Matt reasoned that he can write 9/1000 as 0.9. Is he correct?

Answer:

[tex]\frac{60x^{20}y^{24}}{30x^{10}y^{12}}[/tex] in simplest form would be [tex]2x^{10}y^{12}.[/tex]

Step-by-step explanation:

Given expression [tex]\frac{60x^{20}y^{24}}{30x^{10}y^{12}}[/tex].

Let us simplify given rational expression in smaller parts.

[tex]\frac{60}{30} = 2[/tex]

[tex]\frac{x^{20}}{x^{10}} = x^{20-10} = x^{10}[/tex]

[tex]\frac{y^{24}}{y^{12}} = y^{24-12} = y^{12}.[/tex]

Combining all terms together, we get

[tex]2x^{10}y^{12}.[/tex]

Therefore,

[tex]\frac{60x^{20}y^{24}}{30x^{10}y^{12}}[/tex] in simplest form would be [tex]2x^{10}y^{12}.[/tex]

Final answer:

The probability of a 5-bit binary code word containing exactly one zero, when each bit has a 0.8 probability of being zero, is calculated using the binomial probability formula. The final answer to this problem is 0.0064.

Explanation:

The student is asking about the probability of having exactly one zero in a 5-bit binary code where each bit has a probability of 0.8 of being zero. To calculate this, we will consider the binomial distribution where the number of trials (n) is 5 (since the code has 5 bits) and the probability of success (having a zero) in each trial (p) is 0.8. We want to find the probability of having exactly one success (one zero).

Using the binomial probability formula:

P(X = k) = (n choose k) * p^k * (1-p)^(n-k),

where X is the random variable representing the number of zeros, and k is the number of successes (zeros) we want, which in this case is 1. We plug in our values:

n = 5

k = 1

p = 0.8

P(X = 1) = (5 choose 1) * 0.8^1 * (1-0.8)^(5-1) = 5 * 0.8 * 0.2^4 = 0.0064.

The final answer is that the probability of a 5-bit binary code containing exactly one zero is 0.0064.

Answer:

A pyramid is a three dimensional shape ,having base in the shape of any polygon and other faces in the shape of triangle.There are 8 edges, 5 vertices of this Pyramid.

→→ Sometimes

Why sometimes,because may be Length of edge that is base of Triangle Exceed the side length of Square, or Edge of Square exceed the side length of base of triangle.

The probability that the three marbles drawn are all the same color is 64/3375.

To find the probability that all three marbles drawn are the same color, we need to consider two cases:

1. The probability of drawing three red marbles
2. The probability of drawing three white marbles

For the first case, the probability of drawing one red marble from the bag is 4/15. After replacing the marble, the probability of drawing another red marble is still 4/15. So, the probability of drawing three red marbles is (4/15) * (4/15) * (4/15).

For the second case, the probability of drawing one white marble from the bag is 5/15. After replacing the marble, the probability of drawing another white marble is still 5/15. So, the probability of drawing three white marbles is (5/15) * (5/15) * (5/15).

Adding the probabilities of the two cases, we get (4/15) * (4/15) * (4/15) + (5/15) * (5/15) * (5/15), which simplifies to 64/3375.

Answer:

i think its 50....

To maximize the atrium area with a 756-ft perimeter, the dimensions should be 189 ft x 189 ft, with an area of 35,721 sq ft.

To find the dimensions that will maximize the area of the rectangular atrium, we need to set up an equation based on the given perimeter and then use calculus to find the dimensions that maximize the area.

Let's denote the length of the atrium as "L" and the width as "W."

The perimeter of a rectangle is given by the formula:

Perimeter = 2L + 2W

In this case, the perimeter is 756 ft, so we have:

2L + 2W = 756

Divide both sides of the equation by 2 to solve for L:

L + W = 378

Now, we can express L in terms of W:

L = 378 - W

The area of a rectangle is given by the formula:

Area = L * W

Substitute the expression for L in terms of W:

Area = (378 - W) * W

Area = 378W - W^2

To maximize the area, we need to find the critical points.

Take the derivative of the area formula with respect to W and set it equal to zero:

d(Area)/dW = 378 - 2W = 0

Solve for W:

2W = 378

W = 189

Now, we can find the corresponding value of L using L = 378 - W:

L = 378 - 189

L = 189

So, the dimensions that will maximize the area of the atrium are:

Length (L) = 189 ft

Width (W) = 189 ft

The maximum area of the atrium will be:

Area = L * W

Area = 189 ft * 189 ft

Area = 35,721 square feet.

For similar question on perimeter.

https://brainly.com/question/19819849

#SPJ3

Answer: The correct option is second, i.e., Subtraction Property of Equality.

Explanation:

The given equation is,

[tex]5x+7=2x-1[/tex]

The Subtraction Property of Equality states that if a, b and c are real numbers then, [tex]a=b[/tex] and [tex]a-c=b-c[/tex] are equivalent equations.

Using Subtraction Property of Equality we subtract 2x from both the sides of given equation,

[tex]5x+7-2x=2x-1-2x[/tex]

Then combined like terms,

[tex](5x-2x)+7=(2x-2x)-1[/tex]

[tex]3x+7=-1[/tex]

It is the required equation.

Hence, second option is correct.

To find the height of the cloud cover, we use the tangent of the 45° elevation angle, which equals 1. Since the distance is 590 m, the height of the cloud cover is also 590 m.

To calculate the height of the cloud cover, we can use trigonometry. The problem describes two angles and a distance between the spotlight and the observer, which forms a right-angled triangle on the ground and a triangle from the observer to the light on the cloud. The two angles in use are 75° (the angle of the spotlight from the horizontal) and 45° (the angle of elevation from the observer to the light). Using the tangent function, which is the ratio of the opposite side over the adjacent side in a right-angled triangle, we can find the height of the cloud cover (h).

Since the angle of elevation from the observer to the light on the cloud is 45°, the tangent of this angle is 1. This means that for every unit distance from the observer to the point directly underneath the cloud cover, there is one unit of rise. As we are given that the distance (d) is 590 m ,, using the tangent relationship (tan(45°) = h/d), we find that the height of the cloud cover (h) is equal to the distance to the observer (d).

The calculations are as follows:

Therefore, the height of the cloud cover is 590 meters.

Answer: $56,000

Step-by-step explanation:

Given: The equation [tex]y=1.55x+110,419[/tex] approximates the total amount, in dollars, spent by a household to raise a child in the United States from birth to 17 years, given the household’s annual income, x.

The household’s total cost of raising a child = $197,219

Put the above value in the equation, we get

[tex]197,219=1.55x+110,419\\\Rightarrow\ 1.55x=197219-110419\\\Rightarrow\ x=\frac{86800}{1.55}\\\Rightarrow\ x=56,000[/tex]

Hence, the household’s annual income is $56,000.

Answer:

Step-by-step explanation:

In the depreciation formula V(t) = Vo(b)^t, Vo is the initial value of $408,000, and b is 0.82 (1 minus 18% depreciation). After 8 years, using these values, the assets are valued at approximately $68,818.46 when rounded to the nearest cent.

The student's question relates to the depreciation of business assets over time. In the formula V(t) = Vo(b)^t, Vo represents the initial value of the assets and b represents the depreciation rate expressed as a base for the exponential function. Knowing that the asset is worth $408,000 initially and depreciates at 18% per annum, Vo equals $408,000, and b equals (1 - 0.18) or 0.82, which reflects the remaining value of the asset after one year.

To calculate the value after 8 years, we use the formula with the given values: V(8) = 408,000(0.82)^8. The calculation proceeds as follows:

To answer the question, the value of the assets after 8 years, rounded to the nearest cent, is $68,818.46.

The associative property of addition is exemplified in the equation 3+(x+y)=(x+y)+3.

The equation that is an example of the use of the associative property of addition is:

3+(x+y)=(x+y)+3

This equation demonstrates that when adding three numbers, the grouping of the numbers does not affect the sum, showcasing the associative property of addition.

To drag the coach at a constant speed, each player needs to apply a force of approximately 583.89 N to counteract the friction force of 1140 N on the sled, calculated using trigonometry and Newton's Second Law of Motion.

The key concept in this question involves Newton's Second Law of Motion and the dynamics of forces acting at angles. The friction force on the sled is given as 1140 N, and since the football coach is moving at a steady speed, we can infer that the net force is zero, implying that the pulling forces balance out the friction force.

To find the force each player must exert, we first determine the total force needed to overcome friction. Since the rope's angle is 25.0° (half the angle between the ropes), the total horizontal force exerted by the two players (F-total) needs to match the friction force. Using trigonometry, the force each player exerts (F-player) can be expressed as:

F-player = F-total / (2 × cos(25.0°/2)) = 1140 N / (2 × cos(12.5°))

So, we calculate the force:

F-player = 1140 N / (2 × cos(12.5°)) = 1140 N / (2 × 0.9763) = 1140 N / 1.9526 = 583.89 N

Therefore, each player needs to exert approximately 583.89 N to drag the coach at a steady speed of 2.30 m/s.