Quadrilateral EFGH is an isosceles trapezoid with bases EH and FG. The measure of angle HGF is (9y + 3)°, and the measure of angle EFG is (By + 5)°.
What is the measure of angle HGF?

Answer: [tex](-\infty, 0) \cup (0, \infty)[/tex]

Step-by-step explanation:

The graph of [tex]\frac{\sin x}{x}[/tex] is continuous for all real [tex]x[/tex] except [tex]x=0[/tex], and multiplying this by [tex]1/3[/tex] does not change this.

Answer:

height = 10 ft

Step-by-step explanation:

the area (A) of a triangle is calculated as

A = [tex]\frac{1}{2}[/tex] bh ( b is the base and h the height )

given A = 70 and b = 14 , then

[tex]\frac{1}{2}[/tex] × 14 × h = 70

7h = 70 ( divide both sides by 7 )

h = 10 ft

To calculate the probabilities, we can use the following information:

Total number of student athletes = 204

Number of students who play soccer = 52

Number of students who play volleyball = 31

Probability of a student playing both soccer and volleyball = 5/204

1.  Probability that a student plays both soccer and volleyball:

Let's denote the probability of playing both soccer and volleyball as P(Soccer and Volleyball). From the given information, we know that the number of students who play both soccer and volleyball is 5.

P(Soccer and Volleyball) = Number of students who play both soccer and volleyball / Total number of student athletes

P(Soccer and Volleyball) = 5 / 204

2.  Probability that a student plays volleyball:

We want to find the probability of a student playing volleyball, denoted as P(Volleyball).

P(Volleyball) = Number of students who play volleyball / Total number of student athletes

P(Volleyball) = 31 / 204

Therefore, the probability that a randomly selected student athlete in this school plays both soccer and volleyball is 5/204, and the probability that they play volleyball is 31/204.

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The boundaries of the 99% confidence interval are

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

Selected, x = 28021

Sample, n = 38433

This means that the proportion, p is

p = 28021/38433

p = 0.729

The 99% confidence interval is then calculated as

CI = p ± z * √(p * (1 - p)/n)

Where

z = 2.576 i.e. the z-score at 99% confidence interval

So, we have

CI = 0.729 ± 2.576  * √(0.729 * (1 - 0.729)/38433)

Evaluate

CI = 0.729 ± 0.0058

Evaluate

CI = (0.7232, 0.7348)

Hence, the confidence interval is (0.7232, 0.7348)

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HI Your answer is 42

I have calculated it you can trust me

Well you have marked right in the pic

PLEASE MARK AS BRAINLIEST

The mean of the scores to the nearest tenth is 83.7.

Mean is the average of the given numbers and is calculated by dividing the sum of given numbers by the total number of numbers.

Given the question above, we need to find the mean of the scores to the nearest tenth.

We can find the mean by using the formula below:

[tex]\text{Mean} = \dfrac{\text{Sum of all the observations}}{\text{Total number of observations}}[/tex]

Now,

[tex]\text{Mean} = \dfrac{70(6)+75(3)+80(9)+85(5)+90(7)+95(8)}{6+3+9+5+7+8}[/tex]

[tex]\text{Mean} = \dfrac{420+225+720+425+630+760}{38}[/tex]

[tex]\text{Mean} = \dfrac{3180}{38}[/tex]

[tex]\text{Mean} = 83.7[/tex]

Therefore, the mean of the scores to the nearest tenth is 83.7.

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Answer:

A: the probability that a 1 is received is 0.56.

B:  the probability that a 0 was sent given that a 1 is received is (2/25) * (1 - P(0 sent)).

Step-by-step explanation:

To solve this problem, we can use conditional probabilities and the concept of Bayes' theorem.

a. To find the probability that a 1 is received, we need to consider the two possibilities: either a 1 was sent and remained unchanged, or a 0 was sent and got flipped to a 1 by the random error.

Let's denote:

P(1 sent) = 0.6 (probability a 1 is sent)

P(0→1) = 0.2 (probability a 0 is flipped to 1)

P(1 received) = ?

P(1 received) = P(1 sent and unchanged) + P(0 sent and flipped to 1)

= P(1 sent) * (1 - P(0→1)) + P(0 sent) * P(0→1)

= 0.6 * (1 - 0.2) + 0.4 * 0.2

= 0.6 * 0.8 + 0.4 * 0.2

= 0.48 + 0.08

= 0.56

Therefore, the probability that a 1 is received is 0.56.

b. If a 1 is received, we want to find the probability that a 0 was sent. We can use Bayes' theorem to calculate this.

Let's denote:

P(0 sent) = ?

P(1 received) = 0.56

We know that P(0 sent) + P(1 sent) = 1 (since either a 0 or a 1 is sent).

Using Bayes' theorem:

P(0 sent | 1 received) = (P(1 received | 0 sent) * P(0 sent)) / P(1 received)

P(1 received | 0 sent) = P(0 sent and flipped to 1) = 0.4 * 0.2 = 0.08

P(0 sent | 1 received) = (0.08 * P(0 sent)) / 0.56

Since P(0 sent) + P(1 sent) = 1, we can substitute 1 - P(0 sent) for P(1 sent):

P(0 sent | 1 received) = (0.08 * (1 - P(0 sent))) / 0.56

Simplifying:

P(0 sent | 1 received) = 0.08 * (1 - P(0 sent)) / 0.56

= 0.08 * (1 - P(0 sent)) * (1 / 0.56)

= 0.08 * (1 - P(0 sent)) * (25/14)

= (2/25) * (1 - P(0 sent))

Therefore, the probability that a 0 was sent given that a 1 is received is (2/25) * (1 - P(0 sent)).

A message is coded into the binary symbols 0 and 1 and the message is sent over a communication channel. The probability a 0 is sent is 0.4 and the probability a 1 is sent is 0.6. The channel, however, has a random error that changes a 1 to a 0 with probability 0.2 and changes a 0 to a 1 with probability 0.1. (a) What is the probability a 0 is received? (b) If a 1 is received, what is the probability a 0 was sent?

Answer:   there are 20 people who claimed to be neither CDO partisans nor ANC partisans.

Step-by-step explanation:

To determine the number of people who are none of these two parties, we need to subtract the total number of people who claimed to be CDO partisans, ANC partisans, and those who claimed to be both from the total number of people surveyed.

Total surveyed people = 130

Number claiming to be CDO partisans = 80

Number claiming to be ANC partisans = 60

Number claiming to be both ANC and CDO = 30

To find the number of people who are none of these two parties, we can calculate it as follows:

None of these two parties = Total surveyed people - (CDO partisans + ANC partisans - Both ANC and CDO)

None of these two parties = 130 - (80 + 60 - 30)

None of these two parties = 130 - 110

None of these two parties = 20

Answer: Nope

Step-by-step explanation:

No, receiving a 20% discount followed by an additional 30% discount does not result in a total discount of 50%.

To understand why, let's consider an example with an item priced at $100.

If there is a 20% discount applied initially, the price of the item would be reduced by 20%, which is $100 * 0.20 = $20. So the new price after the first discount would be $100 - $20 = $80.

Now, if there is an additional 30% discount applied to the $80 price, the discount would be calculated based on the new price. The 30% discount would be $80 * 0.30 = $24. So the final price after both discounts would be $80 - $24 = $56.

Comparing the final price of $56 to the original price of $100, we can see that the total discount is $100 - $56 = $44.

Therefore, the total discount received is $44 out of the original price of $100, which is a discount of 44%, not 50%.

Hence, receiving a 20% discount followed by an additional 30% discount does not result in a total discount of 50%.

Answer:

So we have $11.64 in American dollars and £5 in British pound coin

Step-by-step explanation:

To solve this problem, we can use a system of equations. Let x be the number of American dollars and y be the number of British pound coins. Then we have:

x + y/1.65 = 20.25 (since each British pound coin is worth 1.65 American dollars)

x = 17 - y (since there are 17 items altogether)

Substituting the second equation into the first, we get:

(17 - y) + y/1.65 = 20.25

Multiplying both sides by 1.65, we get:

28.05 - y + y = 33.4125

y = 33.4125 - 28.05

y = 5.3625

Therefore, we have 5 British pound coins and:

x = 17 - y = 17 - 5.3625 = 11.6375

Answer:A' = (15, -15), B' = (9, -9), and C' = (15, -9)

Step-by-step explanation:

To dilate triangle ABC with a center of dilation at the origin (0,0) and a scale factor of 3, you need to multiply the coordinates of each vertex by the scale factor.

Let's calculate the coordinates of triangle A'B'C':

For point A:

x-coordinate of A' = scale factor * x-coordinate of A = 3 * 5 = 15

y-coordinate of A' = scale factor * y-coordinate of A = 3 * (-5) = -15

Therefore, A' = (15, -15)

For point B:

x-coordinate of B' = scale factor * x-coordinate of B = 3 * 3 = 9

y-coordinate of B' = scale factor * y-coordinate of B = 3 * (-3) = -9

Therefore, B' = (9, -9)

For point C:

x-coordinate of C' = scale factor * x-coordinate of C = 3 * 5 = 15

y-coordinate of C' = scale factor * y-coordinate of C = 3 * (-3) = -9

Therefore, C' = (15, -9)

Hence, the correct coordinates of triangle A'B'C' are A' = (15, -15), B' = (9, -9), and C' = (15, -9).

Answer:

8/15

Step-by-step explanation:

1/15(4 + 1/20(4)

4/15 + 4/20  I can reduce 4/20 to 1/5

4/15 + 1/5

4/15 + 3/15 = 7/15

Together they can get 7/15 of the job done.  15/15 - 7/15 is 8/15.

That means that they have 8/15 left to do.

Helping in the name of Jesus.

The range of the given graph is expressed as:

Option A: {-∞, ∞}

The range of a function is defined as the set of all the possible output values of y. The formula to find the range of a function is y = f(x).

In a relation, it is only a function if every x value corresponds to only one y value,

Now, looking at the given graph, we see that At x = 0, the function is also y = 0.

However, between 0 and π intervals, we see that the graph approaches positive and negative infinity and as such we can tell that the range is expressed as: {-∞, ∞}

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Answer:

-33x√2

Step-by-step explanation:

[tex]-5x\sqrt{98}+2\sqrt{2x^2}\\\\= -5x\sqrt{2*7^{2} } + 2(x\sqrt{2} )\\\\= -5x(7)(\sqrt{2} ) + 2x\sqrt{2} \\\\= -35x\sqrt{2} +2x\sqrt{2} \\\\= (-35+2)x\sqrt{2}\\ \\=-33x\sqrt{2}[/tex]

Answer:

1. 01= 234, 03= 255 (since the data is

already sorted)

2. I0R = 255 - 234= 21

3. Lower limit = 234- 1.5 * 21= 203.5

Upper limit = 255+ 1.5 * 21= 285.5

4. Outliers: 110, 120, 178 (below the

lower limit), and 273 (above the upper

limit)

(i) Correcting the figures to 3 decimal places:

-0.005627 ≈ -0.006

0.0056 ≈ 0.006

-0.0049327 ≈ -0.005

0.0049 ≈ 0.005

-0.001342 ≈ -0.001

(ii) Correcting the figures to 3 significant figures:

-0.005627 ≈ -0.00563

0.0056 ≈ 0.00560

-0.0049327 ≈ -0.00493

0.0049 ≈ 0.00490

-0.001342 ≈ -0.00134

(i) When rounding to 3 decimal places, we look at the fourth decimal place and round the figure accordingly. If the fourth decimal place is 5 or above, we round up the preceding third decimal place. If the fourth decimal place is less than 5, we simply drop it.

(ii) When rounding to 3 significant figures, we consider the digit in the third significant figure. If the digit in the fourth significant figure is 5 or above, we round up the preceding third significant figure. If the digit in the fourth significant figure is less than 5, we simply drop it.

Rounding to the correct number of decimal places or significant figures is important to maintain precision and accuracy in calculations and measurements. It helps to ensure that the reported values are appropriate for the level of precision required in a given context.

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The length of this line segment is: B. 2√13 units.

In Mathematics and Geometry, the distance between two (2) end points that are on a coordinate plane can be calculated by using the following mathematical equation:

Distance = √[(x₂ - x₁)² + (y₂ - y₁)²]

Where:

x and y represent the data points (coordinates) on a cartesian coordinate.

By substituting the given end points into the distance formula, we have the following;

Distance = √[(4 + 2)² + (1 + 3)²]

Distance = √[(6)² + (4)²]

Distance = √[36 + 16]

Distance = √52

Distance = 2√13 units.

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Answer:

Step-by-step explanation:

Given:

<RPS = 35

<PAQ = 130

Solution:  

If <PAQ = 130 then <QAR =50  

because they are a linear pair that add to 180

<QAR = mQR = 50

mRS = 2(<RPS)                    >inscribed angle

mRS = 2(35)

mRS = 70

<PAQ = 130

<PAQ = mPQ

<PSQ = 1/2 (mPQ)                     >inscribed angle

<PSQ = 1/2 (130)

<PSQ = 65

<PBS = 180 - <RPQ - PSQ             >triangle

<PBS = 180 - 35-65

<PBS = 80

<PBS = <QBR                               >vertical angles

<QBR = 80

<ABQ = 180- <QBR                      >linear pair

<ABQ = 180 - 80

<ABQ= 100

<AQB = 180 - <ABQ - <QAR             >triangle

<AQB = 180 -100 - 50

<AQB = 30

<AQB = <AQS

<AQS =30

mRS = 70

mPS = 180-mRS                         180 for semicircle

mPS = 180 - 70

mPS = 110

The rectangles that are similar are Rectangle 3 and Rectangle 4.

To determine which rectangles are similar, we need to compare their corresponding side lengths.

Rectangle 1:

Length: 3 cm

Height: 5 cm

Rectangle 2:

Length: 2.5 cm

Height: 5.5 cm

Rectangle 3:

Length: 2.5 cm

Height: 2 cm

Rectangle 4:

Length: 5 cm

Height: 4 cm

To determine similarity, we need to compare the ratios of the corresponding side lengths of the rectangles.

Comparing Rectangle 1 with Rectangle 2:

Length ratio: 3 cm / 2.5 cm = 1.2

Height ratio: 5 cm / 5.5 cm ≈ 0.91

The length ratio and height ratio are not equal, so Rectangle 1 and Rectangle 2 are not similar.

Comparing Rectangle 1 with Rectangle 3:

Length ratio: 3 cm / 2.5 cm = 1.2

Height ratio: 5 cm / 2 cm = 2.5

The length ratio and height ratio are not equal, so Rectangle 1 and Rectangle 3 are not similar.

Comparing Rectangle 1 with Rectangle 4:

Length ratio: 3 cm / 5 cm = 0.6

Height ratio: 5 cm / 4 cm = 1.25

The length ratio and height ratio are not equal, so Rectangle 1 and Rectangle 4 are not similar.

Comparing Rectangle 2 with Rectangle 3:

Length ratio: 2.5 cm / 2.5 cm = 1

Height ratio: 5.5 cm / 2 cm = 2.75

The length ratio and height ratio are not equal, so Rectangle 2 and Rectangle 3 are not similar.

Comparing Rectangle 2 with Rectangle 4:

Length ratio: 2.5 cm / 5 cm = 0.5

Height ratio: 5.5 cm / 4 cm = 1.375

The length ratio and height ratio are not equal, so Rectangle 2 and Rectangle 4 are not similar.

Comparing Rectangle 3 with Rectangle 4:

Length ratio: 2.5 cm / 5 cm = 0.5

Height ratio: 2 cm / 4 cm = 0.5

The length ratio and height ratio are equal, so Rectangle 3 and Rectangle 4 are similar.

Therefore, the rectangles that are similar are Rectangle 3 and Rectangle 4.

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