The top stripe is 2 ft tall and is yellow.
The middle stripe is 1 ft tall and is blue.
The bottom stripe is also 1 ft tall and is red.
Create a scale drawing of the Colombian flag
with a scale of 1 cm to 2 ft.
Create a scale drawing of the Colombian flag
with a scale of 2 cm to 1 ft.

$85000 is the prize money is awarded in total.

A ratio is an ordered pair of numbers a and b, written a / b where b does not equal 0.

Given that  top four winners in a golf tournament share the prize money in the ratio 9:5:2:1.

The top prize winner receives $45,000

We need to find the prize money is awarded in total.

9+5+2+1=17

9/17 x = 45000

Divide both sides by 9/17

x=$85000

Hence, $85000 is the prize money is awarded in total.

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Answer: Its C!( 15 units)

Step-by-step explanation:

Answer:

c

Step-by-step explanation:

Step-by-step explanation:

I entered it in and it was .9993319736

Answer:11 kids were at the party

Step-by-step explanation:24 slices in all , 22 were eaten so that means 22 divided by 11 is 2 so there was 11 kids

Answer:

11 children

Step-by-step explanation:

First you have to do 8 times 3 to get the number of slices of pizza there were.

Then you have to divided that by 2. But remember, there are 2 slices left.

So, you have to take away one child to get the answer: 11 children

I feel bad for that last child. I hope this helped ;)

To determine if a data point is considered an outlier for a data set, we need to calculate the interquartile range (IQR) and use it to define the outlier boundaries. The IQR is the difference between the third quartile (Q3) and the first quartile (Q1). The correct option is (B).

We have that Q1 = 9 and Q3 = 17, we can calculate the IQR as follows:

IQR = Q3 - Q1 = 17 - 9 = 8

To identify outliers, we can use the following rule:

- Any data point that is less than Q1 - 1.5 * IQR or greater than Q3 + 1.5 * IQR is considered an outlier.

Using this rule, we can evaluate each data point:

A. 27: This data point is greater than Q3 + 1.5 * IQR = 17 + 1.5 * 8 = 29. It is considered an outlier.

B. 17: This data point is not an outlier because it is equal to the third quartile (Q3).

C. 3: This data point is less than Q1 - 1.5 * IQR = 9 - 1.5 * 8 = -3. It is considered an outlier.

D. 41: This data point is greater than Q3 + 1.5 * IQR = 17 + 1.5 * 8 = 29. It is considered an outlier.

Therefore, the outliers in the data set are A (27) and D (41).

As for when to use the mean, it is generally recommended to use the mean as a measure of central tendency when the data is symmetric and does not have extreme values.

Therefore, the correct option would be B. When the data is symmetric.

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Answer:

0.40390 km

Step-by-step explanation:

1 cm = 1 / 100000 km

40390 cm

= 40390 / 100000

= 0.40390 km

Given, Height of the cliff = 20 m Distance of the car from the foot of the cliff = 10 m.

The time taken by the car to fall from the cliff can be found using the formula:

\(`h = (1/2) g t^2`\)

Where h is the height of the cliff, g is the acceleration due to gravity and t is the time taken by the car to fall from the cliff.

Substituting the given values,`20 = (1/2) × 9.8 × t^2`

Solving for t, `t = sqrt(20/4.9)` = 2.02 s

Let the initial velocity of the car be V0 and the time taken for the car to come to rest after applying brakes be t1.

Distance covered by the car before coming to rest can be found using the formula: `\(s = V0t1 + (1/2) (-5) t1^2\)`

Where s is the distance covered by the car before coming to rest.

Simplifying the above equation,\(`2.5 = V0 t1 - (5/2) t1^2`\)

Substituting the given values,`5 = V0 - 5 t1`

Solving the above two equations,\(`V0 = 32.5/2 t1`\)

Simplifying the above equation,`V0 = 16.25 t1`

Substituting the value o\(f t1,`V0 = 16.25 × 2` = 32.5 m/s\)

Therefore, the speed of the car before the start of braking is 32.5 m/s.

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Answer:

Step-by-step explanation:

Explanation in image

There are 43 small dogs signed up in the competition

Represent the small dogs with x and the large dogs with y.

So, we have the following equations

x + y = 50

x = 36 + y

Make y the subject in x = 36 + y

y = x - 36

Substitute y = x - 36 in x + y = 50

x + x - 36 = 50

Evaluate the like terms

2x = 86

Divide both sides by 2

x = 43

Hence, there are 43 small dogs in the competition

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It takes the pair of gloves approximately 2 seconds to hit the ground.

To find the time it takes for the gloves to hit the ground, we can use the free fall equation, which is d = 1/2 * g * t^2, where d is the distance (64 feet), g is the acceleration due to gravity (approximately 32 ft/s^2), and t is the time in seconds.
1. Rearrange the equation to solve for t: t^2 = 2d/g
2. Plug in the given values: t^2 = 2(64 ft) / 32 ft/s^2
3. Calculate the result: t^2 = 4
4. Find the square root to get the time: t = √4 = 2 seconds

Summary: When Gary drops the pair of gloves from a 64-foot-high balcony, it takes about 2 seconds for them to hit the ground.

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The five-number summary is:

Minimum: 9

First Quartile: 16.5

Median: 25.5

Third Quartile: 39

Maximum: 51

3. Range = 42

4. Interquartile range = 22.5

Given the data for the lengths as, 36, 15, 9, 22, 36, 14, 42, 45, 51, 29, 18, 20, to find the five-number summary of the data set, we would follow the steps below:

1. The numbers in ordered from the smallest to the largest would be:

9, 14, 15, 18, 20, 22, 29, 36, 36, 42, 45, 51

2. The five-number summary for the lengths in minutes would be:

3. Range of the data = max - min = 51 - 9 = 42

4. The interquartile range for the data set = Q3 - Q1 = 39 - 16.5

Interquartile range for the data set = 22.5

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To determine if the polynomial is even, odd, or neither, we substitute -x for x in the polynomial and simplify. -3(-x)² + 6(-x) = -3x² - 6x. Since the polynomial is not equal to its negation, it is neither even nor odd.

To write the equation of the line with an x-intercept at -6 and a y-intercept at 2, we can use the slope-intercept form of a line, y = mx + b, where m is the slope and b is the y-intercept.

In this case, the y-intercept is given as 2, so the equation becomes y = mx + 2. To find the slope, we can use the formula (y2 - y1) / (x2 - x1) with the given points (-6, 0) and (0, 2). We find that the slope is 1/3. Thus, the equation of the line is y = (1/3)x + 2.

For the polynomial f(x) = -3x² + 6x, the degree is 2 and the leading coefficient is -3. The end behavior of the graph is determined by the degree and leading coefficient. Since the leading coefficient is negative, the graph will be "downward" or "concave down" as x approaches positive or negative infinity.

To find the zeros, we set the polynomial equal to zero and solve for x. -3x² + 6x = 0. Factoring out x, we get x(-3x + 6) = 0. This gives us two solutions: x = 0 and x = 2.

The x-intercept is the point where the graph intersects the x-axis, and since it occurs when y = 0, we substitute y = 0 into the polynomial and solve for x. -3x² + 6x = 0. Factoring out x, we get x(-3x + 6) = 0. This gives us two x-intercepts: (0, 0) and (2, 0).

To determine if the polynomial is even, odd, or neither, we substitute -x for x in the polynomial and simplify. -3(-x)² + 6(-x) = -3x² - 6x. Since the polynomial is not equal to its negation, it is neither even nor odd.

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To show that w is in the subspace of R4 spanned by V1, V2, and V3, we need to find constants c1, c2, and c3 such that:

w = c1V1 + c2V2 + c3V3

We can write this as a matrix equation:

| 1 -4 -9 -4 | | c1 | | -2 |

| 6 3 5 1 | x | c2 | = | -1 |

| 2 4 11 -8 | | c3 | | -1 |

| -15 7 22 -14 | | -2 |

We can solve this system of equations using row reduction:

| 1 -4 -9 -4 | | c1 | | -2 |

| 6 3 5 1 | x | c2 | = | -1 |

| 2 4 11 -8 | | c3 | | -1 |

| -15 7 22 -14 | | -2 |

R2 = R2 - 6R1

R3 = R3 - 2R1

R4 = R4 + 15R1

| 1 -4 -9 -4 | | c1 | | -2 |

| 0 27 59 25 | x | c2 | = | 11 |

| 0 12 29 -16 | | c3 | | 3 |

| 0 -23 67 -59 | | -32 |

R4 = R4 + 23R2

| 1 -4 -9 -4 | | c1 | | -2 |

| 0 27 59 25 | x | c2 | = | 11 |

| 0 12 29 -16 | | c3 | | 3 |

| 0 0 174 -294 | | 225 |

R3 = R3 - (12/27)R2

R4 = (1/174)R4

| 1 -4 -9 -4 | | c1 | | -2 |

| 0 27 59 25 | x | c2 | = | 11 |

| 0 0 -1 22/3 | | c3 | | -13/3 |

| 0 0 1 -98/87 | | 25/58 |

R1 = R1 + 4R3

R2 = R2 - 59R3

R4 = R4 + (98/87)R3

| 1 0 -13 -10/3 | | c1 | | 21/29 |

| 0 27 0 -2119/87 | x | c2 | = | 2238/87 |

| 0 0 1 -98/87 | | c3 | | 25/58 |

| 0 0 0 1390/2391 | | 1009/2391 |

R1 = R1 + (13/1390)R4

R2 = (1/27)R2

R3 = R3 + (98/1390)R4

| 1

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Answer:

You have 3 pennies and your friend has 1 penny

Step-by-step explanation:

The given parameters are;

Let friend A = you

friend B = Your friend

The number of pennies with  friend A when he gives friend B 1 penny = Equal number of pennies between friend A and B

The number of pennies with friend A when he is given 1 penny by friend B = 2 times the number of pennies with friend B

Let the number of pennies with friend A = X and the number of pennies with friend B = Y, therefore, we have;

X - 1 = Y + 1.....................(1)

X + 1 = 2 × Y...................(2)

Subtracting equation (1) from equation (2), we get;

X + 1 - (X - 1) = 2·Y - (Y - 1)

X - X + 1 + 1  = 2·Y - Y + 1

2 = Y + 1

Y = 2 - 1 = 1

Y = 1

From equation (1), X - 1 = Y + 1 gives;

X - 1 = 1 + 1

X = 1 + 1 + 1 = 3

Friend A has 3 pennies and friend B has 1 penny.

Therefore, you have 3 pennies and your friend has 1 penny

We can conclude that the linear transformation T(x) = Ax is necessarily the orthogonal projection onto a subspace of R^n since A is a projection matrix that projects vectors onto a subspace that is the direct sum of orthogonal eigenspaces.

The answer to this question is a long one, so let's break it down.

First, let's define what it means for a matrix to be symmetric.

A matrix A is symmetric if it is equal to its transpose, or A = A^T. This means that the entries of A above and below the diagonal are equal, and the matrix is "reflected" along the diagonal.

Now, let's consider what it means for a matrix A to satisfy A^2 = A.

This condition is often called idempotency since squaring the matrix doesn't change it.

Geometrically, this means that the linear transformation T(x) = Ax "squares" to itself - applying T twice is the same as applying it once.

One interpretation of idempotency is that A "projects" vectors onto a subspace of R^n, since applying A to a vector x "flattens" it onto a lower-dimensional subspace.

So, is T(x) = Ax necessarily the orthogonal projection onto a subspace of R^n? The answer is yes but with some caveats.

First, we need to show that A is a projection matrix, meaning it does indeed project vectors onto a subspace of R^n. To see this, let's consider the eigenvectors and eigenvalues of A.

Since A is symmetric, it is guaranteed to have a full set of n orthogonal eigenvectors, denoted v_1, v_2, ..., v_n. Let λ_1, λ_2, ..., λ_n be the corresponding eigenvalues.

Now, let's look at what happens when we apply A to one of these eigenvectors, say v_i. We have:
Av_i = λ_i v_i

But since A^2 = A, we also have:
A(Av_i) = A^2 v_i = Av_i

Substituting the first equation into the second, we get:
A(λ_i v_i) = λ_i (Av_i) = λ_i^2 v_i

So, we see that A(λ_i v_i) is a scalar multiple of λ_i v_i, which means that λ_i v_i is an eigenvector of A with eigenvalue λ_i. In other words, the eigenspace of A corresponding to the eigenvalue λ_i is spanned by the eigenvector v_i.

Now, let's consider the subspace W_i spanned by all the eigenvectors corresponding to λ_i. Since A is symmetric, these eigenvectors are orthogonal to each other. Moreover, we have:
A(W_i) = A(span{v_i}) = span{Av_i} = span{λ_i v_i} = W_i

This means that A maps the subspace W_i onto itself, so A is a projection matrix onto W_i. Moreover, since A has n orthogonal eigenspaces, it is the orthogonal projection onto the direct sum of these spaces.


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Answer:

do the work if you think its easy

Answer: 9/7

Step-by-step explanation:

Girls = 18

Total = 32

32-18 = 14

18/14 simplify

= 9/7 (B)

Answer:

(2,5) is right answer ok

Answer:

C. y=2x-48

Step-by-step explanation:

A. y=48x−2

B. y=48x+2

C. y=2x−48

D. y=2x+48

Profit per hotdog=$2

Quantity=x

Cost of hotdog, buns and mustard=$48

Find the equation profit earned by the hotdog stand for x hot dog sold

Total Profit(y)=(units profits*quantity)-cost

=$2*x-$48

y=2x-48

Answer:

C. y=2x−48

Step-by-step explanation:

did it on edge 2020

Answer:

14.25 hours

Step-by-step explanation:

We know

2.25 hours = 1 yard

13.5 hours = 6 yards

1/3 yard = 2.25 / 3 = 0.75 hours

To find how long it takes him to complete 6 1/3 yards, we take

13.5 + 0.75 = 14.25 hours

So, he spent 14.25 hours mowing yards.

The critical value that should be used in constructing the confidence interval is 1.645.

We are constructing a 90% confidence interval, so the alpha level is 1 - 0.90 = 0.10. The z-score that corresponds to an alpha level of 0.10 is 1.645.

We can find the critical value using the following steps:

1. We can look up the z-score in a z-table.

2. We can use a statistical calculator to find the z-score.

The following is the z-table for a two-tailed test with an alpha level of 0.10:

```

z-score | Probability

------- | --------

1.645 | 0.9500

```

As we can see, the z-score that corresponds to an alpha level of 0.10 is 1.645.

We can also use a statistical calculator to find the z-score. For example, in Excel, we can use the following formula:

```

=NORMSINV(0.95)

```

This will return the value 1.645.

Once we have found the critical value, we can use it to construct the confidence interval.

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Answer:

16.789

Step-by-step explanation:

The end of the number doesnt end in a 0 and all of the numbers after the decimal are significant

Answer:

16.789

Step-by-step explanation:

The end of the number doesn't end in a 0 and all of the numbers after the decimal are significant

Answer: 1 tablespoon

Step-by-step explanation:

The probability that a randomly selected student is enrolled in a history class but not a math class is \(\frac{47}{200}\)

We can solve this problem using the formula: P(History but not Math) = P(History) - P(History and Math)

Where P(History) is the probability of a student being enrolled in history, and P(History and Math) is the probability of a student being enrolled in both history and math.

Given:

P(History) = \(\frac{138}{200}\)

P(Math) = \(\frac{115}{200}\)

P(History and Math) = \(\frac{91}{200}\)

Substituting the values:

\(P(History but not Math) = \frac{138}{200}- \frac{91}{200}\)

Simplifying:

\(P(History but not Math) = \frac{47}{200}\)

Therefore, the probability that a randomly selected student is enrolled in a history class but not a math class is \(\frac{47}{200}\).

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Answer:

16

Step-by-step explanation:

Answer:

9

Step-by-step explanation:

Use Pemdas

6 - 3 = 3

4 + 8 = 12

12 / 2 = 6

6 + 3 = 9

Solution :

Given :

n = 200, p = 80% = 0.8, q = 1 - p = 0.2

\($\mu = np$\)

  \($= 200 \times 0.8 = 160$\)

\($\sigma= \sqrt{npq}$\)

\($\sigma= \sqrt{200 \times 0.8 \times 0.2}$\)

   = 5.6569

a). x = 169.5

∴ \($z = \frac{x- \mu}{\sigma}$\)

  \($z = \frac{169.5- 160}{5.6569}$\)

     = 1.6794

\($P(x \geq 170) = P(z>1.6794) = 0.0465$\)

Therefore, probability that the stolen goods were not recovered in 170 robberies or more is = 0.0465

b). x= 149.5

   \($z = \frac{x- \mu}{\sigma}$\)

   \($z = \frac{149.5- 160}{5.6569} = -1.8561$\)

\($P(x \geq 150) = P(z>-1.8561) = 0.9683$\)

Therefore the probability that the stolen goods were not recovered in the 150 or in more robberies is = 0.9683

Quadrilateral ABCD is rotated 90° clockwise to produce A'B'C'D'. A'B'C'D' is congruent with ABCD.

Quadrilateral:

A quadrilateral is a quadrilateral geometric shape. There are many different shapes such as parallelograms, rectangles, squares, rhombuses, rhombuses, and trapezoids. Assuming these numbers lie in the Cartesian plane, we can apply other transformations such as: B. Move the rectangle along a ruler or along a translation vector called translation. Flipping a square through the axis of symmetry is a reflection. Rotation involves rotating a square clockwise or counterclockwise around a fixed point.

The Axis Collection below shows a chart of rectangles ABCD. If ABCD is rotated 90 degrees clockwise from the origin, quadrilateral A' B' C' D' appears. Rectangle ABCD is congruent with rectangle A' B' C' D'.

So in such a scenario A'B'C'D' will always match when rotated around the origin.

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