the escape speed from the moon is much smaller than from earth, around 2.38 km/s.

Answer: 5

Step-by-step explanation: 21 + 4 = 25

square root of 25  = 5

Answer and Step-by-step explanation:

T = 21

U = 4

[tex]\sqrt{ 21 + 4} \\\\\\\sqrt{25} \\\\\\5[/tex]

5 is the answer to the expression.

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Answer:

5670 yards

Step-by-step explanation:

Length of first race = 2 miles

Since, 1 mile = 1760 yards

Therefore, 2 miles = 1760 × 2

                               = 3520 yards

Length of second race = 1300 yards

Length of third race = 850 yards

Total distance to be run by the horse = 3520 + 1300 + 850

                                                              = 5670 yards

Answer:

900 cm = 9 m

Step-by-step explanation:

9 m = 900 cm

Therefore, 9 m equals 900 cm.

Answer:

. the area decreased by 7 is 14

Answer:

I don't see any equations.

Answer:

circle: ( 1, 0.5 )

square: ( 4.5, 3.5 )

triangle: ( 0, 2.5 )

Step-by-step explanation:

Answer:

option B should be correct

The best estimate for the solution to the system of linear equations among the given ordered pairs is (-4, 2 1/2).

In the context of a system of linear equations, the solution represents the values of the variables that satisfy all the equations simultaneously. To determine the best estimate for the solution, we need to evaluate each ordered pair and see which one satisfies the given system.

By substituting the values of the ordered pairs into the equations of the system, we can determine if they satisfy the equations or not. Among the given options, when substituting (-4, 2 1/2) into the system of linear equations, it is likely to result in a solution that satisfies all the equations.  Therefore, it is important to consider the specific equations and the context of the problem to determine the best estimate for the solution.

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Answer:

what should i answer?

Step-by-step explanation:

The condition imposed on the constants k₁ and k₂ for k₁Θ⁻₁ + k₂Θ⁻₂ to be an unbiased estimator of θ is that their sum must equal 1.

For k₁Θ⁻₁ + k₂Θ⁻₂ to be an unbiased estimator of θ, its expected value should be equal to θ. In other words, we want to find the conditions on k₁ and k₂ such that E(k₁Θ⁻₁ + k₂Θ⁻₂) = θ.

Given that Θ⁻₁ and Θ⁻₂ are unbiased estimators of θ, we have:

E(Θ⁻₁) = θ

E(Θ⁻₂) = θ

Now, let's calculate the expected value of k₁Θ⁻₁ + k₂Θ⁻₂:

E(k₁Θ⁻₁ + k₂Θ⁻₂) = k₁E(Θ⁻₁) + k₂E(Θ⁻₂)

Since E(Θ⁻₁) = θ and E(Θ⁻₂) = θ, we can substitute these values into the equation:

E(k₁Θ⁻₁ + k₂Θ⁻₂) = k₁θ + k₂θ

To make sure this expression is equal to θ, we need:

k₁θ + k₂θ = θ

This implies that k₁ + k₂ = 1. Therefore, the condition imposed on the constants k₁ and k₂ for k₁Θ⁻₁+ k₂Θ⁻₂ to be an unbiased estimator of θ is that their sum must equal 1.

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Multiply the discount percent to the marked price and subtract from the marked price.

As we know that the discount is always given on the marked price of the product.

We will take one example to understand it supposes the marked price of any object is MP= 100 and the discount =15% so to find the SP

First, find the discount amount

[tex]MP\times Discout \ percent =100\times \dfrac{15}{100} =15[/tex]

Now to find the SP subtract the discount from the MP

[tex]SP=MP-Discount[/tex]

[tex]SP=100-15=75[/tex]

So for finding SP multiply the discount percent to the marked price and subtract from the marked price.

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Answer:

92/37

Step-by-step explanation:

Answer:

B and D.

Step-by-step explanation:

A p e x

The following are among the five basic postulates of Euclidean geometry

The five basic postulates of Euclidean geometry

The five (5) basic postulates are:

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a. The total amount owed for the $525 loan at 9.9% interest for 6 months is approximately $556.19 when rounded to the nearest cent.

b. The total amount owed for the $12,460 loan at 5.6% interest for 30 months is approximately $33,504.80 when rounded to the nearest cent.

a. To calculate the total amount owed for the $525 loan at 9.9% interest for 6 months, we can use the formula for simple interest:

Total amount owed = Principal + (Principal × Interest Rate × Time)

Given:

Principal (P) = $525

Interest Rate (R) = 9.9% = 0.099 (converted to decimal)

Time (T) = 6 months

Plugging these values into the formula, we get:

Total amount owed = $525 + ($525 × 0.099 × 6)

Simplifying the equation:

Total amount owed = $525 + ($31.185)

Total amount owed = $556.185

Therefore, the total amount owed for the $525 loan at 9.9% interest for 6 months is approximately $556.19 when rounded to the nearest cent.

b. To calculate the total amount owed for the $12,460 loan at 5.6% interest for 30 months, we'll follow the same formula for simple interest:

Total amount owed = Principal + (Principal * Interest Rate * Time)

Given:

Principal (P) = $12,460

Interest Rate (R) = 5.6% = 0.056 (converted to decimal)

Time (T) = 30 months

Plugging these values into the formula, we get:

Total amount owed = $12,460 + ($12,460 * 0.056 * 30)

Simplifying the equation:

Total amount owed = $12,460 + ($21,044.8)

Total amount owed = $33,504.8

Therefore, the total amount owed for the $12,460 loan at 5.6% interest for 30 months is approximately $33,504.80 when rounded to the nearest cent.

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Using Simpson's [tex]\frac{1}{3}[/tex] rule with 6 strips, the approximate value of the integral ∫[2.0, 2.8] f(x) dx is -3.8492.

To integrate the function [tex]\begin{equation}y = f(x) = \frac{ax^2}{b + x^2}[/tex] using Simpson's 1/3 rule, we need to divide the interval [2.0, 2.8] into an even number of strips (in this case, 6 strips). The formula for approximating the integral using Simpson's 1/3 rule is as follows:

[tex]\begin{equation}\int_a^b f(x) dx \approx \frac{h}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + ... + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n) \right][/tex]

Where:

h is the width of each strip ([tex]\begin{equation}h = \frac{b - a}{n}[/tex], where n is the number of strips)

[tex]x_0[/tex] is the lower limit (2.0)

[tex]x_n[/tex] is the upper limit (2.8)

f(xi) represents the function evaluated at each strip's midpoint

Given the values of a = 1.2 and b = -0.587, we can proceed with the calculations.

Step 1: Calculate the width of each strip (h):

[tex]\begin{equation}h = \frac{b - a}{n} = \frac{-0.587 - 1.2}{6} = \frac{-1.787}{6} \approx -0.2978[/tex]

Step 2: Calculate the function values at each strip's midpoint:

x₀ = 2.0

x₁ = x₀ + h = 2.0 + (-0.2978) = 1.7022

x₂ = x₁ + h = 1.7022 + (-0.2978) = 1.4044

x₃ = x₂ + h = 1.4044 + (-0.2978) = 1.1066

x₄ = x₃ + h = 1.1066 + (-0.2978) = 0.8088

x₅ = x₄ + h = 0.8088 + (-0.2978) = 0.511

x₆ = x₅ + h = 0.511 + (-0.2978) = 0.2132

xₙ = 2.8

Step 3: Evaluate the function at each midpoint:

[tex]f(x_0) = \frac{1.2 \times 2^2}{-0.587 + 2^2} = \frac{4.8}{3.413} \approx 1.406 \\\\f(x_1) = \frac{1.2 \times 1.7022^2}{-0.587 + 1.7022^2} \approx 2.445 \\\\f(x_2) = \frac{1.2 \times 1.4044^2}{-0.587 + 1.4044^2} \approx 2.784 \\\\f(x_3) = \frac{1.2 \times 1.1066^2}{-0.587 + 1.1066^2} \approx 2.853 \\\\[/tex]

[tex]f(x_4) = \frac{1.2 \times 0.8088^2}{-0.587 + 0.8088^2} \approx 2.455 \\f(x_5) = \frac{1.2 \times 0.511^2}{-0.587 + 0.511^2} \approx 1.316 \\f(x_6) = \frac{1.2 \times 0.2132^2}{-0.587 + 0.2132^2} \approx 0.29[/tex]

Step 4: Apply Simpson's 1/3 rule formula:

[tex]\begin{equation}\int_{2.0}^{2.8} f(x) dx \approx \frac{h}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + 4f(x_5) + f(x_6) \right][/tex]

[tex]\begin{equation}\approx \frac{-0.2978}{3} \left[ 1.406 + 4(2.445) + 2(2.784) + 4(2.853) + 2(2.455) + 4(1.316) + 0.29 \right][/tex]

[tex]\begin{equation}= \frac{-0.2978}{3} \left[ 1.406 + 9.78 + 5.568 + 11.412 + 4.91 + 5.264 + 0.29 \right][/tex]

≈ (-0.09926) * 38.63

≈ -3.8492

Therefore, the approximate value of the integral ∫[2.0, 2.8] f(x) dx using Simpson's 1/3 rule with 6 strips is approximately -3.8492.

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The given initial value problem using the method of undetermined coefficients. The final solution to the initial value problem is y = cos(2x) + x - (1/9)*sin(3x).

To solve the given initial value problem, we begin by finding the general solution to the associated homogeneous equation. The homogeneous equation is given by y'' + 4y = 0. The characteristic equation is obtained by substituting y = e^(mx) into the equation, resulting in the quadratic equation m^2 + 4 = 0. Solving this equation yields two distinct roots: m_1 = 2i and m_2 = -2i. Thus, the general solution to the homogeneous equation is y_h = c_1cos(2x) + c_2sin(2x), where c_1 and c_2 are arbitrary constants.

Next, we assume a particular solution in the form of y_p = Ax + B + Csin(3x) + Dcos(3x), where A, B, C, and D are undetermined coefficients. We substitute this particular solution into the given differential equation and solve for the coefficients. Comparing the coefficients of like terms, we find A = 0, B = 1, C = -1/9, and D = 0.

The particular solution is y_p = x - (1/9)*sin(3x), and the complete solution is obtained by adding the particular solution to the homogeneous solution: y = y_h + y_p. Applying the initial condition y(0) = 1, we find that c_1 = 1 and c_2 = 0. Therefore, the final solution to the initial value problem is y = cos(2x) + x - (1/9)*sin(3x).

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Answer:

The answer is 3/4

Step-by-step explanation:

Answer:

There is no fixed number of trials, so X is not a binomial variable

Step-by-step explanation:

mama

There is no fixed number of trials, so X is not a binomial variable.

This is a specific kind of discrete random variable. A binomial random variable counts how regularly a specific event occurs in a fixed variety of attempts or trials.

The binomial is a form of distribution that has possible effects (the prefix “bi” method two, or twice). as an example, a coin toss has only viable effects: heads or tails, and taking a check may want to have viable outcomes: pass or fail. A Binomial Distribution indicates both success and failure.

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Tossing of a fair coin infinitely many times. The process X(t) can be defined as follows:

- X(t) = 1 if a head shows up at time t, where t = nT for some positive integer n.

- X(t) = 0 if a tail shows up at time t.

In the given scenario, we are considering the tossing of a fair coin infinitely many times. We want to define a process X(t) that represents the outcome of each toss at different time points.

The process X(t) is defined as 1 when a head shows up at time t, where t is a multiple of T (the fixed time interval between tosses). In other words, X(t) takes the value 1 when t is of the form nT, where n is a positive integer.

Conversely, X(t) is defined as 0 when a tail shows up at time t. This includes all time points that are not of the form nT.

The process X(t) is a representation of the outcome of the coin tosses over time. It takes the value 1 when a head shows up at time t = nT for some positive integer n, and 0 when a tail shows up. This process allows us to track the occurrences of heads at specific time intervals in the infinite sequence of coin tosses.

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Answer:

The two triangles are related by angles, so the triangles are similar but not proven to be congruent.

Step-by-step explanation:

Because the triangles have the same angles, they are congruent. The definition of congruence is if you take a shape and scale it up or down (or keep it the same) therefore, they are congruent.

Hope this helped, have a nice day

EDIT: I screwed up, I thought it was supposed to be similar.  These triangles are SIMILAR not congruent. The actual answer is they are related by AAA similarity but they are similar, but they are not proven to be congruent. Hope this clears it up, and sorry.

~cloud

Each satisfies the wave equation.

We are given the functions as follows:

(a) (x+at)², (b) 2e-(x-at) ², 7 satisfy the wave equation J²u(x, t) Ət² = (c) 5 sin[3 (x - at)] + (x + at).

₂d²u(x, t) dx²

Let us prove that they satisfy the wave equation using the formula of the wave equation. Wave equation is given by;

J²u(x, t) Ət² = ₂d²u(x, t) dx²

Applying the partial derivative to

(a) with respect to time, t, we obtain:

2a(x+at)

The second partial derivative with respect to x is as follows:

2a

By substituting these results into the wave equation, we have:

J²u(x, t) Ət² = ₂d²u(x, t) dx²

(2a(x+at)) = 2aJ²u(x, t) Ət² = 2a

Ət² = 1/J².

Thus, (a) satisfies the wave equation.  

For part (b), let us begin by taking the partial derivative of the function with respect to time, t. This is given by:

-4a e^-(x-at) ²

By taking the second partial derivative with respect to x, we get:4a e^-(x-at) ²

Similar to above, we substitute these results into the wave equation as follows:

J²u(x, t) Ət² = ₂d²u(x, t) dx²

-4a e^-(x-at) ² = 4aJ²u(x, t) Ət² = -4a e^-(x-at) ²/J²

Ət² = -1/J²e^-(x-at) ².

Thus, (b) satisfies the wave equation.

For part (c), let us calculate the partial derivative with respect to t as follows:

5a cos[3(x-at)] + a

The second partial derivative with respect to x is given by:-

15a sin[3(x-at)]

By substituting these results into the wave equation, we have:

J²u(x, t) Ət² = ₂d²u(x, t) dx²

(5a cos[3(x-at)] + a) = -15a

sin[3(x-at)]J²u(x, t) Ət² = -15a

sin[3(x-at)]/(5a cos[3(x-at)] + a)

Ət² = -3 sin[3(x-at)]/(cos[3(x-at)] + 1/5).

Thus, (c) satisfies the wave equation.

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Answer:

18 units to the right.

Step-by-step explanation:

Hope this helps!

Approximately 0.11 percent of all samples have mean heights of at least 64.76 inches.

Given: 50 years prior, the mean level of ladies in a specific country in their 20s was 62.9 inches. With a standard deviation of 2.87 inches, women in their 20s today have heights that are roughly typical. We must determine the proportion of all 21 samples that have mean heights of at least 64.76 inches among today's women in their 20s. Bit by bit clarification:

Let μ be the mean level of ladies in this day and age and allow n to be the example size. We can find the likelihood of an example mean being more prominent than or equivalent to 64.76 inches utilizing the z-score recipe, as follows: z = (x - μ)/(σ/√n) = (64.76 - 62.9)/(2.87/√21) = 3.06The likelihood that an example of 21 ladies will have a mean level more noteworthy than or equivalent to 64.76 inches is equivalent to the likelihood that a typical irregular variable Z is more noteworthy than or equivalent to 3.06.

This probability is approximately 0.0011, or 0.11 percent, according to a typical normal table or calculator. This means that approximately 0.11 percent of all samples have mean heights of at least 64.76 inches.

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Answer:

8.8

Step-by-step explanation:

Answer:

Explanation:

320.29 × 7 = 2,242.03 m = 1.39313285413187 miles ≈ 1.393

Answer:

A, The probability that it will rain is 0.09 and the probability that the flight will be delayed is 0.18.

Step-by-step explanation:

Answer: 0.154

Step-by-step explanation:

Answer:

3/4 is 0.75 and 1 is 1

Step-by-step explanation:

:\

As per the given values, the after-tax salvage value is $435,200.

Amount required = $3,200,000

Time = 6 years

Calculating the accumulated depreciation -

Amount/ Number of years

=  $3,200,000 / 6

= $533,333.3.

Calculating the accumulated depreciation at project end -

= 6 x $533,333.33

= $3,200,000.

Calculating the book value of the fixed assets -

Book value = Cost of fixed assets - Accumulated depreciation

= $3,200,000 - $3,200,000

= $0

Calculating the taxable gain or loss on the sale of the fixed assets -

Taxable gain/loss = Selling price - Book value

= $640,000 - $0

= $640,000

Calculating the tax liability -

Tax liability = Tax rate x Taxable gain

= 0.32 x $640,000

= $204,800

Calculating the after-tax salvage value -

After-tax salvage value = Selling price - Tax liability

= $640,000 - $204,800

= $435,200

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The sum of the binomial coefficients nCk, where n ranges from 0 to 10 and k varies from 0 to 10, is equal to [tex]2^10[/tex], which is 1024.

The expression ∑10k=010Ck represents the sum of the binomial coefficients for all possible values of k from 0 to 10. The binomial coefficient nCk, also known as "n choose k," represents the number of ways we can choose k objects from a set of n objects.

In this case, we are summing up the binomial coefficients for n ranging from 0 to 10. For each value of n, we calculate the binomial coefficient nCk for k values ranging from 0 to 10. The formula to calculate the binomial coefficient is n! / (k!(n-k)!), where "!" denotes the factorial operation.

When we substitute the values into the formula, we find that all the binomial coefficients are 1, except when k equals 0 or n. In these cases, the binomial coefficient is equal to 1, as there is only one way to choose 0 objects or all n objects from a set.

Since there are 11 values of n (0 to 10), and for each n there is one non-zero binomial coefficient, the sum of all the binomial coefficients from 0 to 10 is equal to 11. Therefore, the expression simplifies to ∑10k=010Ck = 11.

So, the sum of all the binomial coefficients ∑10k=010Ck is equal to 11, which means we have 11 ways to choose k objects from a set of 10 objects. Another way to interpret this is that the sum of the binomial coefficients represents the number of subsets of a set with 10 elements, which is 11. However, if we are considering the case of choosing 0 to 10 objects, the sum becomes 2^10, which is equal to 1024.

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