Catherine walks her dog 3/4 mile everyday how far does she walk each week

Love them or hate them, they do be spitting straight facts

The planet Mercury is larger than the Moon, because the radius of Mercury is greater than the Moon.

Here,

Given that;

The planet Mercury has a diameter of 4,878 km.

And, The Moon has a radius of 1736.5 km.

We have to find the larger planet.

What is the relation between radius and diameter?

The radius is half or the diameter of the object.

Now,

The planet Mercury has a diameter of 4,878 km.

Hence, The radius of Mercury = 4,878/2 = 2,439 km

And,  The radius of the Moon = 1736.5 km

Clearly, 2,439 > 1,736.5

So, The planet Mercury is larger than the Moon, because the radius of Mercury is greater than the Moon.

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you have n/3 = -12

to solve for n, multiply both sides by 3

n = -12 * 3 = -36

answer is A. -36

To find the time it takes for the smaller pump to empty the tank alone, we need to first determine the rate of the smaller pump and solve an equation involving the rates of both pumps.

Let's denote the rate at which the smaller pump can empty the tank as x gallons per hour. Since the larger pump can empty the tank in 4 hours less than the smaller pump, its rate would be x+4 gallons per hour.  

Working together, the total rate at which they can empty the tank is the sum of their individual rates:

x + (x+4) = 1/3 gallons per hour.

Solving this equation will give us the rate of the smaller pump (x), and we can then calculate how long it would take the smaller pump to empty the tank alone.

Final answer:

To find the time it takes for the smaller pump to empty the tank alone, we calculate the combined rate of both pumps and set up an equation based on their separate rates. After solving, we find that it takes the smaller pump 6 hours to empty the tank by itself.

Explanation:

To solve this problem, we can set up a system of equations based on the rates at which the pumps work. Let's denote the time it takes the smaller pump to empty the tank alone as x hours. According to the problem, the larger pump can do this in x - 4 hours. When working together, they can empty the tank in 3 hours.

If the smaller pump's rate is 1/x tanks per hour and the larger pump's rate is 1/(x-4) tanks per hour, the combined rate when they work together is the sum of their individual rates, which is 1/3 tanks per hour.

We can set up the equation as follows:
1/x + 1/(x-4) = 1/3
To solve for x, find a common denominator and combine terms:
(x-4 + x)/(x(x-4)) = 1/3

Multiply all terms by the common denominator x(x-4) to clear the fractions:
2x - 4 = x(x-4)/3

Multiply both sides by 3 to get rid of the fraction:
6x - 12 = x2 - 4x

Rearrange the equation to set it to zero and solve for x:
x² - 10x + 12 = 0

By factoring, we find that x equals 6 or 2. However, since x - 4 must be greater than zero (a pump can't empty a tank in negative time), we discard the solution x = 2 and keep x = 6 hours as the correct answer.

Therefore, the smaller pump will take 6 hours to empty the tank alone.

A normal distribution with a mean of 0 and a standard deviation of 1 called The Standard normal distribution or z - distribution.

A normal distribution or Gaussian distribution is a type of continuous probability distribution for a real-valued random variable.

Given that a normal distribution with a mean of 0 and a standard deviation of 1

In the given condition, the normal distribution is called Standard normal distribution or z - distribution or Gaussian distribution.

Hence, the normal distribution with a mean of 0 and a standard deviation of 1 called Standard normal distribution or z - distribution or Gaussian distribution.

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Final answer:

To solve the equation 3/4x + 5/6 = 5x - 125/3 for x, you can start by getting rid of the fractions. Multiply every term in the equation by the least common multiple (LCM) of the denominators, which in this case is 12. Combine like terms and solve for x.

Explanation:

To solve the equation 3/4x + 5/6 = 5x - 125/3 for x, you can start by getting rid of the fractions. To do this, multiply every term in the equation by the least common multiple (LCM) of the denominators, which in this case is 12. This gives you:

9x + 10 = 60x - 500

Next, combine like terms by subtracting 9x from both sides of the equation:

10 = 51x - 500

Then, add 500 to both sides of the equation:

510 = 51x

Finally, divide both sides of the equation by 51 to solve for x:

x = 10

it would be 50 and 37

Explains how many times per second the magnitude of the voltage equals 220 V in a European outlet at 50 Hz frequency.

The frequency at which the magnitude of the voltage is equal to 220 V is:

The given function is f(x)=2x^3+2x^2-x. In this case, the highest degree among the terms is 3. According to the Fundamental Theorem of Algebra, the number of zeros or roots of the equation is equal to the highest degree among the terms. Hence for every value of y, there are 3 equivalent x's. The answer thus is B. It is a many-to-one function.

Since the company’s profits drop by $1,500 per month for every increase of $450 per year in the worker’s average pay, the slope of the graph of the equation is -40.

In Mathematics and Geometry, the slope of any straight line can be determined by using the following mathematical equation;

Slope (m) = (Change in y-axis, Δy)/(Change in x-axis, Δx)

Slope (m) = rise/run

[tex]Slope(m)=\frac{y_2-y_1}{x_2-x_1}[/tex]

Based on the information provided, we can logically deduce that the rise and run are as follows;

Rise = 12 × -1500 = 18,000 units (its negative because the profit drops)

Run = 450 units.

By substituting the given data into the formula for the slope of a line, we have the following:

Slope (m) = rise/run

Slope (m) = -18000/450

Slope (m) = -40.

Answer:  Joseph's residual is 1.86 (option d).

Step-by-step explanation:

Hi, to answer this question we have to replace the variable "x" for Joseph’s age (23.5 months) in the equation given:

y = 63.80 + 0.61x.

y = 63.80 + 0.61 (23.5)

y = 63.80 + 14.335 = 78.135

Now, we subtract the amount obtained to Joseph's height ( 80 centimeters).

Mathematically speaking:

80 - 78.135 = 1.865  

So, Joseph’s residual is 1.86 (option d).

Actually in this case we will make use of the z test statistic. In which we calculate for the z score and then use the standard normal distribution tables to find for the P value at the calculate z score.

z = (x – u) / s

z = (97 – 98) / 2

z = -0.5

The z value should be greater than 1.65 to for the null hypothesis to be accepted. Hence rejected.

Answer:

Net income of Kaycie is $1,195

Step-by-step explanation:

We are given a monthly Income statement of September 2013.

First we count total income of Kaycie's

Income = Wages

Total Income = $2,700

Now we calculate total expenses

Kaycie's expenses are on food, rent clothing, transportation, medical and Insurance.

Expenses = Food+Rent+Clothing+Transportation+Medical+Insurance

Expenses=210+975+90+120+60+50

Total Expenses = $1,505

Net Income = Total Income - Total Expenses

Net Income = 2700 - 1505

Net Income = $1,195

Thus, Net income of Kaycie is $1,195.

Answer:

In so many words its "1,195"

Step-by-step explanation:

Final answer:

The correct O scale model train track spacing should be 1.17 inches. However, the actual O gauge spacing is 1.25 inches. The relative difference in the rail spacing is approximately 6.8%.

Explanation:

To solve for the correct O scale model train track spacing:

Convert the real train track gauge to inches: 4 feet x 12 inches/foot + 8 inches = 56 inches.

Divide the real gauge by the scale factor of 48 to find the model gauge: 56 inches / 48 = 1.1667 inches.

Round the result to the nearest hundredth of an inch: 1.17 inches.

Now to calculate the relative difference between O scale and the actual O gauge:

Determine the actual O gauge spacing: 1.25 inches.

Calculate the difference in spacing: 1.25 inches - 1.17 inches = 0.08 inches.

Find the relative difference by dividing the spacing difference by the O scale spacing and then multiply by 100 to get the percentage: (0.08 inches / 1.17 inches) x 100 = 6.8%.

Round the result to the nearest tenth of a percent: 6.8%.

The Laplace transform of  f(t) is [tex]\( \frac{4}{s}e^{-4s} \)[/tex].

Sure, let's find the Laplace transform of the piecewise function [tex]\( f(t) = \begin{cases} 0, & 0 \leq t < 4 \\ t - 4, & t \geq 4 \end{cases} \).[/tex]

Define the function for all ( t ):

[tex]\[ f(t) = \begin{cases} 0, & 0 \leq t < 4 \\ t - 4, & t \geq 4 \end{cases} = (t - 4)u(t-4) \][/tex]

where  u(t) is the unit step function.

Use the definition of the Laplace transform:

[tex]\[ \mathcal{L}\{f(t)\} = \int_0^\infty f(t)e^{-st} dt \][/tex]

Break the integral into two parts based on the piecewise definition of  f(t) :

[tex]\[ \mathcal{L}\{f(t)\} = \int_0^4 0 \cdot e^{-st} dt + \int_4^\infty (t - 4)e^{-st} dt \][/tex]

Solve each integral separately:

[tex]\[ \int_0^4 0 \cdot e^{-st} dt = 0 \][/tex]

[tex]\[ \int_4^\infty (t - 4)e^{-st} dt = \int_4^\infty te^{-st} dt - 4\int_4^\infty e^{-st} dt \][/tex]

Integrate:

[tex]\[ = \left[ -\frac{1}{s}te^{-st} \right]_4^\infty - \left[ -\frac{4}{s}e^{-st} \right]_4^\infty \][/tex]

[tex]\[ = \left(0 - \left(-\frac{4}{s}e^{-4s}\right)\right) - \left(0 - \left(-\frac{4}{s}e^{-4s}\right)\right) \][/tex]

[tex]\[ = \frac{4}{s}e^{-4s} \][/tex]

Combine the results:

[tex]\[ \mathcal{L}\{f(t)\} = \frac{4}{s}e^{-4s} \][/tex]

So, the Laplace transform of  f(t) is [tex]\( \frac{4}{s}e^{-4s} \)[/tex].

The Laplace transform of the given piecewise function f(t) is found using the Heaviside step function and shifting theorem, yielding[tex]\[ \mathcal{L}\{f(t)\} = \frac{6}{s^4} \][/tex]

To find the Laplace transform of the function f(t), we'll break it down into two parts: one for  t < 4  and another for [tex]\( t \geq 4 \).[/tex]

For  t < 4 ,  f(t) = 0 , so its Laplace transform is simply 0.

For [tex]\( t \geq 4 \), \( f(t) = (t - 4)^3 \)[/tex]. We'll use the property of the Laplace transform that states [tex]\( \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} \). So, for \( f(t) = (t - 4)^3 \)[/tex], we get:

[tex]\[ \mathcal{L}\{(t - 4)^3\} = \frac{3!}{s^{3+1}} \\\[ = \frac{6}{s^4} \][/tex]

Now, we can combine both parts:

[tex]\[ \mathcal{L}\{f(t)\} = 0 \quad \text{for } t < 4 \\\[ \mathcal{L}\{f(t)\} = \frac{6}{s^4} \quad \text{for } t \geq 4 \\\text{Therefore, the Laplace transform of}\; \( f(t) \) is:\[ \mathcal{L}\{f(t)\} = \frac{6}{s^4} \][/tex]