Four white, one black, and three striped tiles are place on a table. Each time a tile is drawn, it is replaced.


Determine the probability of drawing two striped tiles in a row.


ANSWER FAST PLEASE

The surveyor is located about 239.6 feet away from the base of the Willis Tower.

The height of the Willis Tower is 165 feet.

The angle of elevation from the surveyor to the top of the antenna is about 3.41°.

The height of the antenna is about 135.9 feet.

Let's call the distance from the surveyor to the base of the Willis Tower "x", and let's call the height of the Willis Tower "h".

We can use trigonometry to solve for x and h. First, let's find x:

tan(34°) = h/x

x = h/tan(34°)

Now we can use the distance from the surveyor to the top of the building to solve for h:

h + 2595 = 2760

h = 165

So the height of the Willis Tower is 165 feet. Now we can solve for x:

x = 165/tan(34°) ≈ 239.6 feet

So the surveyor is located about 239.6 feet away from the base of the Willis Tower.

To find the angle of elevation from the surveyor to the top of the antenna, we can use trigonometry again:

tan(θ) = h/2760

θ = tan^(-1)(h/2760)

θ ≈ 3.41°

So the angle of elevation from the surveyor to the top of the antenna is about 3.41°.

Finally, we can use the height of the Willis Tower and the distance from the surveyor to the top of the antenna to solve for the height of the antenna:

tan(34°) = (h + a)/2760

a ≈ 135.9

So the height of the antenna is about 135.9 feet.

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The marked points are (-2,-4), (-2,-1), (0,-2), (2,-1), and (2,-4), under the condition that they are on the same coordinate plane having (A) y=x-2, (B) y=-x-2, (C) y=|x|-2.

In the given graph points on the coordinate plane, we have to plot the points (x,y)
Here
x = horizontal axis
y = vertical axis.

In the given point A, y=x-2, we can continue at the origin (0,0) and move 2 units go down on the y-axis and 2 units right on the x-axis to plot point A at (2,0).

In the given point B, y=-x-2, we can continue at the origin (0,0) and transfer 2 units down on the y-axis and 2 units left on the x-axis to plot point B at (-2,0).

In the given point C, y=|x|-2, we can continue plotting two points for this equation.
When x is considered negative, we can procees at the origin (0,0) and transfer 2 units down on the y-axis and 2 units left on the x-axis to plot point C at (-2,0).
When x is positive, we can start at the origin (0,0) and move 2 units down on the y-axis and 2 units right on the x-axis to plot point C at (2,0).

Then, all points (x,y) such that (A) y=x-2, (B) y=-x-2, (C) y=|x|-2 are (-2,-4), (-2,-1), (0,-2), (2,-1), and (2,-4).
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The number of milligrams of sugar his spoon holds is 1250 milligrams.

To find out how many milligrams of sugar Chad's spoon holds, we first need to know how much sugar he removed in total. To do this, we can subtract the amount of sugar he needed (280 grams) from the amount he poured (295 grams).

295 grams - 280 grams = 15 grams

Next, we need to divide the total amount of sugar Chad removed (15 grams) by the number of spoonfuls he used (12).

15 grams ÷ 12 = 1.25 grams per spoonful

Finally, we can convert grams to milligrams by multiplying by 1000.

1.25 grams x 1000 = 1250 milligrams

Therefore, Chad's spoon holds 1250 milligrams of sugar.

It's important to note that when cooking or baking, precise measurements are crucial to the success of the recipe. Even small changes can greatly affect the outcome. While it's great that Chad was able to remove the excess sugar, it's best to be as accurate as possible from the start.

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Answer:

B, A, C

Step-by-step explanation:

The rate is the another name for the slope.

A:

Change in y over the change in x.  You find the change by subtracting

[tex]\frac{7-3}{5-3}[/tex] = [tex]\frac{4}{2}[/tex] = 2

The rate is 2.

B:

Change in y over the change in x. You find the change by subtracting.

[tex]\frac{0-3}{-5-0}[/tex] = [tex]\frac{-3}{-5}[/tex] = [tex]\frac{3}{5}[/tex]

The rate is [tex]\frac{3}{5}[/tex].

C:

The rate is the number before the x in the equation.

The rate is 3.

Helping in the name of Jesus.

Sorry for bad handwriting

if i was helpful Brainliests my answer ^_^

The conditions for inference are indeed met. The correct option is:
Yes, the conditions for inference are met.

The conditions for inference are met when conducting a confidence interval for a proportion if the following conditions are satisfied:

Random Sample: The sample should be a simple random sample or a random sample from a well-defined sampling frame. This ensures that the sample is representative of the population of interest.

Large Counts Condition: The sample size should be large enough so that both the number of successes (defective chips) and failures (non-defective chips) in the sample are at least 10. This ensures that the sampling distribution of the proportion is approximately normal.

Independence: The individual observations in the sample should be independent of each other.

In this scenario, the quality control specialist took a random sample of 100 chips from the day's production, which satisfies the random sample condition.

Now, let's check the Large Counts Condition.

The quality control specialist found 12 defective chips in the sample. To satisfy the Large Counts Condition, both the number of defective chips and the number of non-defective chips should be at least 10.

In this case, the number of defective chips is 12, and the number of non-defective chips is 100 - 12 = 88.

Both numbers are greater than 10, so the Large Counts Condition is met.

Since both the random sample condition and the Large Counts Condition are met, the conditions for inference are indeed met. Therefore, the answer is:

Yes, the conditions for inference are met.

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Yes, Mark is correct he believes that doubling one of the dimensions would double the volume of his toolbox.

Volume refers to the space occupied by a 3-Dimensional space. The volume of a cuboid is given by:

V = l * b * h

where l is the length

b is the breadth

h is the height

V = 9 * 6 * 7

= 378 cubic inches

If we double any of the dimensions, like

By doubling the 9 we get

V = 18 * 6 * 7

= 756 cubic inches

By doubling the 6 we get

V = 9 * 12 * 7

= 756 cubic inches

By doubling the 7 we get

V = 9 * 6 * 14

= 756 cubic inches

Then the volume of the toolbox is doubled as shown above.

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The clockwise rotation about the origin from the pre-image to the image​ include the following: a rotation of 90° clockwise.

In Mathematics, a rotation is a type of transformation which moves every point of the object through a number of degrees around a given point, which can either be clockwise or counterclockwise (anticlockwise) direction.

Additionally, the rotation of a point 90° about the center (origin) in a clockwise direction would produce a point that has these coordinates (y, -x).

Next, we would apply a rotation of 90° clockwise to the coordinate of triangle ABC as follows;

(x, y)                               →            (y, -x)

Coordinate A = (1, 1) → Coordinate A' = (1, -(1)) = (1, -1)

Coordinate B = (1, 4) → Coordinate B' = (4, -(1)) = (4, -1)

Coordinate C = (5, 1) → Coordinate C' = (1, -(5)) = (1, -5)

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Ans. Correct option in (c) 12!/11!

we know that

the formula of nPr is n! / (n−r)!.

in this question

n = 12 and r = 1

so putting values we get ,

= 12! / (12-1)!

=  12!/11!

(a) To minimize the area of the right triangle formed by the cable, ground, and building, we need to minimize the length of the cable. To do this, we can use the Pythagorean theorem:

c^2 = a^2 + b^2

where c is the length of the cable, a is the distance from the guidepost to the point where the cable is attached to the building, and b is the distance from that point to the ground.

Since we want to minimize c, we can differentiate the equation with respect to a and set the derivative equal to zero:

dc/da = 2a/c = 0

Solving for a, we get a = c/2. This means that the point where the cable is attached to the building should be halfway up the building, or 10 feet high.

(b) To minimize the length of the cable, we can use the principle of least action, which states that the path taken by the cable is the one that minimizes the integral of the tension along the cable.

Assuming that the tension in the cable is constant, we can use the law of sines to find the angle θ:

sin θ / 20 = sin (90° - θ) / c

where c is the length of the cable.

We want to minimize c, so we can differentiate the equation with respect to θ and set the derivative equal to zero:

d(c)/d(θ) = -20cos(θ) / sin^2(θ) + cos(θ) / sin(θ) * dc/d(θ) = 0

Solving for dc/d(θ), we get:

dc/d(θ) = 20c * tan(θ)

Substituting this into the original equation, we get:

-20cos(θ) / sin^2(θ) + cos(θ) / sin(θ) * 20c * tan(θ) = 0

Simplifying, we get:

cos(θ) / sin(θ) = tan(θ)

Solving for θ, we get:

θ = 45°

Therefore, to minimize the length of the cable, it should make an angle of 45° with the face of the building.

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The problem presents a scenario where a square patio is placed in the centre of a 10m x 10m square backyard. The side length of the patio is given by x, and the remaining area of grass is expressed as A=-x^2+100.

To find the area of grass remaining, we can substitute different values of x into the equation.

For instance, if the patio is 5m x 5m, then x = 5 and the area of grass remaining is[tex]A = -5^2 + 100 = 75[/tex] square metres. Similarly, if the patio is 8m x 8m, then x = 8 and the area of grass remaining is [tex]A = -8^2 + 100 = 36[/tex] square metres.

As we can see, the area of grass remaining decreases as the size of the patio increases.

This problem illustrates the concept of content loaded and content remaining, where the initial content is the entire area of the square backyard, and the loaded content is the area of the square patio.

The remaining content is what is left after the loaded content is subtracted from the initial content. In this case, the loaded content is the patio area, and the remaining content is the grass area.

In summary, the area of grass remaining in the backyard after a square patio is placed in the centre can be calculated using the equation [tex]A=-x^2+100,[/tex] where x is the side length of the patio in metres.

The concept of content loaded and content remaining is also illustrated in this problem, where the loaded content is the patio area and the remaining content is the grass area.

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The final values is (fog)(x) = f(g(x)) = f(0) = -8.

In the given problem, we are given two functions, f(x) and g(x). The function f(x) is a polynomial function, and g(x) is a constant function equal to 0. We are asked to find the composition of these two functions, that is, (fog)(x) and (gof)(x).

The composition of two functions f(x) and g(x) is denoted by (fog)(x) and is defined as follows:

(fog)(x) = f(g(x))

This means that we first evaluate g(x) and then use the output of g(x) as the input of f(x) to get the final output of (fog)(x).

In this case, since g(x) = 0, we have:

(fog)(x) = f(g(x)) = f(0)

To evaluate f(0), we substitute x = 0 in the expression for f(x):

[tex]f(x) = 2x^2 - 6x^2 + 4x - 8[/tex]

[tex]f(0) = 2(0)^2 - 6(0)^2 + 4(0) - 8[/tex]

f(0) = -8

Therefore, (fog)(x) = f(g(x)) = f(0) = -8.

Now, to find (gof)(x), we need to evaluate g(f(x)). Since f(x) is a polynomial function, we can find its value for any value of x. However, since g(x) is a constant function equal to 0, its output is always 0 for any input x. Therefore, g(f(x)) = 0 for all values of f(x).

This means that (gof)(x) = g(f(x)) = 0 for all x.

In summary, we found that (fog)(x) = -8 and (gof)(x) = 0.

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The perimeter is indeed 26 feet, which means they have used all the extra fencing.

Let's start by assuming that the length and width of the garden patch are whole numbers of feet, since we are asked to use only whole numbers.

Let's call the length of the garden patch "L" and the width "W".

We know that Liam has 26 feet of fencing left over. This fencing will be used to make the perimeter of the garden patch, which is given by:

Perimeter = 2L + 2W

We can substitute the value of the perimeter with the amount of fencing that Liam has:

26 = 2L + 2W

Simplifying this equation, we get:

13 = L + W

Since we want to use all the extra fencing, we know that the perimeter of the garden patch must be 26 feet. We can use this information to write another equation:

Perimeter = 2L + 2W = 26

We can substitute the value of 13 for L + W in this equation:

2L + 2W = 26

2L + 2(13-L) = 26

2L + 26 - 2L = 26

26 = 26

This equation is true, which means that our assumption that L and W are whole numbers is correct.

Therefore, the dimensions of the garden patch that Liam and his brother can make for their sister are 6 feet by 7 feet.

To check, we can calculate the perimeter:

Perimeter = 2L + 2W = 2(6) + 2(7) = 12 + 14 = 26

So the perimeter is indeed 26 feet, which means they have used all the extra fencing.

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The distance between Town A and Town B is 550 km.

Let's denote the distance between Town A and Town B as D.

When the two cars first passed each other, let's assume that car A traveled a distance of x km and car B traveled a distance of D - x km.

Let's also denote the initial speeds of car A and car B as 6s and 5s, respectively, where s is some constant representing the speed of the slower car.

The time it took for the two cars to pass each other can be calculated using the formula:

time = distance / speed

For car A, the time it took to travel x km was:

x / (6s)

For car B, the time it took to travel D - x km was:

(D - x) / (5s)

Since the two cars traveled the same amount of time until they passed each other, we can set these two expressions equal to each other:

x / (6s) = (D - x) / (5s)

Solving for x, we get:

x = 6Ds / (11s)

After the speeds of both cars were reduced, car A's speed was (5/6) * 6s = 5s, and car B's speed was (3/4) * 5s = (15/4)s.

Let's denote the time it took for car A to travel the remaining distance from x to D as t.

Then, the time it took for car B to travel a distance of (D - x - 54) km is also t.

Using the new speeds, we can write the equation:

[tex](D - x - 54) = (15/4)s * t[/tex]

Solving for t, we get:

[tex]t = (4/15)(D - x - 54) / s[/tex]

The distance car A traveled after the two cars passed each other is:

D - x = D - 6Ds / (11s) = (5/11)D

The time it took for car A to travel this distance is:

[tex]t + x / (6s) = (4/15)(D - x - 54) / s + 6Ds / (66s)[/tex]

Setting these two expressions equal to each other and solving for D, we get:

D = 550 km

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Regarding the second question, the management of Neat Upholsterers may determine whether a person has worked overtime by tracking their hours of work and comparing them to the standard working hours or the overtime policy defined in the employment contract or labor laws. This could involve using time cards, electronic systems, or other methods of tracking employee hours.

Whether this policy is fair or not depends on various factors, such as the specific overtime policy, the industry norms, the labor laws, and the bargaining power of the employees. If the overtime policy is reasonable, transparent, and consistent with the labor laws and the industry standards, and if the employees are compensated fairly for their extra work, then the policy could be considered fair. However, if the policy is exploitative, discriminatory, or violates the legal or ethical standards, then it could be considered unfair.

Answer:

The answer is A

Step-by-step explanation:

Answer:

38

Step-by-step explanation:

Formula

Inscribed angle = Central angle/2

Here

Inscribed angle = x

Central angle = 76

x = 76/2

x = 38

The result should be interpreted with caution and cannot be directly extrapolated to the real-world Fender's blue butterfly populations, and the simulation does not take these factors into account.

The result of the simulation suggests that in a hypothetical scenario where the Fender's blue butterfly population is restricted to stepping stones, and the leave prairie probability is set to 0.9, the population is likely to increase over time. However, it is important to note that the simulation represents an idealized scenario and may not reflect the complexity of real-world butterfly populations.

Furthermore, the absence of fires in the simulation may not reflect the natural habitat of Fender's blue butterfly, as fire is a crucial factor in maintaining prairie habitats. In the real world, fire suppression and habitat fragmentation are major threats to the survival of Fender's blue butterfly populations, and the simulation does not take these factors into account.

In summary, while the simulation result may provide insights into the potential effectiveness of using stepping stones to conserve butterfly populations, it should be interpreted with caution and cannot be directly extrapolated to the real-world Fender's blue butterfly population. Further research and monitoring of butterfly populations in their natural habitats are necessary to fully understand their dynamics and inform conservation efforts.

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The options that are statistical are:

how many pets do you have?

what time do you go to sleep?

how many siblings do you have?

The options that are non-statistical are:

what kind of dog does bruno have?

what is clara's brother's name?

what time did javed go to sleep?

A statistical question is defined as one that will obtain the data that will vary from one particular response to another. However, a non-statistical question is defined as one that will obtain data that is basically exact and then has only one response.

A question that will not provide a variety of different answers is referred to as not a statistical question. For example, we can say that 'how many siblings do I have?' is not referred to as statistical. The answer will definitely have just one response, and not many.

A non-statistical question in math is defined as a question that will not provide a variety of answers. Finally, non-statistical questions provide us with exact answers that do not change.

Thus, the options that are statistical are:

how many pets do you have?

what time do you go to sleep?

how many siblings do you have?

The options that are non-statistical are:

what kind of dog does bruno have?

what is clara's brother's name?

what time did javed go to sleep?

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Ali's Gulf club account will have a future value of BD 101.93 at the end of the term on October 7. The calculation was done using the formula for future value of an annuity with monthly payments, interest rate of 8% per year, and a term of 2.9 months.

We can first calculate the number of months from July 11 to October 7: 2 months and 27 days (or approximately 2.9 months).

Then, we can use the formula for future value of a present sum with simple interest

FV = P(1 + rt)

where FV is the future value, P is the present sum (in this case, BD 100), r is the annual interest rate (8% = 0.08), and t is the time in years (2.9/12 = 0.2417 years).

Substituting the values, we get

FV = 100(1 + 0.08*0.2417)

= 100(1.01934)

= BD 101.93

Therefore, the future value of Ali's golf club account is BD 101.93.

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The greatest common what is the greatest common factor of 6a2b2 and 15a4b37 is option d.

To find the greatest common factor (GCF) of 6a^2b^2, 15a^4b^3, 7a^3b, and 3a^2b^2, follow these steps:

Step 1: Find the GCF of the numerical coefficients: The GCF of 6, 15, 7, and 3 is 1.

Step 2: Find the GCF of the 'a' terms: The lowest power of 'a' is a^2, so the GCF is a^2.

Step 3: Find the GCF of the 'b' terms: The lowest power of 'b' is b, so the GCF is b.

Combine the results from steps 1, 2, and 3: The GCF of 6a^2b^2, 15a^4b^3, 7a^3b, and 3a^2b^2 is 1a^2b.

Therefore, the GCF of 6a^2b^2 and 15a^4b^3 is 3a^2b^2, which is option (d).

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The statement "if point b is on ac and between points a and c, then mab + mbc = mac" describes the angle addition postulate in geometry.

In geometry, an angle is formed by two rays that share a common endpoint called a vertex. The measure of an angle is the amount of rotation between the two rays, usually measured in degrees or radians. The angle addition postulate states that if point B is on line segment AC and between points A and C, then the sum of the measures of angles MAB and MBC is equal to the measure of angle MAC. This postulate is used in various proofs and constructions in geometry, and it is also useful in real-world applications such as navigation, surveying, and engineering. The postulate is based on the fact that a straight angle measures 180 degrees, so if we know the measures of two angles that share a common ray, we can find the measure of the third angle by subtracting the sum of the first two angles from 180 degrees.

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Answer:

9, 10, 11, 12, 13.

Step-by-step explanation:

All possible values for the number of tables that the store must sell in order to meet the requirements are 9, 10, 11, 12, 13

The radius of the base of the cone is approximately 12.65 feet if The cone is formed from 3,200 ft of gravel.

Height of cone =  24 feet

The volume of the cone = [tex]3,200 ft^3[/tex]

To find the volume of the cone, the formula used here is:

V =π* [tex]r^2h[/tex]

Here, the values of V and H are known terms. we need to calculate the radius r of the cone.

π = 3.14 constant value

Substituting the values in the above equation, we get:

[tex]3,200 = (1/3)^2*(24)*[/tex] π

[tex]r^2[/tex]= 3,200 / (8π)

[tex]r^2[/tex] = 400 / π

r = 12.65

Therefore, we can conclude that the radius of the base of the cone is  12.65 feet.

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The farmer would have made a profit of 5% if he had sold the goat for 2100 shillings.

Let's use the given terms and find out the percentage profit if the farmer had sold the goat for 2100 shillings.
Calculate the cost price of the goat
We know that the farmer made a loss of 28% by selling the goat for 1440 shillings. Let's represent the cost price as "CP".

We can write the equation:
[tex]CP \times  (1 - loss% ) = selling price (SP)[/tex]
[tex]CP \times  (1 - 0.28) = 1440[/tex]
Solve for CP
[tex]CP \times 0.72 = 1440[/tex]
CP = 1440 / 0.72
CP = 2000 shillings
Calculate the percentage profit
Now we want to find out the percentage profit if the farmer had sold the goat for 2100 shillings.

We can write the equation:
[tex](SP_{new - CP)} / CP \times  100 = profit[/tex]
[tex](2100 - 2000) / 2000 \times  100 = profit%[/tex]
[tex]100 / 2000 \times  100 = profit%[/tex]
5% = profit%.

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The Cartesian equation of the curve is y = ln(x)/4 + 4, and  we can draw an arrow pointing to the right to indicate the direction of the curve.

(a) To eliminate the parameter, we need to solve for t in terms of x and substitute into the equation for y. From the equation x = e⁴ᵗ, we have t = ln(x)/4. Substituting into y = t + 4, we get y = ln(x)/4 + 4. Therefore, the Cartesian equation of the curve is y = ln(x)/4 + 4.

(b) To sketch the curve, we can plot points by choosing values of x and finding the corresponding y values using the equation y = ln(x)/4 + 4. As x increases, y increases but at a slower rate. This means that the curve is increasing but is becoming less steep.

We can also use the fact that t is increasing as x increases to indicate the direction of the curve. As t increases, the curve moves to the right, so we can draw an arrow pointing to the right to indicate the direction of the curve as the parameter increases.

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The bearing of U from T in the image is 32 degrees South by West of T

In mathematics, bearings refer to the direction of an object or location in relation to two points. It is determined by the angle between the line joining them and that of the north.

Measured typically in degrees, it follows a cardinal system where 0° or 360° signifies North; East stands for 90°, South denotes at 180° while West represents 270°.

Thus, it can be seen that the bearing of U from T in the image is 32 degrees South by West of T

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Answer:

12

Step-by-step explanation

60/100 to get the probability of success

0.6 * 20 attempts = 12

B. Whole number enlarges, fraction reduces.

When a figure is dilated by a whole number, the image is enlarged by a factor of that whole number. For example, if a figure is dilated by a scale factor of 2, the image will be twice as large as the original.

On the other hand, when a figure is dilated by a fraction, the image is reduced. For example, if a figure is dilated by a scale factor of 1/2, the image will be half as large as the original.

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