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katovenus [111]
3 years ago
8

College administrators noticed that students who had higher high school GPAs tend to have higher college GPAs. The data in the t

able show various high school GPAs and college GPAs for a sample of students. A 2-column table with 16 rows. Column 1 is labeled High School G P A with entries 3.25, 3.5, 3.5, 3.25, 2.5, 4, 4, 3.5, 3.25, 3.5, 3.5, 4, 3, 3.5, 2.25, 2.5 Column 2 is labeled College G P A with entries 3.2, 3.0, 2.9, 3.0, 3.0, 3.0, 3.1, 3.1, 3.1, 3.0, 3.0, 3.1, 3.0, 3.1, 2.9, 2.8. Which scatterplot represents the student data? A graph titled high school and college G P A has H S G P A on the x-axis, and College G P A on the y-axis. Points are at (3.25, 3.2), (3.5, 3.0), (3.5, 2.9), (3.25), (3.0), (2.5, 3.0), (4, 3.0), (4, 3.1), (3.5, 3.1), (3.25, 3.1), (3.25, 3.1), (3.5, 3.0), (3.5, 3.0), (4, 3.1), (3, 3.0), (3.5, 3.1), (2.25, 2.9), (2.5, 2.8). A graph titled high school and college G P A has College G P A on the x-axis, and H S G P A on the y-axis. Points are at (2.8, 2.5), (2.9, 2.25), (2.9, 3.5), (3, 2.5), (3, 3), (3, 3.25), (3, 3.5), (3, 4), (3.1, 3.25), (3.1, 3.5), (3.1, 4), (3.2, 3.25). A graph titled high school and college G P A has H S G P A on the x-axis, and College G P A on the y-axis. Points are at (3.25, 3.2), (3.5, 3.0), (3.5, 2.9), (3.25), (3.0), (2.5, 3.0), (4, 3.0), (4, 3.1), (3.5, 3.1), (3.25, 3.1), (3.25, 3.1), (3.5, 3.0), (3.5, 3.0), (4, 3.1), (3, 3.0), (3.5, 3.1), (2.25, 2.9), (2.5, 2.8), (4, 3.2). A graph titled high school and college G P A has College G P A on the x-axis, and H S G P A on the y-axis. Points are at (2.8, 2.5), (2.9, 2.25), (2.9, 3.5), (3, 2.5), (3, 3), (3, 3.25), (3, 3.5), (3, 4), (3.1, 3.5), (3.1, 4), (3.2, 3.25).
Mathematics
1 answer:
riadik2000 [5.3K]3 years ago
4 0

Answer:

Points are at (3.25, 3.2), (3.5, 3.0), (3.5, 2.9), (3.25), (3.0), (2.5, 3.0), (4, 3.0), (4, 3.1), (3.5, 3.1), (3.25, 3.1), (3.5, 3.0), (3.5, 3.0), (4, 3.1), (3, 3.0), (3.5, 3.1), (2.25, 2.9), (2.5, 2.8)

Step-by-step explanation:

Given the data :

Column 1 High School G P A:

3.25, 3.5, 3.5, 3.25, 2.5, 4, 4, 3.5, 3.25, 3.5, 3.5, 4, 3, 3.5, 2.25, 2.5

Column 2 College G P A :

3.2, 3.0, 2.9, 3.0, 3.0, 3.0, 3.1, 3.1, 3.1, 3.0, 3.0, 3.1, 3.0, 3.1, 2.9, 2.8.

To prepare a scatter plot ; with x - axis plotted against y - axis ;

Each point on the x - axis is paired with a corresponding point on the y - axis ;

With high school GPA on the x axis and College GPA on y :

Values are paired in the form:

(X1, Y1),.. (Xn, Yn)

(3.25, 3.2)

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find the equation of a circle which passes through the point (2,-2) and (3,4) and whose centre lies on the line x+y=2
Nadusha1986 [10]

Answer:

Equation of the circle

(x - 0.7)² + (y - 1.3)² = 12.58

Step-by-step explanation:

The formula for the equation of a circle is given as:

(x - a)² + (y - b)² = r²,

where(a, b) is the center of the circle and r = radius of the circle.

a) We are told in the question that the equation of the circle passes through point(2, -2)

Hence,

Substituting 2 for x and -2 for y in the equation of the circle.

(x - a)² + (y - b)² = r²

(2 - a)² +(-2 - b)² = r²

Expanding the bracket

(2 - a) (2 - a) + (-2 - b)(-2 - b) = r²

4 - 2a - 2a +a² +4 +2b +2b +b² = r²

4 - 4a + a² + 4 + 4b + b² = r²

a² + b² -4a + 4b + 4 + 4 = r²

a² + b² -4a + 4b + 8 = r²............Equation 1

We are also told that the equation of the circle also passes through point (3,4) also, where 3 = x and 4 = y

Hence,

Substituting 3 for x and 4 for y in the equation of the circle.

(x - a)² + (y - b)² = r²

(3 - a)² +(4 - b)² = r²

Expanding the bracket

(3 - a) (3 - a) + (4 - b)(4- b) = r²

9 - 3a - 3a +a² +16 -4b -4b +b² = r²

9 -6a + a² + 16 -8b + b² = r²

a² + b² -6a -8b + 9 + 16 = r²

a² + b² -6a -8b + 25 = r²..........Equation 2

The next step would be to subtract Equation 1 from Equation 2

a² + b² -4a + 4b + 8 - (a² + b² -6a -8b + 25) = r² - r²

a² + b² -4a + 4b + 8 - a² - b² +6a +8b - -25= r² - r²

Collecting like terms

a² - a² + b² - b² - 4a + 6a + 4b + 8b +8- 25 = 0

2a + 12b -17 = 0

2a + 12b = 17...........Equation 3

Step 2

We are going to have to find the values of a and b in other to get our equation of the circle.

Since the center of the circle(a, b) lies on x + y = 2

Therefore, we have

a + b = 2

a = 2 - b

2a + 12b = 17 ..........Equation 3

Substituting 2 - b for a in

2(2 - b) + 12b = 17

4 - 2b + 12b = 17

4 + 10b = 17

10b = 17 - 4

10b = 13

b = 13/10

b = 1.3

Substituting 1.3 for b in

a + b = 2

a + 1.3 = 2

a = 2 - 1.3

a = 0.7

hence, a = 0.7, b = 1.3

Step 3

We have to find the value of r using points (2, -2)

(x - a)² + (y - b)² = r²

Where x = 2 and y = -2

(-2 - 0.7)² + (-2 - 1.3)² = r²

(-2.7)² + (-3.3)² = r²

1.69 + 10.89 = r²

r² = 12.58

r = √12.58 = 3.55

Step 4

The formula for the equation of a circle is given as:

(x - a)² + (y - b)² = r²,

where(a, b) is the center of the circle and r = radius of the circle

a = 0.7

b = 1.3

r² = 12.58

Equation of the circle =

(x - 0.7)² + (y - 1.3)² = 12.58

7 0
3 years ago
Help pls I attached a picture
seropon [69]

Answer:

x= 7\sqrt{2}

Step-by-step explanation:

It's a 45 45 90 triangle

The legs are = x while the hypotenuse is x \sqrt{2}.

Therefore the answer is 7\sqrt{2}

8 0
3 years ago
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