Please answer these... please

Answer:

Yes

Step-by-step explanation:

y is the litters of water in the tank

x is the time in minutes

m is the change in litters

b is the initial amount of water

so the formula is y = mx + b

whereas m is -14 and b is 770

y = -14x + 770 or y = 770 - 14x

Answer:

40,000 songs, I think

Step-by-step explanation:

The image position is approximately 1111.11 cm behind the mirror for first case and 25.68 cm in front of the mirror for second case, and the image height is approximately 138.9 cm and 6.848 cm.

To solve these problems, we can use the mirror equation:

1/f = 1/\(d_o\) + 1/\(d_i,\)

where:

f is the focal length of the mirror,

\(d_o\) is the object distance (distance from the object to the mirror), and

\(d_i\) is the image distance (distance from the image to the mirror).

We can also use the magnification equation:

m = -\(d_i\)/\(d_o\),

where m is the magnification of the mirror.

Let's solve the problems step by step:

Problem 1:

Object height \((h_o)\) = 2.0 cm

Object distance \((d_o)\) = -16 cm (negative because it is in front of the mirror)

Focal length (f) = -25 cm (negative because it is a concave mirror)

Using the mirror equation:

1/f =\(1/d_o + 1/d_i\)

Substituting the given values:

1/-25 = 1/-16 + 1/\(d_i\)

Simplifying the equation:

-1/25 = -1/16 + 1/\(d_i\)

Finding a common denominator:

-1/25 = (-16 + 25)/(-16 * 25) + 1/\(d_i\)

Simplifying further:

-1/25 = 9/(-400) + 1/\(d_i\)

Combining terms:

-1/25 = 9/(-400) + 1/\(d_i\)

Multiplying through by (-25) to get rid of the fraction:

1 = 9/400 + (-25/\(d_i\))

Rearranging the equation:

-9/400 = -25/\(d_i\)

Cross-multiplying:

-9 * \(d_i\) = -25 * 400

Simplifying:

-9 * \(d_i\) = -10,000

Dividing by -9:

\(d_i\) = 10,000 / 9 ≈ 1111.11 cm

The image position is approximately 1111.11 cm (or 11.11 m) behind the mirror.

Using the magnification equation:

m =\(-d_i/d_o\)

Substituting the given values:

m = -1111.11 / -16

Simplifying:

m ≈ 69.45

The magnification is approximately 69.45.

To find the image height \((h_i),\) we can use the magnification equation:

m =\(h_i/h_o\)

Substituting the values:

69.45 =\(h_i / 2\)

Simplifying:

\(h_i\) ≈ 138.9 cm

The image height is approximately 138.9 cm.

Therefore, the image position is approximately 1111.11 cm behind the mirror, and the image height is approximately 138.9 cm.

Problem 2:

Object height (h₀) = 4.0 cm

Object distance (d₀) = -15 cm (negative because it is in front of the mirror)

Focal length (f) = 25 cm (positive because it is a convex mirror)

Using the mirror equation:

1/f =\(1/d_o + 1/d_i\)

Substituting the given values:

1/25 = 1/-15 + 1/\(d_i\)

Simplifying the equation:

1/25 = -1/15 + 1/\(d_i\)

Finding a common denominator:

1/25 = (-15 + 25)/(15 * 25) + 1/\(d_i\)

Simplifying further:

1/25 = 10/(375) + 1/\(d_i\)

Combining terms:

1/25 = 10/(375) + 1/\(d_i\)

Multiplying through by (25) to get rid of the fraction:

1 = 10/375 + (25/\(d_i\))

Rearranging the equation:

1 - 10/375 = 25/\(d_i\)

Simplifying:

365/375 = \(25/d_i\)

Cross-multiplying:

\(365 * d_i\) = 25 * 375

Simplifying:

\(365 * d_i\) = 9375

Dividing by 365:

\(d_i\) = 9375 / 365 ≈ 25.68 cm

The image position is approximately 25.68 cm in front of the mirror.

Using the magnification equation:

m = -\(d_i/d_o\)

Substituting the given values:

m = -25.68 / -15

Simplifying:

m ≈ 1.712

The magnification is approximately 1.712.

To find the image height \((h_i)\), we can use the magnification equation:

m = \(h_i/h_o\)

Substituting the values:

1.712 = \(h_i\)/ 4

Simplifying:

\(h_i\) ≈ 6.848 cm

The image height is approximately 6.848 cm.

Therefore, the image position is approximately 25.68 cm in front of the mirror, and the image height is approximately 6.848 cm.

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The solution to this problem is the p-value must be higher than 0.05.

A confidence interval in frequentist statistics is a range of estimates for an unobserved parameter. The most common confidence level for computing confidence intervals is 95%, however other levels, including 90% or 99%, are sporadically employed.

Here ,

In this given problem we can see confidence interval has 0 in it.

So,

Since the confidence interval comprises 0, the null hypothesis should be rejected.

When the p-value exceeds the level of significance, the null hypothesis is only rejected.

Since alpha=0.05 in this case, the p-value must be higher than 0.05.

Thus, the p-value must be higher than 0.05.

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Answer:

x over y , but you'll have to do y over x to get the answer .

Answer:

x > 1/2

Step-by-step explanation:

Let [r, s] denote the least common multiple of positive integers r and s. (m+1)(n+1)2ᵏ-1 if k > 1. (If k = 1, then r is prime and c = r, so there are no restrictions on  [a, b].

Consider an ordered triple (a, b, c) of positive integers such that [a, b] = c. Let c = r₁r₂r₃ ... rₖ be the prime factorization of c. Suppose that rᵢ divides a and rⱼ divides b for some i, j (1 ≤ i ≤ j ≤ k). Without loss of generality, suppose that i is the smallest index for which such a condition is met. Since c is the least common multiple of a and b, it must be the case that a / rᵢ and b / rⱼ are coprime.

Since a / rᵢ is a positive integer, it must have a factorization of the form a / rᵢ = p₁p₂p₃ ... pₘ, where each pᵢ is a prime. It follows that a = p₁p₂p₃ ... pₘrᵢ. Likewise, b = q₁q₂q₃ ... qₙrⱼ, where each qᵢ is a prime such that qᵢ ≠ rᵢ. (Note that we have the strict inequality here since a / rᵢ and b / rⱼ must be coprime.)Since c = r₁r₂r₃ ... rₖ is the least common multiple of a and b, it must be the case that each prime factor in the prime factorization of c occurs to the highest power in a and b. This means that each prime factor in the prime factorization of c must be present in p₁p₂p₃ ... pₘ and q₁q₂q₃ ... qₙ except for rᵢ and rⱼ.

There are  \(2^(k-2)\) ways to choose the set of prime factors (aside from rᵢ and rⱼ) that will appear in p₁p₂p₃ ... pₘ and q₁q₂q₃ ... qₙ. This includes the empty set. Note that there is no choice for the prime factors of rᵢ and rⱼ. Also, each prime factor in the prime factorization of rᵢrⱼ must be either in p₁p₂p₃ ... pₘ or in q₁q₂q₃ ... qₙ, but not both. Since there are k-2 such prime factors, there are \(2^(k-2)\) ways to choose the prime factors of rᵢrⱼ. (There is only one way if k = 2.)Finally, there are m + 1 ways to choose the exponents of rᵢ, since it must divide a.

There are n + 1 ways to choose the exponents of rⱼ, since it must divide b. There is only one way to choose the exponent of each of the other primes. This gives us a total of (m+1)(n+1) \(2^(k-2)\)· \(2^(k-2)\) ordered triples (a, b, c) such that [a, b] = c. The answer is therefore: (m+1)(n+1)2ᵏ-1 if k > 1. (If k = 1, then r is prime and c = r, so there are no restrictions on a and b.)

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Using proportions, the costs are given as follows:

a) Total Labor Cost: $2,505,600.

b) Total Product Cost: $1,950,000.

c) Total Cost = $4,455,600.

d) Cost per home = $461.77.

e) Cost per week = $99,013.33.

A proportion is a fraction of a total amount, and the measures are related using a rule of three. Due to this, relations between variables, either direct(when both increase or both decrease) or inverse proportional(when one increases and the other decreases, or vice versa), can be built to find the desired measures in the problem, or equations to find these measures.

For item a, the labor cost is found using the earnings of the workers, as follows:

45 weeks x 5 days x 8 hours x 58 per hour x (12 + 8 + 4 workers)

Hence:

Total Labor Cost = 45 x 5 x 8 x 58 x 24 = $2,505,600.

For item b, the product cost is the cost of the boards, hence:

(175 + 150 + 50 boards) x 5,200

Total Product Cost = 375 x 5,200 = $1,950,000.

For item c, the total cost is the sum of the product cost and the labor cost, hence:

Total Cost = 2,505,600 + 1,950,000 = $4,455,600.

For item d, the cost per home is found dividing the total cost by the 9,649 homes, hence:

Cost per home = 4455600/9649 = $461.77.

For item e, the cost per week is found dividing the total cost by the 45 weeks, hence:

Cost per week = 4455600/45 = $99,013.33.

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Personal car would be the best option to manage both tasks efficiently. While the train is a reliable mode of transportation, it may have fixed schedules that might not align perfectly with Patrick's needs.

On the other hand, using his personal car provides more flexibility and allows him to tailor the departure time according to his daughter's coding class schedule.

If Patrick decides to take the train, he would need to check the train schedule to see if there are convenient departure and arrival times for both the soccer game and the coding class. This option would require planning and coordination to ensure he arrives at the game on time and can pick up his daughter afterward.

Using his personal car gives Patrick the freedom to leave at a time that accommodates both the soccer game and the coding class. He can drop off his daughter at her coding class, attend the soccer game, and then pick her up afterward without being restricted by train schedules.

Considering the circumstances, Patrick might find it more convenient to use his personal car to manage both tasks effectively and ensure he can attend his son's soccer game while also ensuring his daughter arrives at her coding class on time.

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Answer:

x = 80

Step-by-step explanation:

x + .1x = 88

or

x(1+.1) = 88

which ever makes more sense to you

1.1 x = 88

x = 88/1.1

x = 80

Answer:C

Step-by-step explanation: 100 x 23/100=23

The proven that [δ(G)*n]/2 <= m <= [Δ(G)*n]/2. Given:  Let G be an undirected graph with n vertices.

Let Δ(G) be the maximum degree of any vertex in G, δ(G) be the minimum degree of any vertex in G, and m be the number of edges in G.Prove: [δ(G)*n]/2 <= m <= [Δ(G)*n]/2 Proving : [δ(G)*n]/2 <= m.

We know that, The sum of degrees of vertices of an undirected graph is equal to twice the number of edges.

Σ deg(V) = 2m Also, δ(G) <= deg(V) <= Δ(G)So, δ(G) <= Σ deg(V) /n <= Δ(G)δ(G)*n <= Σ deg(V) <= Δ(G)*n2m = Σ deg(V)≥ δ(G)*n and m ≥ [δ(G)*n]/2 Proving : m <= [Δ(G)*n]/2Also,δ(G) <= deg(V) <= Δ(G)δ(G)*n <= Σ deg(V) <= Δ(G)*n2m = Σ deg(V)So, m <= Σ deg(V) /2 <= Δ(G)*n /2and m <= [Δ(G)*n]/2

Therefore, we have proven that [δ(G)*n]/2 <= m <= [Δ(G)*n]/2.

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To find the roots of a third-degree polynomial, also known as a cubic polynomial, we can use a method called factoring or apply the cubic formula.

The first step is to check if there are any common factors that can be factored out. Next, we can use the rational root theorem to determine potential rational roots. By applying synthetic division or long division, we can divide the polynomial by the potential roots to see if they are indeed roots.

If a rational root is found, we can then use synthetic division to factor out the corresponding quadratic equation. Finally, we can solve the quadratic equation using methods like factoring, completing the square, or using the quadratic formula to find the remaining roots.

It's important to note that not all cubic polynomials can be easily factored or solved algebraically. In such cases, numerical methods or approximation techniques may be used to find the roots.

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To find the roots of a third degree polynomial, you can follow these steps: 1) Check for rational roots using the Rational Root Theorem. 2) Use synthetic division to divide the polynomial by a linear factor. 3) Factor the resulting quadratic equation. 4) Solve for the roots by setting each factor equal to zero.

To find the roots of a third degree polynomial, we can follow these steps:

Remember, the Fundamental Theorem of Algebra states that every polynomial equation of degree n has exactly n complex roots, counting multiplicities.

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Weighted Mean = (90 × 25% + 79 × 20% + 82 × 10% + 70 × 15% + 85 × 15%) / (25% + 20% + 10% + 15% + 15%)

Unweighted Mean = (90 + 79 + 82 + 70 + 85) / 5

One student, Kyle, hasn't taken his final exam yet, but his marks in the first five categories are 90, 79, 82, 70, and 85. This problem requires determining the weighted mean for Kyle before the final exam and comparing it to the unweighted mean.

To calculate the weighted mean for Kyle before the final exam, we need to multiply each category's mark by its corresponding weight, sum them up, and divide by the total weight. For Kyle, the weighted mean would be calculated as follows:

Weighted Mean = (Knowledge × 25% + Application × 20% + Thinking × 10% + Culminating Project × 15% + Final Exam × 15% + Communication × 15%) / (Total Weight)

However, since Kyle hasn't written his final exam yet, we can exclude the Final Exam mark and its weight from the calculation. The weighted mean would then be:

Weighted Mean = (90 × 25% + 79 × 20% + 82 × 10% + 70 × 15% + 85 × 15%) / (25% + 20% + 10% + 15% + 15%)

To find the difference between the weighted mean and the unweighted mean, we need to calculate the unweighted mean by simply taking the average of the marks in the first five categories:

Unweighted Mean = (90 + 79 + 82 + 70 + 85) / 5

By comparing the weighted mean and the unweighted mean, we can evaluate how much the inclusion of weights for different categories affects Kyle's overall mark.

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Answer:

A) \(6x^8y^5\)

B) \(4x^5z^8\)

C) \(12a^9b^7\)

D) \(6t^3s^9\)

Step-by-step explanation:

A) \(3x^2y^4\) ×\(2x^6y\)

Multiply the numbers

\(6x^2y^4x^6y\)

Combine exponents

\(6x^8y^4y\)

Combine exponents

\(6x^8y^5\)

===================================================================

B) \(xz^3x 4x^4z^5\)

Step-by-step explanation:

\(xz^3\) ×\(4x^4z^5\)

\(4x^1^+^4z^3^+^5\)

\(4x^5z^8\)

===================================================================

C) \(4a^3b^2 X 3a^6b^5\)

Multiply the numbers

\(12a^3b^2a^6b^5\)

Combine exponents

\(12a^9b^2b^5\)

Combine exponents

\(12a^9b^7\)

===================================================================

D)

To multiply powers of the same base, add their exponents. Add 5 and 4 to get 9.

\(6s^9tt^2\)

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

\(6s^9t^3\)

Answer

260 kilograms

Step-by-step explanation:

the correct answer is 260 kg

Answer: 12.6 kg

Step-by-step explanation: add the amounts of food for her farm, and just search for how many kg are in 1,260 grams

In  the inequality of x >= 13, meets the goal of selling at least 13 cases of candy on the second week.

In mathematics, an inequality is a statement that compares two values and shows that one value is greater than, less than, or not equal to the other value. Inequalities are represented using mathematical symbols such as ">," "<," "≥," and "≤."

For example, the statement "x > 3" is an inequality that says that the value of x is greater than 3. The symbol ">" is read as "is greater than." Similarly, the statement "x < -2" means that x is less than -2. The symbol "<" is read as "is less than."

In some cases, an inequality can involve multiple terms, such as "2x + 3 > 7." In this example, the inequality says that the value of 2x + 3 is greater than 7.

Inequalities can also be used to describe relationships between sets of numbers, such as "x belongs to the set of real numbers greater than or equal to 2." This means that x is a real number that is greater than or equal to 2.

To solve the graph for the inequality that can be used to find the possible amount of candies sold on the second week, you can start by representing the total number of candies sold as an equation. Let x be the number of cases of candy sold on the second week. Then the total number of candies sold on both weeks is given by:

7 cases * 40 candies/case + x cases * 40 candies/case = 280 + 40x candies

To meet the goal of selling at least 13 cases during the second week, you can write an inequality as follows:

x >= 13

This inequality represents all possible values of x that satisfy the condition of selling at least 13 cases of candy on the second week.

To graph this inequality on a coordinate plane, you can use the x-intercept and y-intercept method. The x-intercept of the graph is the point where the line crosses the x-axis, which occurs when the y-value is 0. The y-intercept of the graph is the point where the line crosses the y-axis, which occurs when the x-value is 0.

Since x >= 13, the x-intercept is (13,0), and the y-intercept is (0,520) (since 280 + 40x = 520 when x = 0). You can now plot these two points and draw a line through them to get the graph of the inequality x >= 13. The region above the line represents all possible values of x that satisfy the inequality and meet the goal of selling at least 13 cases of candy on the second week.

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Answer:5

Step-by-step explanation:

The area of the rectangle is increasing at a rate of \(16 cm^2\) per minute when the length is 8.94 cm and increasing at 2 cm per minute.

Let's use the formula for the area of a rectangle, which is A = l*w, where A is the area, l is the length and w is the width.

We are given that one side of the rectangle (width) is 8 cm, and we want to find the rate of change of the area when the other side (length) is 12 cm and increasing at 2 cm per minute.

We can start by finding the length (l) of the rectangle using the Pythagorean theorem, which states that for a right triangle, the square of the length of the hypotenuse (c) is equal to the sum of the squares of the other two sides (a and b). In our case, one of the sides (b) is the width (8 cm), and the other side (a) is the length we want to find. The hypotenuse (c) is the other side of the rectangle (12 cm), so we have:

\(c^2 = a^2 + b^2\\12^2 = a^2 + 8^2\\144 = a^2 + 64\\a^2 = 80\\a = \sqrt{80} = 8.94 cm\)

Now we can use the formula for the area of a rectangle to find the area (A) of the rectangle when the length is 8.94 cm:

A = l × w

A = 8.94 cm × 8 cm

A ≈ 71.52\(cm^2\)

To find the rate of change of the area (dA/dt) when the length is increasing at 2 cm per minute, we can use the product rule of differentiation:

dA/dt = d/dt(l × w)

dA/dt = w × (dl/dt) + l × (dw/dt)

We know that w is constant at 8 cm, so dw/dt = 0. We also know that dl/dt = 2 cm/min, since the length is increasing at 2 cm per minute. So we have:

dA/dt = w × (dl/dt) + l × (dw/dt)

dA/dt = 8 cm × (2 cm/min) + 8.94 cm × 0

dA/dt = 16 \(cm^2\)/min

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The probability that X is between 57 and 105 is 0.8914.

Given:

* X is normally distributed with mean=90 and standard deviation=12

* P(57 < X < 105)

Solution:

* Convert the given values to z-scores:

   * z = (X - μ) / σ

   * z = (57 - 90) / 12 = -2.50

   * z = (105 - 90) / 12 = 1.25

* Use the z-table to find the probability:

   * P(Z < -2.50) = 0.0062

   * P(Z < 1.25) = 0.8944

* Add the probabilities to find the total probability:

   * P(57 < X < 105) = 0.0062 + 0.8944 = 0.8914

Therefore, the probability that X is between 57 and 105 is 0.8914.

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Answer:

x = -12/5

Step-by-step explanation:

hope this helps you

Answer:

Integers: -25, 3, -5, 0

Not Integers: 1/2, 8.1

Step-by-step explanation:

Integers are whole numbers that appear on a number line. Any number that is not a whole number is not an integer.

Answer:

Integers::3,-5,0    

Not Integers:-2.5,1/2,8.1

Step-by-step explanation:

Integers are WHOLE numbers both + and - they cannot be fractions or decimals

Answer:

In linear algebra, an augmented matrix is a matrix obtained by appending the columns of two given matrices, usually for the purpose of performing the same elementary row operations on each of the given matrices.

Step-by-step explanation:

What is the question for this so I can answer it?

Answer:

The answer for the first part is 2/5

Step-by-step explanation:

Because if you add the total number of bags it equals 10 and then you do 2/10 because if 2 bags of funyuns and another 2/10 because its 2 bags of hot cheetos so 2/10+2/10=4/10 and simplified it is 2/5

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Answer:

x(2,3), y(2,5) z(7,3)

Step-by-step explanation: