is  +2

We observe that the degree of the function is even

         Also the leading coefficient is positive.

For even degree and positive leading coefficient, the end behavior of a graph is:

x → ∞ , f(x) = +∞

x → -∞ , f(x) = +∞

The graph is similar to the attached image

Learn more:

End behavior brainly.com/question/3097531

#learnwithBrainly

6 0
3 years ago
A statistician chooses 27 randomly selected dates, and when examining the occupancy records of a particular motel for those date
Anni [7]

Answer:

Confidence interval variance [21.297 ; 64.493]

Confidence interval standard deviation;

4.615, 8.031

Step-by-step explanation:

Given :

Variance, s² = 34.34

Standard deviation, s = 5.86

Sample size, n = 27

Degree of freedom, df = 27 - 1 = 26

Using the relation for the confidence interval :

[s²(n - 1) / X²α/2, n-1] ; [s²(n - 1) / X²1-α/2, n-1]

From the chi distribition table :

X²α/2, n-1 = 41.923 ; X²1-α/2, n-1 = 13.844

Hence,

[34.34*26 / 41.923] ; [34.34*26 / 13.844]

[21.297 ; 64.493]

The 95% confidence interval for the population variance is :

21.297 < σ² < 64.493

Standard deviation is the square root of variance, hence,

The 95% confidence interval for the population standard deviation is :

4.615 < σ < 8.031

3 0
2 years ago
What percent of 76 is 19?
olga_2 [115]
7.5 is the answer to your question 
8 0
3 years ago
Read 2 more answers
What is the answer i need plsss
earnstyle [38]
Do 2-1 that should work

5 0
3 years ago
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Imagine that you would like to purchase a $275,000 home. Using 20% as
vfiekz [6]

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

8 0
3 years ago
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