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Katen [24]
2 years ago
13

Which regression line properly describes the data relationship in the scatterplot? On a graph, a trend line has a positive slope

. There are 4 points above the line, and 2 points below. On a graph, a trend line has a positive slope. There is 1 point above the line, and 5 points below. On a graph, a trend line has a positive slope. There are 3 points above the line, and 3 points below. On a graph, a trend line has a positive slope. There are 5 points above the line, and 1 point below.

Mathematics
2 answers:
Elena-2011 [213]2 years ago
7 0

Answer:

Graph C

Step-by-step explanation:

I just took the quiz on edge

nevsk [136]2 years ago
4 0

Answer:

I believe it is C

Step-by-step explanation

I'm taking it on edg 2020

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The diameter of the sun is about 1.4x10^6 km. The diameter of the planet Mercury is about 5000 km. What is the approximate ratio
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2 years ago
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Point \blue{A}Astart color #6495ed, A, end color #6495ed is at \blue{(-2, -7)}(−2,−7)start color #6495ed, left parenthesis, minu
Lina20 [59]

Answer:

(7,4)

Step-by-step explanation:

The midpoint formula is expressed as:

M(X, Y) = {(x2+x1)/2, (y2+y1)/2

Given the coordinates of the midpoint

M(2.5, -1.5)

Endpoint A(-2, -7)

Required

Endpoint B

From the formula;

X = x2+x1/2

Given

X = 2.5, x1 = -2

Get x2

2.5 = -2+x2/2

5 = -2+x2

x2 = 5+2

x2 = 7

Also

Y = y2+y1/2

-1.5 = -7+y2/2

-3 =-7+y2

y2 = -3+7

y2 = 4

Hence the coordinate of the endpoint B (x2,y2) is (7,4)

3 0
2 years ago
A group of 5 friends decides to share the cost of a party together. If the party costs $1,100, how much does each friend pay?
seraphim [82]

Answer:

1,100 dollars divided by 5 friends is 220 dollars so each friend would have to pay 220 dollars

Step-by-step explanation:


7 0
2 years ago
A circle is centered at the point -3 2 and passes through the point 1 and 5 the radius of the circle is blank units
photoshop1234 [79]

Answer:

The radius of the circle is 5 units

Step-by-step explanation:

Given in the question,

centered point = (-3 , 2)

a point in the circle = (1,5)

To find the radius of circle we will use the circle equation

(x – h)² + (y – k)² = r²,

with the centre being at the point (h, k) and the radius being r.

Plug values in the equation

(1 – (-3))² + (5 – 2)² = r²

(1+3)² + (3)² = r²

4² + 3² = r²

25 = r²

r = √25

r = 5

7 0
2 years ago
You are traveling down a country road at a rate of 95 feet/sec when you see a large cow 300 feet in front of you and directly in
emmainna [20.7K]

Answer:

1) You can rely solely on your brakes because when doing so the car will just travel 250ft from the point you hit your brakes till the point the car stopped completely, leaving you 50ft away from the cow.

2) See attached picture.

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in ft/s^{2}

3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) j(t)=\left \{ {{95t-9t^{2}; 0\le t

j'(t)=\left \{ {{95-18t; 0\leq t

(see attached picture for graph)

Step-by-step explanation:

1) In this part of the problem we need to find the time when the speed of the car is 0. Gets to a complete stop. For this we will need to take the derivative of the position function so we get:

j(t)=95t-9t^2

j'(t)=95-18t

and we set the first derivative equal to zero so we get:

95-18t=0

and solve for t

-18t=-95

t=\frac{95}{18}

t=5.28s

so now we calculate the position of the car after 5.28 seconds, so we get:

j(5.28)=95(5.28)-9(5.28)^{2}

j(5.28)=250.69ft

so we have that the car will stop 250.69ft after he hit the brakes, so there will be about 50ft between the car and the cow when the car stops completely, so he can rely just on the breaks.

2) For answer 2 I take the second derivative of the function so I get:

j(t)=95t-9t^{2}

j'(t)=95-18t

j"(t)=-18

and then we graph them. (See attached picture)

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in ft/s^{2}

3)  yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) j(t)=\left \{ {{95t-9t^{2}; 0\le t

j'(t)=\left \{ {{95-18t; 0\leq t

(see attached picture for graph)

5 0
2 years ago
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