Compare one third to sixths

$3000 is invested at 3% annual interest for 3 years. a. Simple Interest b. Compound Interest

stich3 [128]

Answer:

a. $270

b. $3,278.18

Step-by-step explanation:

Given that

The principal amount is $3,000

Annual rate of interest is 3%

And, the time period is 3 years

We need to find out the simple interest & compound interest

The following formulas should be used

a. For simple interest

= Principal × rate of interest × time period

= $3,000 × 3% × 3 years

= $270

b. For compound interest

= Principal × (1 + rate of interest)^time period

= $3,000 × (1 + 0.03)3

= $3,000 × 1.03^3

= $3,278.18

8 0

2 years ago

A regression analysis of 117 homes for sale produced the following model, where price is in thousands of dollars and size in squ

Vsevolod [243]

Answer:

A) m = 0.062

B) 202.87 thousand dollars

C) Original price = 116.07 thousand dollars. $6000 is called the error term.

Step-by-step explanation:

We are given the following in the question:

Price = 47.87+0.062(size)

The above equation is regression equation where price is in thousand dollars and size in square feet.

Let p be the prize and s be the size.

$p(s) = 47.87 + 0.062s$

Comparing it to linear equation, we have,

$y= mx + c$

where m is the slope and c is the y-intercept.

m = 0.062

c = 47.87

A) slope of the line

The slope of line tells the rate of change. Thus, it tells the change in price if the size of house is increased by 1 square feet.

The price of house increases by 0.062 thousand dollars if the size of house is increased by 1 square feet.

B) Price of house

We are given s = 2500. Putting this value in the equation.

$p(2500) = 47.87 + 0.062(2500) =202.87$

Thus, the price of a 2500 square feet house is 202.87 thousand dollars.

C) We are given s = 1100

Putting the value in the equation:

$p(1100) = 47.87 + 0.062(1100) =116.07$

Original price = 116.07 thousand dollars

Asking price =

$116.07 - 6 = 110.07$

The buyer is asking for 110.07 thousand dollars.

$6000 is called the error term.

3 0

3 years ago

The coordinates of the vertices of the triangle are (–8, 8), (–8, –4), and . Consider QR the base of the triangle. The measure o

Mademuasel [1]

The coordinates of the vertices of the triangle are
(–8, 8), (–8, –4), and (10, –4).

Consider QR the base of the triangle. The measure of the base is b = 18 units, and the measure of the height is h = 12 units.

The area of triangle PQR is108 square units.

4 0

3 years ago

Read 2 more answers

A cheetah's speed was timed over a set distance. The cheetah was clocked running 60 miles per hour. Which equation is the relati

Troyanec [42]

Answer:

D. y=70x

Step-by-step explanation:

The cheetah was clocked running 70 miles per hour.

i.e. in 1 hour cheetah covered 70 miles

in 2 hours cheetah covered 140 miles

in 3 hours cheetah covered 210 miles

in general, in n hour cheetah covered 70n miles

If y denote the distance covered

and x denotes the time cheetah runs

Then, y=70x

Hence, the equation which shows the relationship between the distance, y, and time, x, the cheetah runs is:

D y=70x

8 0

3 years ago

find the parametric equations for the line of intersection of the two planes z = x + y and 5x - y + 2z = 2. Use your equations t

Kaylis [27]

Answer:

You didn't give the points in which you want the parametric equations be filled, but I have obtained the parametric equations, and they are:

x = (1/3 + t)

y = (-1/3 - 7t)

z = -6t

Step-by-step explanation:

If two planes intersect each other, the intersection will always be a line.

The vector equation for the line of intersection is given by

r = r_0 + tv

where r_0 is a point on the line and v is the vector result of the cross product of the normal vectors of the two planes.

The parametric equations for the line of intersection are given by

x = ax, y = by, and z = cz

where a, b and c are the coefficients from the vector equation

r = ai + bj + ck

To find the parametric equations for the line of intersection of the planes.

x + y - z = 0

5x - y + 2z = 2

We need to find the vector equation of the line of intersection. In order to get it, we’ll need to first find v, the cross product of the normal vectors of the given planes.

The normal vectors for the planes are:

For the plane x + y - z = 0, the normal vector is a⟨1, 1, -1⟩

For the plane 5x - y + 2z = 2, the normal vector is b⟨5, -1, 2⟩

The cross product of the normal vectors is

v = a × b =

|i j k|

|1 1 -1|

|5 -1 2|

= i(2 - 1) - j(2 + 5) + k(-1 - 5)

= i - 7j - 6k

v = ⟨1, -7, -6⟩

We also need a point on the line of intersection. To get it, we’ll use the equations of the given planes as a system of linear equations. If we set z = 0 in both equations, we get

x + y = 0

5x - y = 2

Adding these equations

5x + x + y - y = 2 + 0

6x = 2

x = 1/3

Substituting x = 1/3 back into

x + y = 0

y = -1/3

Putting these values together, the point on the line of intersection is

(1/3, -1/3, 0)

r_0= (1/3) i - (1/3) j + 0 k

r_0 = ⟨1/3, -1/3, 0⟩

Now we’ll plug v and r_0 into the vector equation.

r = r_0 + tv

r = (1/3)i - (1/3)j + 0k + t(i - 7j - 6k)

= (1/3 + t) i - (1/3 + 7t) j - 6t k

With the vector equation for the line of intersection in hand, we can find the parametric equations for the same line. Matching up r = ai + bj + ck with our vector equation,

r = (1/3 + t) i + (-1/3 - 7t) j + (-6t) k

a = (1/3 + t)

b = (-1/3 - 7t)

c = -6t

Therefore, the parametric equations for the line of intersection are

x = (1/3 + t)

y = (-1/3 - 7t)

z = -6t

3 0

3 years ago