Identify the coordinates of the vertices after the transformation

Answer:

8% of $15,000 = .08 × $15,000 = $1,200

For a probability distribution to be represented, it is needed that P(X = 0) + P(X = 1) = 0.44. Hence one possible example is:

P(X = 0) = 0.40.

P(X = 1) = 0.04.

Discrete random variable to represent a probability distribution =

The sum of all the probabilities must be of 1, hence:

P(X = 0) + P(X = 1) + P(X = 3) + P(X = 4) + P(X = 5) = 1.

Then, considering the table:

P(X = 0) + P(X = 1) + 0.15 + 0.17 + 0.24 = 1

P(X = 0) + P(X = 1) + 0.56 = 1

P(X = 0) + P(X = 1) = 0.44.

Hence one possible value is:

P(X = 0) = 0.40.

P(X = 1) = 0.04.

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The slopes of the lines connecting all the points are distinct. Hence, they form a triangle.

To show that the points (2, 5), (5, 2), and (6, 6) form a triangle, we can calculate the slopes of the lines connecting these points. If the slopes are all distinct, then the points form a triangle.

Let's calculate the slopes:

The slope between (2, 5) and (5, 2):

m₁ = (y₂ - y₁) / (x₂ - x₁)

= (2 - 5) / (5 - 2)

= -3/3

= -1

The slope between (2, 5) and (6, 6):

m₂ = (y₂ - y₁) / (x₂ - x₁)

= (6 - 5) / (6 - 2)

= 1/4

The slope between (5, 2) and (6, 6):

m₃ = (y₂ - y₁) / (x₂ - x₁)

= (6 - 2) / (6 - 5)

= 4/1

= 4

Since the slopes of the lines connecting these points are all distinct (-1, 1/4, and 4), the points (2, 5), (5, 2), and (6, 6) form a triangle.

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The measure of angle BGF from the given parallel lines is 108°.

Given that, measure of angle BGF=4x+28 and the measure of angle EHD = 2x+32.

1) From the given figure, angle BGF and angle EHC are interior alternate angles.

Angle FHC and angle EGA are corresponding angles.

Angle FHC and angle EGA are exterior alternate angles.

2) angle BGF+angle EHD=180°

3) 4x+28+2x+32=180°

6x+60=180

6x=120

x=20

Angle BGF=4x+28=108 °

Angle EHD = 2x+32

= 112°

Angle FHD=180-112=68°

Angle EGB=180-108=72°

Therefore, the measure of angle BGF from the given parallel lines is 108°.

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The surface area of Grace's shipping box is 664 square inches.

We have,

The box can be divided into six rectangles, each of which represents the face of the box.

To find the surface area of the box, we need to find the area of each face and then add them all up.

The top and bottom of the box are both 10 inches by 14 inches, so the area of each of these faces is:

= 10 x 14

= 140 square inches

The sides of the box are both 14 inches by 8 inches, so the area of each of these faces is:

= 14 x 8

= 112 square inches

The front and back of the box are both 8 inches by 10 inches, so the area of each of these faces is:

8 x 10

= 80 square inches

To find the surface area of the whole box, we add up the areas of all six faces:

= 2(140) + 2(112) + 2(80)

= 280 + 224 + 160

= 664 square inches

Therefore,

The surface area of Grace's shipping box is 664 square inches.

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62. The slope of the line is 1/8. the equation of line is y = 1/8x - 3/8

63. It has an infinite slope.  The equation of line is x = 3

64. The slope of the line is -5. The slope of the new line is 1/5. The equation of line is  y = 1/5x - 1

65. The slope of the line is 2/9. The slope of the new line is -9/2. The equation of line is y = -9/2x - 4

62) through: (3,0), parallel to y = 1/8x - 3. If a line is parallel to another, they have the same slope. Therefore the slope is 1/8.

For line equation⇒ (y - y1 = m(x - x1))

Substituting (3,0) and slope 1/8 gives us:

(y = mx + b) ⇒ y - 0 = 1/8 × (x - 3)

First, substitute the point and the slope into the equation: 0 = 1/8×3 + b

Then solve for b, the y-intercept: b = -3/8

⇒ y = 1/8x - 3/8

63) through: (3,-1), parallel to x = 0. A line parallel to x = 0 is a vertical line. A vertical line has the equation x = a, where a is the x-coordinate of any point on the line. A line through (3, -1) and parallel to x = 0 would simply be x = 3 or y = -1

64) through: (5, 0), perp. to y= -5x - 4. If a line is perpendicular to another, their slopes are negative reciprocals. The slope of y = -5x - 4 is -5, so the slope of the line we are looking for is -1/(-5) = 1/5

65) through: (-2, 5), perp. to y=2/9x - 1.  perpendicular  line to y = 2/9x - 1 has the slope that is the negative reciprocal of 2/9, which is -9/2. y = -9/2x - 4.

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The  value of function f (α + β) is determined as  (3√3  +  1)/3,  (2√2 - 3)/3.

The value of f (α + β) is calculated by applying the following formula as follows;

The given functions;

for α:  x²  +  y² = 4

y is given as -1, the value of x is calculated as;

x²  +  (-1)² = 4

x²  +  1 = 4

x² = 3

x = √3

α = (√3, - 1)

for β:  x²  +  y² = 1

x is given as ¹/₃, the value of y is calculated as;

(¹/₃)² + y² = 1

¹/₉ + y² = 1

y² = 1 - ¹/₉

y² = ⁸/₉

y = √ (8/9)

y = 2√2/3

β = (¹/₃, 2√2/3)

The  value of function f (α + β) is calculated as;

f(α + β) =  (√3, - 1) + (¹/₃, 2√2/3)

f(α + β) = (3√3  +  1)/3,  (2√2 - 3)/3

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Out of the following 30 samples, you may expect that about 12 will be portions of vanilla cake.

To solve this problem, use the concept of probability and make a prediction based on the past data.

In this case, out of the 10 pieces of cake that have been taken, 4 are vanilla and 6 are coconut.

Therefore, the probability of selecting a piece of vanilla cake from the tray is 4/10, or 2/5.

To predict how many of the next 30 samples will be pieces of vanilla cake, multiply the probability of selecting a vanilla cake by the total number of samples:

(2/5) x 30 = 12

So expect that out of the next 30 samples taken, approximately 12 of them will be pieces of vanilla cake.

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The number of 1/3 inch cubes required to fill the box is 62.

Volume = length × width × height

Height = 7/3

Length = 10/3

width = 8/3

Hence ,

Volume = 7/3 × 10/3 × 8/3 = 560/27

Size of cube = 1/3

Number of cubes required = Volume of box / size of cube

Number of cubes = 560/27 × 3/1

Number of cubes = 1680/27 = 62.22

Therefore, it will take 62 1/3 cubes to fill the box

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The sample size needed to be 99​% confident that the sample proportion will not differ from the true proportion by more than ​4% is given as follows:

n = 1037.

A confidence interval of proportions has the bounds given by the rule presented as follows:

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which the variables used to calculated these bounds are listed as follows:

The margin of error is given as follows:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

The confidence level is of 99%, hence the critical value z is the value of Z that has a p-value of [tex]\frac{1+0.99}{2} = 0.995[/tex], so the critical value is z = 2.575.

As we have no estimate, the parameter is given as follows:

[tex]\pi = 0.5[/tex]

Then the sample size for M = 0.04 is obtained as follows:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]0.04 = 2.575\sqrt{\frac{0.5(0.5)}{n}}[/tex]

[tex]0.04\sqrt{n} = 2.575 \times 0.5[/tex]

[tex]\sqrt{n} = \frac{2.575 \times 0.5}{0.04}[/tex]

[tex](\sqrt{n})^2 = \left(\frac{2.575 \times 0.5}{0.04}\right)^2[/tex]

n = 1037.

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The equivalent sine and cosine values are as follows:

The relationship between sine and cosine angles is:

sinθ = cos(90 - θ)

This equation is known as the complementary angle identity for sine and cosine. It states that the sine of an angle is equal to the cosine of its complementary angle (90 degrees minus the original angle).

For example, if we have an angle theta = 30 degrees, its complementary angle would be 90 - 30 = 60 degrees.

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Answer:

[tex]3 {cos}^{2} x - {sin}^{2} x = 0[/tex]

[tex]3 - 3 {sin}^{2} x - {sin}^{2} x = 0[/tex]

[tex]3 - 4 {sin}^{2} x = 0[/tex]

[tex]4 {sin}^{2} x = 3[/tex]

[tex] {sin}^{2} x = \frac{3}{4} [/tex]

sin(x) = +(1/2)√3

x = π/3 + 2kπ or x = 2π/3 + 2kπ

x = -π/3 + 2kπ or x = -2π/3 + 2kπ

(k is an integer)

The number of weeks that the puppies weigh the same amount will be A. 6 weeks.

The equation that has no solution is A. 13 - 7x = -7x + 13.

The option determined from the graph is B. The ordered pair (25, 6) is a solution to y = x + 40 and y = 3x + 15.

Puppy A weighs 4 1/5 pounds at birth and gains 3/4 pound each week. Puppy B weighs 5 2/3pounds at birth and gains 1/2 pound each week.

After how many weeks will the puppies weigh the same amount will be calculated thus:

4 1/5 + 3/4x = 5 2/3 + 1/2x

4.2 + 0 75x = 5.67 + 0.5x

Collect the like terms

5.67 - 4.2 = 0.25x

Divide

x = 1.47 / 0.25

x = 6

Hence ,the number of weeks that the puppies weigh the same amount will be A. 6 weeks.

It should be noted that the equation that has no solution is A. 13 - 7x = -7x + 13. This is because the values gives 0 all through.

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Answer:

5

Step-by-step explanation:

Just do as the problem says:

xy-y = (2)(5)-(5) = 10-5 = 5

Answer: 5

Step-by-step explanation: All you have to do is substitute x & y for their respective numbers. In this case, change x for 2 & y for 5, So xy-y becomes 2(5)-5, which equals 10-5, which is 5.

Answer:

23.35 minutes

Step-by-step explanation:

Brendan finished in 22.5 minutes. Tobias finished 85 seconds after. Just add 85 seconds to 22.5 minutes. There are 100 second in 1 minute. Since Brendan is already at 22.5, there is only 0.5 left to get to 23. 0.5 in this case means 50 seconds. You do 85 - 50 = 35. Now, you are at 23 minutes. Add 35 seconds to that to get to 23.25 minutes.

Answer: 0

Step-by-step explanation:

Put 0 in the front box.

this is because you cant divide 11 apples equally into 6 people; 11 isn't a multiple of 6

The reference angle of 13π/8 is 3π/8 and the sketch of the angle is attached

From the question, we have the following parameters that can be used in our computation:

Angle = 13π/8

Rewrite as

θ = 13π/8

The above angle is in the fourth quadrant

So, we subtract it from 2π to calculate the reference angle

Using the above as a guide, we have the following:

Reference angle = 2π - 13π/8

Evaluate the difference

Reference angle = 3π/8

Hence, the reference angle is 3π/8 and the sketch of the angle is added as an attachment

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In triangle ABC, side BC (a) is 15 cm, side AC (b) is 13.30 cm, and side AB (c) is 12.23 cm. Also angle A is 70°, B is 50°, and C is 60°

Using the Law of Sines and the fact that the sum of the angles in a triangle is 180 degrees.

The Law of Sines states that the ratio of the length of a side of a triangle to the sine of its opposite angle is constant. Using this, we can find the lengths of the other sides of triangle ABC.

Let's denote side AB as c, BC as a, side AC as b, and angle C as ∠C.

Using the Law of Sines, we have:

sin(∠A) / a = sin(∠B) / b

Substitute in the known values, we get:

sin(70°) / 15 = sin(50°) / b

Make b the subject of the formula:

b = (15 * sin(50°)) / sin(70°)

Using a calculator, we can find the value of b:

b = (15 * 0.766) / 0.9397

  = 12.23 cm

Therefore side AC = b = 12.23 cm

Now, to find side AB (c), we can use the Law of Sines again:

sin(∠B) / b = sin(∠C) / c

But ∠C is unknown. To find ∠C, we apply the sum of angle in a triangle rule

∠A + ∠B + ∠C = 180

70 + 50 + ∠C = 180

∠C = 180 - 70 - 50

∠C = 60°

Substitute the values in the Sine rule equation

sin(50°) / 12.233 = sin(60) / c

Make c the subject of the formula

c = (12.233 * sin(60)) / sin(50°)

c = (12.233 * sin(60°)) / sin(50°)

  = 13.30 cm

Therefore, the lengths of the sides of triangle ABC are:

Side AB = c = 12.23 cm

Side BC = a = 15 cm

Side AC = b = 13.30 cm

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Answer:

4x = 100, so x = 25

The correct answer is C.

The answers to all parts are shown below.

The law of uninhibited decay is a first-order chemical reaction, which means that the rate of the reaction is proportional to the concentration of compound A.

The rate law can be written as:

rate = k[A]

where k is the rate constant and [A] is the concentration of compound A.

From the given data, we can use the following equation to determine the rate constant:

ln([A]t/[A]0) = -kt

where [A]t is the concentration of compound A at time t, [A]0 is the initial concentration of compound A, k is the rate constant, and t is the time.

Using the given data:

[A]t = 0.37 M

[A]0 = 0.40 M

t = 30 minutes = 0.5 hours

ln(0.37/0.40) = -k(0.5)

k = 0.0486 h⁻¹

(a) How much of compound A will remain after 2 hours?

Using the same equation and substituting the known values:

ln([A]t/[A]0) = -kt

ln([A]t/0.40) = -0.0486 x 2

[A]t/0.40 = e⁻(⁹⁷² /¹⁰⁰⁰⁰)

[A]t = 0.25 M

Therefore, after 2 hours, 0.25 M of compound A will remain.

(b) How long will it take until 0.10 M of compound A remains?

Substituting the known values into the equation again:

ln([A]t/[A]0) = -kt

ln(0.10/0.40) = -0.0486t

t = 4.26 hours

Therefore, it will take approximately 4.26 hours until 0.10 M of compound A remains.

(c) After 2 hours, the amount of compound A remaining will be 0.25 M.

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You are approximately 3.4 miles from your starting point.

We  have,

To solve this problem, we need to use the Pythagorean theorem, which states that the sum of the squares of the two shorter sides of a right triangle is equal to the square of the length of the hypotenuse.

 |\

8 | \

 |  \

 |___\

    3

We can see that the three blocks north and 10 blocks south cancel each other out, so we only need to consider the 8 blocks west.

Using the distance formula, we can find that the distance traveled west is:

8 blocks × 2/10 mile per block = 1.6 miles

Now we can use the Pythagorean theorem to find the distance from the starting point:

distance = √(3² + 1.6²)

distance ≈ 3.4 miles

Therefore,

You are approximately 3.4 miles from your starting point.

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The surface area of the prism is 248.88 cm²

The area occupied by a three-dimensional object by its outer surface is called the surface area.

The surface area of a prism is expressed as;

SA = 2B + ph

where h is the height of the prism, B is the base area of the prism and P is the perimeter of the base.

Base area = 61.94 cm²

height = 5cm

perimeter = 5 × 5 = 25cm

SA = 2 × 61.94 + 25 × 5

SA = 123.88+ 125

SA = 248.88 cm²

Therefore the surface area of the prism is 258.88 cm²

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Diego is correct as the number of groups result to 6/5 or 1 1/5.

We have to find the number of groups of 5/6 in 1.

So, we need to perform the division as

= 1/ (5/6)

= 1 x 6/5

= 6/5

= 1 1/5

Thus, Diego is correct as the number of groups result to 6/5 or 1 1/5.

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To find the length of arc LM, we need to use the formula for the circumference of a circle:

C = 2πr

In this case, the arc LM is a portion of the circumference of the circle. To find the length of the arc, we need to determine the angle at the center of the circle that corresponds to the arc.

Looking at the image you provided, we can see that angle LOM is a right angle (90 degrees). The angle at the center of the circle, LOCM, is half of the right angle, which is 45 degrees.

To calculate the length of arc LM, we need to find the fraction of the total circumference represented by the central angle LOCM. The fraction is given by:

Fraction of circumference = Angle at center / 360 degrees

Fraction of circumference = 45 degrees / 360 degrees

Fraction of circumference = 1/8

Now, we can calculate the length of arc LM:

Length of arc LM = Fraction of circumference * Circumference

Length of arc LM = (1/8) * (2πr)

Since the radius (r) is not given in the image, we cannot calculate the exact length of arc LM. However, we can provide the formula for the length of arc LM in terms of the radius:

Length of arc LM = (1/8) * (2πr) = πr/4

So, the length of arc LM is πr/4, where r is the radius of the circle.

If you have the value of the radius, you can substitute it into the formula to find the specific length of arc LM.

On performing , the "row-operation" 2R₁ + R₂ → R₂ on the matrix "M", the resulting new-matrix is [tex]\left[\begin{array}{ccc}5&1&-5\\10&0&-8\end{array}\right][/tex] .

The matrix "M" is given as : [tex]\left[\begin{array}{ccc}5&1&-5\\0&-2&2\end{array}\right][/tex],

We have to apply the Row-Operation : 2R₁ + R₂ → R₂, on the matrix "M",

Performing the row-operation "2R₁ + R₂ → R₂" on a matrix "M" means multiplying the first-row of "M" by 2, and then adding the resulting values to the second-row of "M" to produce a new value for each element in the second row.

This operation does not affect the first row of "M" and changes the values in the second row of "M".

⇒ [tex]\left[\begin{array}{ccc}5&1&-5\\2\times5+0&2\times 1-2&2\times(-5)+2\end{array}\right][/tex],

⇒ [tex]\left[\begin{array}{ccc}5&1&-5\\10+0&2-2&-10+2\end{array}\right][/tex],

⇒ [tex]\left[\begin{array}{ccc}5&1&-5\\10&0&-8\end{array}\right][/tex].

Therefore, the new-matrix is [tex]\left[\begin{array}{ccc}5&1&-5\\10&0&-8\end{array}\right][/tex].

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Answer:

A

Step-by-step explanation:

i do this all the time math is my thing

Step-by-step explanation:

(35-24.8) / 6.2  = z-score of  + 1.645  corresponding to   .9495  =

           ~ 95th percentile

There are 23 boys and 20 girls in the class who cannot read a letter or a three-letter word.

The Venn diagram is drawn as shown in the attached image, and it shows that the number of boys and girls who cannot read a letter or a three-letter word is marked as (c+d) which represents 13 boys and 8 girls.

Given:

The class has 42 students.

12 boys and 15 girls are said to know how to read three-letter words in English.

19 boys and 17 girls are said to know how to identify all letters of the alphabet.

None of the students in the class is able to write a sentence of five words if dictated to them.

There are in total 22 boys and 20 girls in the class.

Read three-letter words    |

     |____________|______________|  

     |            |              |

     |  Boys (12) | Girls (15)   |   Can read three-letter words

     |____________|______________|

     |            |              |

     |  Boys (10) | Girls (13)   |   Cannot read three-letter words

_______|____________|______________|

     |            |              |

     |  Boys (9)  | Girls (13)   |   Can identify all letters

     |____________|______________|

     |            |              |

     |  Boys (13) | Girls (7)    |   Cannot identify all letters

     |____________|______________|

To find the number of boys and girls who cannot read a letter or a three-letter word, we need to add up the numbers in the two bottom circles:

Boys who cannot read a letter or a three-letter word: 10 + 13 = 23

Girls who cannot read a letter or a three-letter word: 13 + 7 = 20

Therefore, there are 23 boys and 20 girls in the class who cannot read a letter or a three-letter word.

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