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3 years ago

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George is drafting his dissertation paper for his doctoral program. The graph below shows the line of best fit for the data reco

rded on the number of
pages of the dissertation written as a function of the number of hours George spends on the draft each week.
Answers and Graph are in the picture(s)

2 answers:

Vsevolod [243]3 years ago

4 0

Answer: the answer to your question is D i graphed them all that one is correct D

sweet [91]3 years ago

3 0

$a. \: y = \frac{3}{2} x + 4 \\ \\ 1. \: \frac{3}{2} x + 4 = y \\ \\ 2. \: \frac{3}{2} x = y - 4 \\ \\ 3. \: \frac{ \frac{3}{2} x}{\frac{3}{2} } = \frac{y - 4}{2} \\ \\ x = \frac{2y - 8}{3}$

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Jared made 4 bird houses in 3 days. How many days will it take him to make 28 bird houses.

pogonyaev

4 times 7 = 28

3 times 7 = 21

It will take him 21 days to make 28 bird houses.

6 0

3 years ago

WILL GIVE BRAINLIEST Mr. and Mrs. Bailey hope to send their son to college in eleven years. How much money should they invest no

Ronch [10]

Mr. and Mrs. Bailey need to invest $2906.50 so as to send their son to college.

<h3>Compound interest</h3>

Compound interest is given by:

$A=P(1+\frac{r}{n})^{nt}$

where A is the amount after t years, P is initial amount, r is the rate and n is the times compounded per period

Given that n = 1, r = 9% = 0.09, A = $7500 t = 11. Hence:

$7500=P(1+\frac{0.09}{1} )^{1*11}\\\\P=\$2906.50$

Mr. and Mrs. Bailey need to invest $2906.50 so as to send their son to college.

Find out more on Compound interest at: brainly.com/question/24924853

7 0

2 years ago

A used automobile dealership recently reduced the price of a used compact car by 6%. If the price of the car before discount was

Alchen [17]

Answer:

The discount = $1,128

New price = $17,672

Step-by-step explanation:

Price of car before discount = $18,800

Discount given = 6% of price of car

Discount value in dollars = 6% of $18,800 = 6/100*$18,800 = $1,128

Thus, the discount is $1,128

New price = Price of car before discount - discount = $18,800- $1,128

New price = $17,672

7 0

3 years ago

6. Acceleration due to gravity varies according to places of the earth, why?

valentina_108 [34]

Answer:

The variation in apparent gravitational acceleration (g) at different locations on Earth is caused by two things.The distance between the centers of mass of two objects affects the gravitational force between them, so the force of gravity on an object is smaller at the equator compared to the poles.

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

A quadrilateral has vertices at $(0,1)$, $(3,4)$, $(4,3)$ and $(3,0)$. Its perimeter can be expressed in the form $a\sqrt2+b\sqr

seraphim [82]

Answer:

a + b = 12

Step-by-step explanation:

Given

Quadrilateral;

Vertices of (0,1), (3,4) (4,3) and (3,0)

$Perimeter = a\sqrt{2} + b\sqrt{10}$

Required

$a + b$

Let the vertices be represented with A,B,C,D such as

A = (0,1); B = (3,4); C = (4,3) and D = (3,0)

To calculate the actual perimeter, we need to first calculate the distance between the points;

Such that:

AB represents distance between point A and B

BC represents distance between point B and C

CD represents distance between point C and D

DA represents distance between point D and A

Calculating AB

Here, we consider A = (0,1); B = (3,4);

Distance is calculated as;

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$(x_1,y_1) = A(0,1)$

$(x_2,y_2) = B(3,4)$

Substitute these values in the formula above

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$AB = \sqrt{(0 - 3)^2 + (1 - 4)^2}$

$AB = \sqrt{( - 3)^2 + (-3)^2}$

$AB = \sqrt{9+ 9}$

$AB = \sqrt{18}$

$AB = \sqrt{9*2}$

$AB = \sqrt{9}*\sqrt{2}$

$AB = 3\sqrt{2}$

Calculating BC

Here, we consider B = (3,4); C = (4,3)

Here,

$(x_1,y_1) = B (3,4)$

$(x_2,y_2) = C(4,3)$

Substitute these values in the formula above

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$BC = \sqrt{(3 - 4)^2 + (4 - 3)^2}$

$BC = \sqrt{(-1)^2 + (1)^2}$

$BC = \sqrt{1 + 1}$

$BC = \sqrt{2}$

Calculating CD

Here, we consider C = (4,3); D = (3,0)

Here,

$(x_1,y_1) = C(4,3)$

$(x_2,y_2) = D (3,0)$

Substitute these values in the formula above

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$CD = \sqrt{(4 - 3)^2 + (3 - 0)^2}$

$CD = \sqrt{(1)^2 + (3)^2}$

$CD = \sqrt{1 + 9}$

$CD = \sqrt{10}$

Lastly;

Calculating DA

Here, we consider C = (4,3); D = (3,0)

Here,

$(x_1,y_1) = D (3,0)$

$(x_2,y_2) = A (0,1)$

Substitute these values in the formula above

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$DA = \sqrt{(3 - 0)^2 + (0 - 1)^2}$

$DA = \sqrt{(3)^2 + (- 1)^2}$

$DA = \sqrt{9 + 1}$

$DA = \sqrt{10}$

The addition of the values of distances AB, BC, CD and DA gives the perimeter of the quadrilateral

$Perimeter = 3\sqrt{2} + \sqrt{2} + \sqrt{10} + \sqrt{10}$

$Perimeter = 4\sqrt{2} + 2\sqrt{10}$

Recall that

$Perimeter = a\sqrt{2} + b\sqrt{10}$

This implies that

$a\sqrt{2} + b\sqrt{10} = 4\sqrt{2} + 2\sqrt{10}$

By comparison

$a\sqrt{2} = 4\sqrt{2}$

Divide both sides by $\sqrt{2}$

$a = 4$

By comparison

$b\sqrt{10} = 2\sqrt{10}$

Divide both sides by $\sqrt{10}$

$b = 2$

Hence,

a + b = 2 + 10

a + b = 12

3 0

3 years ago