Find the regression line associated with the set of points. (Round all coefficients to four decimal places.) HINT [See Example 2.] (4, 6), (6, 10), (10, 14), (12, 2) y(x) =

The lengths of the other two sides of the right triangle are 36 and 85, respectively.

To find the lengths of the other two sides of a right triangle when one leg has a length of 11, we can use the Pythagorean theorem.

The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

Let's denote the lengths of the other leg and the hypotenuse as x and y, respectively.

According to the Pythagorean theorem, we have:

x² + 11² = y²

To find the values of x and y, we need to find a pair of whole numbers that satisfy this equation.

We can start by checking for perfect squares that differ by 121 (11^2). One such pair is 36 and 85.

If we substitute x = 36 and y = 85 into the equation, we have:

36² + 11² = 85²

1296 + 121 = 7225

This equation is true, so the lengths of the other two sides are:

The other leg = 36

The hypotenuse = 85

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The predicted sales in month 5 is -2778.

Obtaining the linear equation which models the data :

b = slope = (y2-y1)/(x2-x1)

b = (926-7408)/(4-1)

b = -2160.67

c = intercept ;

taking the points (x = 2 and y = 3704)

Inserting into the general equation:

3704 = -2160.67(2) + c

3704 = -4321.33 + c

c = 3704 + 4321.33

c = 8025.33

General equation becomes : y = -2160.67x + 8025.33

To obtain sales in month 5:

y = -2160.67(5) + 8025.33

y = -2778

Hence, the predicted sales in month 5 is -2778.

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At the end of her fifth round, Charimaya would have covered a distance of 1500 meters.

To find the distance covered by Charimaya at the end of her fifth round, we need to calculate the total distance covered in one round and then multiply it by five.

Given that the track is square-shaped with a length of 75 m, we know that all four sides of the track are equal in length.

To calculate the distance covered in one round, we need to find the perimeter of the square track. Since all sides are equal, we can simply multiply the length of one side by 4.

The length of one side of the square track is 75 m. Therefore, the perimeter of the track is:

Perimeter = 4 × 75 m = 300 m

So, Charimaya covers a distance of 300 m in one round.

To find the distance covered at the end of her fifth round, we multiply the distance covered in one round by 5:

Distance covered in 5 rounds = 300 m × 5 = 1500 m

Therefore, at the end of her fifth round, Charimaya would have covered a distance of 1500 meters.

It's worth noting that since the track is square-shaped, each round consists of running along all four sides of the track.

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Answer:

[tex](x+5)^2+(y+5)^2=25[/tex]

Step-by-step explanation:

Recall that the equation of a circle with center (h,k) and radius "r" is [tex](x-h)^2+(y-k)^2=r^2[/tex]

Since the center of the circle is (h,k)=(-5,-5) and the radius is r=5, then our equation will be [tex](x-(-5))^2+(y-(-5))^2=5^2[/tex] which can be simplified into [tex](x+5)^2+(y+5)^2=25[/tex]

Hola!

-4 - 6

= -10

the answer is -10

Answer:

x = 8

Step-by-step explanation:

The given diagram shows a triangle with the length of two sides and its included angle.

To find the value of the missing variable x, we can use the Law of Cosines.

[tex]\boxed{\begin{minipage}{6 cm}\underline{Law of Cosines} \\\\$c^2=a^2+b^2-2ab \cos C$\\\\where:\\ \phantom{ww}$\bullet$ $a, b$ and $c$ are the sides.\\ \phantom{ww}$\bullet$ $C$ is the angle opposite side $c$. \\\end{minipage}}[/tex]

From inspection of the given triangle:

Substitute the values into the formula and solve for x:

[tex]\begin{aligned}x^2&=18^2+21^2-2(18)(21)\cos 22^{\circ}\\x^2&=324+441-756\cos 22^{\circ}\\x^2&=765-756\cos 22^{\circ}\\x&=\sqrt{765-756\cos 22^{\circ}}\\x&=8.00306228...\\x&=8\end{aligned}[/tex]

Therefore, the value of the missing variable x is x = 8, rounded to the nearest hundredth.

Answer:

Step-by-step explanation:

If {0, 3, 6, 9} are are your x's or domain  or input and there are no repeats, then yes TRUE it is a function.

Answer:

D. A pair of intersecting lines

Step-by-step explanation:

A conic section is a fancy name for a curve that you get when you slice a double cone with a plane. Imagine you have two ice cream cones stuck together at the tips, and you cut them with a knife. Depending on how you cut them, you can get different shapes. These shapes are called conic sections, and they include circles, ellipses, parabolas and hyperbolas. If you cut them right at the tip, you get a point. If you cut them slightly above the tip, you get a line. If you cut them at an angle, you get two lines that cross each other. That's what happened in your question. The plane cut the cone at an angle, so the curve is two intersecting lines. That means the correct answer is D. A pair of intersecting lines.

I hope this helps you ace your math question.

17x = 425

x = 25

8x = 200 boys

9x = 225 girls

Answer:

127

Step-by-step explanation:

Angle C+Angle D=Angle ABC

Since C+D+CBD=180 and ABC+CBD=180

subtract getting C+D-CBD=0 and C+D=CBD

so 67+60=127 which is your answer

*Just to clarify, when i said C and D, i meant angle C and angle D

Answer:

(2, 7) and (-6, -1)

Step-by-step explanation:

y = x + 5

y = x² + 5x − 7

Equatig the above,

x² + 5x − 7 = x + 5

⇒ x² + 4x −12 = 0

⇒ x² + 6x - 2x - 12 = 0

⇒ x(x + 6) - 2(x + 6) = 0

⇒  (x - 2)(x + 6) = 0

⇒ x = 2 or x = -6

Eq(1) : y = x + 5 (given)

When x = 2

y = 2 + 5 = 7

Point : (2, 7)

When x = -6

y = -6 + 5 = -1

Point: (-6, -1)

The amount of snow accumulated between 2 am and 5 am is: 1.25 inches

The general formula for the equation of a line in slope intercept form is:

y = mx + c

where:

m is slope

c is y-intercept

From the given graph attached, we see that the y-axis gives the amount of snow at different specific times.

Meanwhile the x-axis gives the time in hours after midnight

At 2am, the y-axis value is 1.25 inches, and as such at 2am snow accumulation was 1.25 inches.

At 5 am, the y-axis value reads 2.5 inches, and as such at 5am snow accumulation was 2.5 inches.

The difference in both snow accumulations is:  2.5 - 1.25 = 1.25

Hence, 1.25 inches snow accumulated between 2 am and 5 am.

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Answer:

b. 160°

d. 55°

Step-by-step explanation:

The Inscribed Angle Theorem states that an inscribed angle is half of the central angle that subtends the same arc.

In other words, if an angle is inscribed in a circle and it intercepts an arc, then the measure of the inscribed angle is equal to half the measure of the central angle that also intersects that arc.

For question:

b.

By using above theorem:

m arc XW=2* m arc XYW

m arc XW= 2*80=160°

d.

m arc WV=125°

The Inscribed Angle Diameter Right Angle Theorem states that any angle inscribed in a circle that intercepts a diameter is a right angle.

By using this theorem:

m arc WV+m arc XV =180°

Now

m arc XV =180°-m arc WV

m arc XV=180°-125°

n arc XV=55°

Answer:

[tex]\text{b.} \quad m\overset{\frown}{XW}=160^{\circ}[/tex]

[tex]\text{d.} \quad m\overset{\frown}{XV}=55^{\circ}[/tex]

Step-by-step explanation:

An inscribed angle is the angle formed (vertex) when two chords meet at one point on a circle.

An intercepted arc is the arc that is between the endpoints of the chords that form the inscribed angle.

[tex]\hrulefill[/tex]

From inspection of the given circle:

According to the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of the intercepted arc. Therefore:

[tex]m \angle WRX = \dfrac{1}{2}\overset{\frown}{XW}[/tex]

         [tex]80^{\circ}= \dfrac{1}{2}\overset{\frown}{XW}[/tex]

   [tex]\boxed{m\overset{\frown}{XW}=160^{\circ}}[/tex]

[tex]\hrulefill[/tex]

From inspection of the given circle:

According to the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of the intercepted arc. Therefore:

[tex]m \angle WVX= \dfrac{1}{2}\overset{\frown}{WX}[/tex]

         [tex]90^{\circ}= \dfrac{1}{2}\overset{\frown}{WX}[/tex]

    [tex]m\overset{\frown}{WX}=180^{\circ}[/tex]

The sum of the measures of the arcs in a circle is 360°.

[tex]m\overset{\frown}{VW}+m\overset{\frown}{WX}+m\overset{\frown}{XV}=360^{\circ}[/tex]

Therefore, so find the measure of arc XV, substitute the found measures of arcs VW and WX, and solve for arc XV:

[tex]125^{\circ}+180^{\circ}+m\overset{\frown}{XV}=360^{\circ}[/tex]

          [tex]305^{\circ}+m\overset{\frown}{XV}=360^{\circ}[/tex]

                    [tex]\boxed{m\overset{\frown}{XV}=55^{\circ}}[/tex]

Answer:

Step-by-step explanation:To calculate the value of x, we can use the formula for the area of a trapezoid:

Area = (1/2) * (sum of the parallel sides) * height

Given that the area of the trapezoid ABCD is 70 square units, we can set up the equation as follows:

70 = (1/2) * (AB + DC) * Height

Substituting the given values:

70 = (1/2) * ((2x + 4) + (x - 2)) * 4

Simplifying the equation:

70 = (1/2) * (3x + 2) * 4

Multiplying both sides by 2 to remove the fraction:

140 = (3x + 2) * 4

Dividing both sides by 4:

35 = 3x + 2

Subtracting 2 from both sides:

33 = 3x

Dividing both sides by 3:

x = 11

Therefore, the value of x is 11.

The exponential decay function that satisfies the given conditions is:

[tex]f(x) = 4 * (1/2)^x[/tex].

In this equation, the y-intercept is 4, which means that when x = 0, the function value is 4. As x increases by 1, the function decreases by a factor of one-half. This behavior is captured by raising 1/2 to the power of x in the equation.

The base of the exponent, 1/2, ensures that the function decreases exponentially. When x = 1, the exponent becomes 1, and[tex]1/2^1[/tex] equals 1/2. This means that the function value decreases to half of its previous value. Similarly, when x = 2, the exponent becomes 2, and[tex]1/2^2[/tex] equals 1/4. The function value decreases to one-fourth of its previous value, and so on.

By multiplying the exponential term by 4, we ensure that the y-intercept is 4. This scaling factor allows us to control the initial value of the function and match the given condition.

The exponential decay function[tex]f(x) = 4 * (1/2)^x[/tex] represents a decaying process where the y-values decrease exponentially as x increases, while starting at a y-intercept of 4.

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10 and 5 are appropriate x-axis and y-axis unit scales given the coordinates (50, 40). Option B.

To determine the appropriate x-axis and y-axis unit scales given the coordinates (50, 40), we need to consider the relationship between the units on the axes and the corresponding values in the coordinate system.

The x-axis represents the horizontal dimension, while the y-axis represents the vertical dimension. The x-coordinate (50) represents a position along the x-axis, and the y-coordinate (40) represents a position along the y-axis.

To determine the appropriate scales, we need to consider how many units on the x-axis are needed to span the distance from 0 to 50 and how many units on the y-axis are needed to span the distance from 0 to 40.

In this case, the x-coordinate is 50, which means we need the x-axis to span a distance of 50 units. However, we don't have enough information to determine the scale for the x-axis accurately. Therefore, options A (50; 10) and D (50; 1) cannot be definitively chosen.

Similarly, the y-coordinate is 40, which means we need the y-axis to span a distance of 40 units. Considering the given options, option B (10; 5) would be a suitable scale, as it allows for the y-axis to span the necessary distance of 40 units.

In summary, given the coordinates (50, 40), the appropriate unit scales would be 10 units per increment on the x-axis and 5 units per increment on the y-axis (Option B).

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Answer:

7.80 ounces can be bought for $1.95

Step-by-step explanation:

Step 1:  Determine how many ounces is in a pound:

Thus, 16 ounces cost $4.

Step 2:  Create a proportion to determine how many ounces can be bought for $1.95.

Since you can get 16 ounces for $4, we can create a proportion to determine how many ounces can be bought for $1.95:

16 ounces / $4 = x ounces / $1.95

Step 3:  Simplify on the left-hand side of the equation:

16/4 = x/1.95

4 = x/1.95

Step 4: multiply both sides by 1.95 to determine how many ounces can be bought for $1.95:

(4 = x/1.95) * 1.95

7.80 = x

Thus, 7.80 ounces can be bought for $1.95.

Answer:

-21

Step-by-step explanation:

2+7-8-13-4-5

9-30

-21

Hope it's helpful.

The name of the Platonic solid that resembles a cuboid is the hexahedron, or more commonly known as a cube.

The correct answer is option C.

The name of the Platonic solid that resembles a cuboid is the hexahedron, also known as a cube. The hexahedron is one of the five Platonic solids, which are regular, convex polyhedra with identical faces, angles, and edge lengths. The hexahedron is characterized by its six square faces, twelve edges, and eight vertices.

The term "cuboid" is often used in general geometry to describe a rectangular prism with six rectangular faces. However, in the context of Platonic solids, the specific name for the solid resembling a cuboid is the hexahedron.

The hexahedron is a highly symmetrical three-dimensional shape. All of its faces are congruent squares, and each vertex is formed by three edges meeting at right angles. The hexahedron exhibits symmetry under several transformations, including rotations and reflections.

Its regularity and symmetry make the hexahedron an important geometric shape in mathematics and design. It has numerous applications in architecture, engineering, and computer graphics. The cube, as a special case of the hexahedron, is particularly well-known and widely used in everyday life, from dice and building blocks to cubic containers and architectural structures.

Therefore, the option which is the correct is C.

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The question probable may be:

What is the name of the Platonic solid which resembles a cuboid?

A. Dodecaheron  

B. Tetrahedron

C. Hexahedron

D. Octahedron  

Answer:

-2097152 is the 20th term

Step-by-step explanation:

Write geometric sequence as an explicit formula

[tex]-4,8,-16,32\rightarrow-4(-2)^0,-4(-2)^1,-4(-2)^2,4(-2)^3\\a_n=a_1r^{n-1}\rightarrow a_n=-4(-2)^{n-1}[/tex]

Find the n=20th term

[tex]a_{20}=-4(-2)^{20-1}=-4(-2)^{19}=4(-524288)=-2097152[/tex]

We have created three different linear systems using the equation 20x + 12y = 24.

System 1 has infinitely many solutions, System 2 has no solution, and System 3 has a unique solution.

Let's create three different linear systems using the equation 20x + 12y = 24 and ensure that each system has a different number of solutions.

System 1:

Equation 1: 20x + 12y = 24 (given)

Equation 2: 40x + 24y = 48

Explanation: In this system, we multiplied both sides of the given equation by 2 to create Equation 2.

By doing so, we have essentially created two equations that are multiples of each other.

Since the equations are equivalent, they represent the same line, and the system has infinitely many solutions.

Any values of x and y that satisfy the first equation will automatically satisfy the second equation as well.

System 2:

Equation 1: 20x + 12y = 24 (given)

Equation 2: 20x + 12y = 48

Explanation: In this system, we changed the constant term in Equation 2 to 48.

By doing so, we have created two parallel lines with the same slope. Since the lines are parallel, they will never intersect, and the system has no solution.

There are no values of x and y that satisfy both equations simultaneously.

System 3:

Equation 1: 20x + 12y = 24 (given)

Equation 2: 40x + 24y = 48

Explanation: In this system, we multiplied both sides of Equation 2 by 2 to create Equation 2.

By doing so, we have created two equations that have the same slope but different y-intercepts.

Since the lines are not parallel and have different y-intercepts, they will intersect at a single point, and the system has a unique solution.

There will be one specific pair of values for x and y that satisfy both equations simultaneously.

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The angle the slide forms with the tower is approximately 41 degrees (rounded to the nearest degree).

To find the angle the slide forms with the tower, we can use trigonometric ratios. Let's consider the right triangle formed by the height of the tower (18 m), the length of the slide (20 m), and the angle we want to find.

Using the tangent function, we have:

tan(angle) = opposite/adjacent

In this case, the opposite side is the height of the tower (18 m) and the adjacent side is the length of the slide (20 m). Therefore:

tan(angle) = 18/20

To find the angle, we can take the inverse tangent (arctan) of both sides:

angle = arctan(18/20)

Using a calculator, we find that arctan(18/20) is approximately 40.56 degrees.

Therefore, the angle the slide forms with the tower is approximately 41 degrees (rounded to the nearest degree).

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Answer:

1 - 3√(5/35) = 1 - 3√(1/7) = 1 - 3*(1/sqrt(7)) ≈ 0.0755
0.0755 * 100 = 7.55%

Step-by-step explanation:

To find the percentage of 1 - 3√(5/35), we need to first evaluate the expression.

1 - 3√(5/35) = 1 - 3√(1/7) = 1 - 3*(1/sqrt(7)) ≈ 0.0755

To convert this decimal to a percentage, we simply multiply by 100:

0.0755 * 100 = 7.55%

A. To determine the least and greatest percentage of students who could be in favor of year-round school, we can use the error given in the survey, which is +/5%. Let's denote the actual percentage of students in favor of year-round school as x.

The least percentage can be found by subtracting 5% from the reported percentage of 32%:

32% - 5% = 27%

So, the least percentage of students in favor of year-round school is 27%.

The greatest percentage can be found by adding 5% to the reported percentage of 32%:

32% + 5% = 37%

Therefore, the greatest percentage of students in favor of year-round school is 37%.

Hence, the least percentage is 27% and the greatest percentage is 37%.

B. A classmate claiming that ⅓ of the student body is actually in favor of year-round school conflicts with the survey data. According to the survey, the reported percentage in favor of year-round school is 32%, which is not equal to 33.3% (⅓). Therefore, the classmate's claim contradicts the survey results.

It's important to note that the survey provides specific data regarding the percentages of students in favor and opposed to year-round school. The claim of ⅓ being in favor does not align with the survey's findings and should be evaluated separately from the survey data.

The given sequence can be generated by multiplying each term by 2, starting from the initial term of 6.

The pattern that fits the given sequence 6, 12, 24, 48, 96, ... is that each term is 2 times the previous term.

In the sequence 6, 12, 24, 48, 96, ... there are multiple possible patterns, each resulting from a different rule applied to generate the next term. Let's examine each of the proposed patterns:

Each term is 6 more than the previous term:

Starting with 6, if we add 6 to each term, we get:

6 + 6 = 12

12 + 6 = 18

18 + 6 = 24

24 + 6 = 30

30 + 6 = 36

...

This pattern does not match the given sequence since it does not produce the subsequent terms.

Each term is 12 more than the previous term:

Starting with 6, if we add 12 to each term, we get:

6 + 12 = 18

18 + 12 = 30

30 + 12 = 42

42 + 12 = 54

54 + 12 = 66

...

This pattern also does not match the given sequence.

Each term is 1/2 the previous term:

Starting with 6, if we multiply each term by 1/2, we get:

6 [tex]\times[/tex] 1/2 = 3

3 [tex]\times[/tex] 1/2 = 1.5

1.5 [tex]\times[/tex] 1/2 = 0.75

0.75 [tex]\times[/tex] 1/2 = 0.375

0.375 [tex]\times[/tex] 1/2 = 0.1875

...

This pattern does not match the given sequence.

Each term is 2 times the previous term:

Starting with 6, if we multiply each term by 2, we get:

6 [tex]\times[/tex] 2 = 12

12 [tex]\times[/tex] 2 = 24

24 [tex]\times[/tex]2 = 48

48 [tex]\times[/tex]2 = 96

96 [tex]\times[/tex]2 = 192

This pattern perfectly matches the given sequence. Each term is indeed 2 times the previous term, resulting in the next term.

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Using the concept of perimeter of polygon, the perimeter of figure C is 27cm shorter than total perimeter of A and B

To solve this problem, we have to know the perimeter of the polygon C.

The perimeter of a polygon is the sum of all the lengths of the outer edges of the figure, that is, we must find the length of all the edges of the polygon, and then add these lengths to obtain the perimeter.

The perimeter of the figures are;

Using the concept of perimeter of a rectangle;

a. figure A = 2(4 + 11) = 30cm

b. figure B = 2(8 + 4) = 24cm

c figure C = 11 + 4 + 8 + 4 = 27cm

Now, we can add A and B and then subtract c from it.

30 + 24 - 27 = 27cm

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Answer:

volume=60cm3, surface area=96cm2

Step-by-step explanation:

volume=1/3×(6×6)×5

=60cm3

surface area= 4(1/2×6×5)+(6×6)

=96cm2

Answer:

44

Step-by-step explanation:

The sum of the series [tex]\sum_{k=1}^4[/tex] (2k²−4) is 44.

The series is: [tex]\sum_{k=1}^4[/tex] (2k²−4)

Let's find the value of each term for k=1, k=2, k=3, and k=4, and then add them up:

For k=1:

2(1)² - 4 = 2(1) - 4 = 2 - 4 = -2

For k=2:

2(2)² - 4 = 2(4) - 4 = 8 - 4 = 4

For k=3:

2(3)² - 4 = 2(9) - 4 = 18 - 4 = 14

For k=4:

2(4)² - 4 = 2(16) - 4 = 32 - 4 = 28

Now, let's add all the terms:

-2 + 4 + 14 + 28 = 44

So, the sum of the series [tex]\sum_{k=1}^4[/tex] (2k²−4) is 44.

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By using the change of base formula: The solution to the equation log(base y) (X-2) = 5 is [tex]X = y^5 + 2.[/tex]

To solve the equation log(base y) (X-2) = 5 using the change of base formula, we can rewrite the equation as log(base b) (X-2) / log(base b) y = 5.

Using the change of base formula, we can choose any base for b.

Let's choose base 10 for simplicity.

So the equation becomes log(base 10) (X-2) / log(base 10) y = 5.

We know that log(base 10) (X-2) represents the logarithm of (X-2) to the base 10, and log(base 10) y represents the logarithm of y to the base 10.

Now, to solve for X, we can isolate it by multiplying both sides of the equation by log(base 10) y:

log(base 10) (X-2) = 5 [tex]\times[/tex] log(base 10) y.

This simplifies to:

log(base 10) (X-2) [tex]= log(base 10) y^5.[/tex]

Since the logarithms on both sides have the same base, we can remove the logarithm and equate the arguments:

[tex]X - 2 = y^5.[/tex]

Now we can solve for X by adding 2 to both sides:

[tex]X = y^5 + 2.[/tex]

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