Answer:
See it in the pic
Step-by-step explanation:
See it in the pic
Steve determined the number of states each of his friends has visited. A number line goes from 0 to 44. The whiskers range from
2 answers:
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X, y - the lengths of the sides of the pathway
The area of a rectangle is the product of the lengths of its sides. The area is 350 m².
The perimeter of a rectangle is two times the sum of the lengths of its sides. The perimeter is 90 m.
Substitute 45-x for y in the first equation:
The dimensions of the pathway are 10 m by 35 m.
The length of the pathway is 35 m.
The area of a rectangle is the product of the lengths of its sides. The area is 350 m².
The perimeter of a rectangle is two times the sum of the lengths of its sides. The perimeter is 90 m.
Substitute 45-x for y in the first equation:
The dimensions of the pathway are 10 m by 35 m.
The length of the pathway is 35 m.
Answer:
The selling price of house is approximately 270.4264 thousand dollars.
Step-by-step explanation:
We are given the following in the question:
A linear model gives the relation between natural logarithm of price, in thousands of dollars, and size, in 100 square feet.
Let p be the price and s be the size.
We have to approximate the selling price for a house with a size of 3,200 square feet.
Thus, we put s = 32
Thus, the selling price of house is approximately 270.4264 thousand dollars.
Answer:
(a) The linear depreciation function of the car which gives the worth of the car y after a number of years, t is given as follows;
y = 25,425 - 2,739 × t
(b) The value of the car 7 years from now is $6,252
(c) $2,739 per year
Step-by-step explanation:
The manufacture's suggested retail price (MSRP) for the car = $25,425
The amount the car is expected to be worth in 5 years = $11,730
(a) The linear depreciation is given as follows;
Where;
Cost of Asset = $25,425
Salvage Value = $11,730
Life of Asset in use = 5 years
We get;
Therefore, the linear depreciation function, can be written as follows;
y - 25,425 = -2,739×(t - 0)
y = -2,739·t + 0 + 25,425 = 25,425 -2,739·t
y = 25,425 - 2,739 × t
Where;
y = The expected worth of the car after a given number of years
t = The number of years used for the calculation of the depreciation
(b) The value of the car 7 years from now is given by substitution as follows;
Whet t = 7, we have;
y = 25,425 -2,739·t = y = 25,425 -2,739 × 7 = $6,252
The value of the car 7 years from now = $6,252
(c)
The car is depreciating at a rate of $2,739 per year.