A truck can be rented from Company A for ​$140 a day plus ​$0.80 per mile. Company B charges ​$80 a day plus ​$0.90 per mile to rent the same truck. Find the number of miles in a day at which the rental costs for Company A and Company B are the same.

The smallest number of eggs that could have been in the box is 1134

The problem is to find the smallest number of eggs that could have been in the box, given the remainder when taking them out by different numbers. Here are the moves toward tackling it:

Allow n to be the quantity of eggs in the container. Then we have the accompanying arrangement of congruences:

n ≡ 1 (mod 2)

n ≡ 1 (mod 3)

n ≡ 1 (mod 4)

n ≡ 1 (mod 5)

n ≡ 1 (mod 6)

n ≡ 0 (mod 7)

For this problem, we have k = 6 k = 6, a i = {1,1,1,1,1,0} a_i = {1,1,1,1,1,0}, M i = {1260,840,630,504,420,720} M_i = {1260,840,630,504,420,720}, and y i = {−1,−2,−3,-4,-5,-6} y_i = {-1,-2,-3,-4,-5,-6}.

Plugging these values into the formula and simplifying modulo 5040, we get:

n = (−1260 + −1680 + −1890 + −2016 + −2100 + 0) mod 5040

n = (−8946) mod 5040

n = (−3906) mod 5040

n = 1134 mod 5040

Therefore, the smallest number of eggs that could have been in the box is 1134

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Answer:It is irrational since it will never stop going on

Step-by-step explanation:

The linear model that represents the cost of any number of candy bars can be written as: C = $1.90B + $0.95

To write the linear model that models the cost of any number of candy bars, we need to find the equation of a line that best fits the given data points. We'll use the variables B for the number of candy bars purchased and C for the total cost of the candy bars.

Looking at the given data, we can see that there is a linear relationship between the number of candy bars and the total cost. As the number of candy bars increases, the total cost also increases.

To find the equation of the line, we need to determine the slope and the y-intercept. We can use the formula for the equation of a line: y = mx + b, where m is the slope and b is the y-intercept.

First, let's find the slope (m) using two points from the given data, for example, (3, $6.65) and (25, $48.45):

m = (C2 - C1) / (B2 - B1)

 = ($48.45 - $6.65) / (25 - 3)

 = $41.80 / 22

 ≈ $1.90

Now, let's find the y-intercept (b) using one of the data points, for example, (3, $6.65):

b = C - mB

 = $6.65 - ($1.90 * 3)

 = $6.65 - $5.70

 ≈ $0.95

Therefore, the linear model that represents the cost of any number of candy bars can be written as:

C = $1.90B + $0.95

This equation represents a linear relationship between the number of candy bars (B) and the total cost (C). For any given value of B, you can substitute it into the equation to find the corresponding estimated total cost of the candy bars.

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Answer:

y=8x+4

Step-by-step explanation:

I am assuming you are asking for an equation.

Total cost = $8*(number of people) + $4

The volume of a cylinder of radius r and height h is given by the equation presented as follows:

V = πr²h.

The volume of a cone is a third of the volume of a cylinder, hence:

V = πr²h/3.

For both cases, the height variable h has an exponent of one, hence:

If we had doubled the radius for example, the volume would be multiplied by 2² = 4, as the radius is squared.

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The manufacturer should set the price on the new blender at $400 for a maximum profit of $31,590.

To find the price that produces the maximum profit, we can use the given profit model and construct a one-way data table in a spreadsheet. In this case, the profit model is represented by the equation:

Total Profit \(= -17,490 + 2520P - 2P^2\)

We input the price values ranging from $200 to $700 in the data table and calculate the corresponding total profit for each price. By analyzing the data table, we can determine the price that yields the maximum profit.

In this scenario, the price that produces the maximum profit is $400, and the corresponding maximum profit is $31,590. Therefore, the manufacturer should set the price on the new blender at $400 to maximize their profit.

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Mass (g) = 1.75 mol/L x 5.25 L x 58.44 g/mol, Volume (L) = 75.00 g / (0.635 M x 58.33 g/mol)

Question 6: What mass of NaCl (in g) is necessary for 5.25 L of a 1.75 M solution?

To find the mass of NaCl needed for the solution, we need to use the formula:

Mass (g) = Concentration (M) x Volume (L) x Molar Mass (g/mol)

Given:
Concentration (M) = 1.75 M
Volume (L) = 5.25 L

First, let's convert the concentration from M to mol/L:
1 M = 1 mol/L

So, 1.75 M = 1.75 mol/L

Now, let's calculate the mass:
Mass (g) = 1.75 mol/L x 5.25 L x Molar Mass (g/mol)

Since we're dealing with NaCl (sodium chloride), the molar mass is 58.44 g/mol.

Mass (g) = 1.75 mol/L x 5.25 L x 58.44 g/mol

Calculating the above expression will give us the mass of NaCl in grams needed for the solution.

Question 7: You have measured out 75.00 g of Mg(OH)2 (formula weight: 58.33 g/mol) to make a solution. What must your final volume be (in L) if you want a solution made from this mass of Mg(OH)2 to have a concentration of 0.635 M?

To find the final volume of the solution, we need to rearrange the formula:

Volume (L) = Mass (g) / (Concentration (M) x Molar Mass (g/mol))

Given:
Mass (g) = 75.00 g
Concentration (M) = 0.635 M
Molar Mass (g/mol) = 58.33 g/mol

Plugging in the given values, we get:

Volume (L) = 75.00 g / (0.635 M x 58.33 g/mol)

Calculating the above expression will give us the final volume of the solution in liters.

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When Ken and Hamid race around a track, Ken completes a lap in 80 seconds while Hamid completes a lap in 60 seconds. ken will run 3 laps where as Hamid will run 4 laps.

Given that,

When Ken and Hamid race around a track, Ken completes a lap in 80 seconds while Hamid completes a lap in 60 seconds.

We have to find if they had faced off on the starting line, how many laps would they have won.

We have to find the LCM of the 80 and 60.

The prefix "LCM" stands for "Least Common Multiple." The smallest multiple that two or more numbers share is known as the least common multiple.

LCM of 80 and 60 is 240

240=80×3

ken will complete 3 laps.

240=60×4

Hamid will complete 4 laps.

Therefore, ken will complete 3 laps where as Hamid will complete 4 laps.

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Answer:

A. $3.96

Step-by-step explanation:

$2.20 × 1.8 pounds = $3.96

To determine the quadrant in which u/2 lies, we need to first find the quadrant where u lies. Given that tan(u) = -3/4 and 3/2 < u < 2, u is in the third quadrant. Now, to find the quadrant of u/2, we can divide the bounds by 2:

(3/2)/2 < u/2 < 2/2
3/4 < u/2 < 1

These conditions indicate that u/2 lies in the first quadrant.

Given conditions are :tan(u) = -3/4, 3/2 < u < 2We need to determine the quadrant in which u/2 lies.Tan is negative in 2nd and 4th quadrants. So, angle lies in either 2nd or 4th quadrants as 3/2 < u < 2 (i.e., second quadrant).But if angle lies in second quadrant then u/2 would lie in third quadrant, which is not possible. Therefore, angle lies in 4th quadrant. Hence, angle u = tan⁻¹(-3/4) ≈ -36.87°So, u/2 = -36.87/2 ≈ -18.43°The quadrant in which u/2 lies is the fourth quadrant.

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                            "Complete questions"

Use the given conditions. tan(u) = -3/4, 3π/2 < U <2n (a) Determine the quadrant in which u/2 lies. Quadrant I Quadrant II Quadrant III Quadrant IV Nice! (b) Find the exact values .

 A suggested variable control chart in the education field is a Student Attendance Rate Control Chart. This chart tracks the attendance metric and calculates it as the percentage of students present in a given time period. The information would be tracked regularly, and variations in attendance rates would be interpreted to identify trends and address potential issues.

The Student Attendance Rate Control Chart is a valuable tool in monitoring and managing attendance in educational institutions. The metric tracked on this chart is the attendance rate, which is calculated as the percentage of students present in a specific time period, such as daily, weekly, or monthly.
To track the attendance rate, the educational institution would collect attendance data for each class or session and calculate the percentage of students present. This data can be collected manually or through an automated attendance tracking system. The attendance rate for each time period is plotted on the control chart, allowing for visual representation and tracking over time.
Interpreting the information on the control chart involves analyzing the patterns and variations in the attendance rate. A stable attendance rate would indicate consistent student attendance, while fluctuations or downward trends could indicate potential issues that need attention. By monitoring the control chart, educational institutions can identify periods of low attendance, investigate the reasons behind it, and take appropriate actions such as implementing interventions or improving communication with students and parents to improve attendance rates. The control chart provides a visual representation of attendance patterns and enables proactive measures to be taken to address any attendance-related concerns.

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Answer:

£ 6,564.70

Step-by-step explanation:

Henry places £6000 in an account which pays 4.6% compound interest each year. Calculate the amount in his amount after 2 years

Compound Interest formula =

A = P(1 + r/n)^nt

A = Final Amounrt

P = Principal = £6,000

r = Interest rate = 4.6%

t = Time in years = 2 years

n = Compounding frequency = Yearly = 1

First, convert R percent to r a decimal

r = R/100

r = 4.6%/100

r = 0.046 per year,

Then, solve our equation for A

A = P(1 + r/n)^nt

A = 6,000.00(1 + 0.046/1)^(1×2)

A = £ 6,564.70

The amount in his account after 2 years = £ 6,564.70

Answer:

The missing number is 9.

226 integers are present between 100 and 999, inclusive, and have the property that some permutation of its digits is a multiple of 11 between 100 and 999. Hence, option A is the correct option.

The problem statement is to find the number of multiples of 11 between 100 and 999 inclusive, where some multiples may have digits repeated twice and some may not.

To solve this problem, we can first count the number of multiples of 11 between 100 and 999 inclusive, which is 81. Some of these multiples may have digits repeated twice, and each of these can be arranged in 3 permutations. Other multiples of 11 have no repeated digits, and each of these can be arranged in 6 permutations. However, we must account for the fact that switching the hundreds and units digits of these multiples also yields a multiple of 11, so we must divide by 2, giving us 3 permutations for each of these multiples.

Thus, we have a total of 81 × 3 = 243 permutations. However, we have overcounted because some multiples of 11 have 0 as a digit. Since 0 cannot be the digit of the hundreds place, we must subtract a permutation for each of these multiples. There are 9 such multiples (110, 220, 330, ..., 990), yielding 9 extra permutations. Additionally, there are 8 multiples (209, 308, 407, ..., 902) that also have 0 as a digit, yielding 8 more permutations.

Therefore, we must subtract these 17 extra permutations from the total of 243, giving us 226 permutations in total. Hence, option A is the correct option.

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Answer:

Step-by-step explanation:

The area of a square is x²

This means it is  x² = 64cm²

x= 8cm <------ One side

Since it is a square, all sides are the same length.

8*4 = 32cm

Side of the square: 8 cm

Perimeter: 32cm

Answer:

Step-by-step explanation:

To do this we need to show that the midpoints of the diagonals PR and QS are the same point.

The midpoint of PR = (-1+4)/2, (4-1)/2 = (3/2, 3/2).

The midpoint of QS = (-2+5)/2, )/2, (1 + 2)/2 = (3/2. 3/2).

We see that the points are identical so the diagonals do bisect each other.

The box plot would show the distribution of Tori's training runs and provide insights into the Central tendency and variability of the data.

A box plot, also known as a box-and-whisker plot, is a visual representation of the distribution of a dataset. It shows the median, quartiles, and range of the data. The median is the middle value of the dataset, the first quartile is the value that separates the lowest 25% of the data from the rest, and the third quartile is the value that separates the lowest 75% of the data from the rest. The range is the distance between the minimum and maximum values of the data.

To create a box plot for Tori's training runs, you would first order the data from smallest to largest: 6, 7, 8, 8, 11, 12, 13, 15, 16, 16, 20. Then, you would find the median (the middle value), which is 12.5. Next, you would find the first quartile (the value that separates the lowest 25% of the data from the rest), which is 8, and the third quartile (the value that separates the lowest 75% of the data from the rest), which is 16. Finally, you would calculate the range (the distance between the minimum and maximum values), which is 14 (20 - 6).

To draw the box plot, you would draw a rectangle with the bottom edge at the first quartile and the top edge at the third quartile. Inside the rectangle, draw a line at the median. Then, draw whiskers (lines) from the edges of the rectangle to the minimum and maximum values of the data (6 and 20). Any outliers (values that are more than 1.5 times the interquartile range away from the nearest quartile) can be plotted as individual points outside of the whiskers.

Without a visual representation of the data, it is difficult to say which box plot represents the data. However, the box plot would show the distribution of Tori's training runs and provide insights into the central tendency and variability of the data.

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Yes, I can help them determine how they did within their respective groups. A higher Z-score indicates a faster time compared to the average for their group, while a lower Z-score indicates a slower time.

To do this, we will calculate the standard score or Z-score for each of them, which measures how many standard deviations above or below the mean their finishing time was. Calculation is as follows:

For Leo:

Z = (4948 - 4313) / 583 = 1.01

For Mary:

Z = (5513 - 5261) / 807 = 0.33

So, Leo finished 1.01 standard deviations above the mean for the Men, Ages 30 - 34 group, while Mary finished 0.33 standard deviations below the mean for the Women, Ages 25 - 29 group.

As for your second question, if the distributions of finishing times were not nearly normal, then the values of the standard score would change. This is because the standard score is calculated using the mean and standard deviation, which are important parameters of a normal distribution, these parameters may not accurately represent the data, and therefore the standard score would not be an accurate measure of relative performance.

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The answer of the given question based on the the number of students using the cafeteria's healthy lunch line can be found by solving the inequality 1/5s - 51 < 20, the correct answer is (1) 142 , (1) or more.

An equation is  mathematical statement that shows  equality of two expressions, typically separated by  equal sign. Equations are used to express relationships between variables and to solve problems in various fields of mathematics, science, and engineering. In essence, an equation is representation of  real-world scenario or problem in mathematical terms.

The variables in equation can take on different values, and  equation remains true regardless of the values chosen for  variables. Solving  equation means finding  values of the variables that make  equation true.

Equations can be linear, quadratic, polynomial, exponential, logarithmic, or trigonometric, among other types. They are fundamental tool in mathematics and have numerous applications in fields like physics, engineering, economics, and finance.

There are 155 students or more students that use this lunch line.

The correct answer is 1. or more.

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Answer:

Simultaneous equations are ​equations involving two or more unknowns that are to have the same values in each equation.

 The total debt of the government at the end of year three is $7.2 billion.

Starting with a total debt of $5.5 billion, the government's deficits of $600 million in year one and $1.5 billion in year two contribute to an increase in the debt. However, in year three, the government runs a surplus of $400 million, which helps reduce the debt.

Thus, the total debt at the end of year three is calculated by adding the initial debt and deficits, and subtracting the surplus. In this case, the total debt at the end of year three is $7.2 billion.

Assuming the government runs a deficit of $200 million in year three, the calculation changes. With the initial debt of $5.5 billion, the deficits of $600 million in year one and $1.5 billion in year two, and the additional deficit of $200 million in year three, the total debt at the end of year three would be $7.8 billion. This demonstrates how the deficit amounts impact the overall debt accumulation.

If a government runs a budget deficit of $5 billion each year for ten years, resulting in a total deficit of $50 billion, followed by a surplus of $2 billion per year for five years, reducing the debt by $10 billion, and then maintaining a balanced budget for another ten years, the government debt would amount to $40 billion. The deficits over the initial ten years lead to a significant accumulation of debt, but the subsequent surpluses help mitigate the debt by reducing the deficit. Finally, the balanced budget in the later years ensures that there is no further increase in debt. Understanding the interplay between deficits, surpluses, and a balanced budget is crucial to grasp the long-term implications on government debt and fiscal stability.

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The probability that a student makes more than 7 mistakes on the exam is 0.39.

In science, the probability of an event is a number that indicates how likely the event is to occur. It is expressed as a number in the range from 0 and 1, or, using percentage notation, in the range from 0% to 100%. T

Now, let X be the number of mistakes a student makes on the exam. Then, the X follows a binomial distribution with n=16 and some unknown probability of success p (i.e., the probability of answering a question correctly).

We know that P(X<5) = 0.47 and P(5<=X<=7) = 0.14. Using the complement rule, we can find P(X>7) as follows:

P(X>7) = 1 - P(X<=7)

= 1 - P(X<5) - P(5<=X<=7)

= 1 - 0.47 - 0.14

= 0.39. Therefore, the probability that a student makes more than 7 mistakes on the exam is 0.39.

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Answer:

a. 15

Step-by-step explanation:

Because when you add 40 by 35 is 75.

Then 90 - 75 = 15.

40 + 35 + 15 = 90 Degrees

Duncan is putting icing on the top of a rectangular cake if the top of the cake is 6 inches wide by 10 inches long what is the area of the top of the cake

A. 80 square inches

B.60 square inches

C.30 square inches

D. 120 square inches

Answer:

B- 60 square inches

Answer:

I know this aint no clever bro cmon that stuff is easy

Step-by-step explanation:

To determine the average number of customers in line at the convenience store, we can use the concept of the queuing theory and apply the M/M/1 queuing model.

In the M/M/1 model: "M" represents Markovian arrivals, which means that arrivals occur in a random and independent manner. "M" also represents Markovian service times, which means that service times are random and independent. "1" represents a single server. Given that customers arrive every 3 minutes on average (λ = 1/3 arrivals per minute) and the clerk can ring up a customer in 2.5 minutes on average (μ = 1/2.5 customers served per minute), we can calculate the average number of customers in line (Lq) using the formula:

Lq = (λ^2) / (μ * (μ - λ))

Substituting the values, we have:

Lq = ((1/3)^2) / ((1/2.5) * ((1/2.5) - (1/3)))

= 1/12

Therefore, on average, there is 1/12 or approximately 0.083 customers in line, exclusive of the customer being served.

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The derivative of the function g(x) = (x² - x + 1\()^1^0\) * (tan(x))³ is g'(x) = 10(x² - x + 1)⁹ * (2x - 1) * (tan(x))³ + 3(x² - x + 1\()^1^0\) * (tan(x))² * sec²(x).

To find the derivative of the given function g(x), we can apply the product rule and the chain rule. Let's break down the function into its constituent parts: f(x) = (x² - x + 1\()^1^0\) and h(x) = (tan(x))³.

Using the product rule, the derivative of g(x) can be calculated as g'(x) = f'(x) * h(x) + f(x) * h'(x).

First, let's find f'(x). We have f(x) = (x² - x + 1\()^1^0\), which is a composite function. Applying the chain rule, f'(x) = 10(x² - x + 1\()^9\) * (2x - 1).

Next, let's determine h'(x). We have h(x) = (tan(x))³. Applying the chain rule, h'(x) = 3(tan(x))² * sec²(x).

Now, we substitute these derivatives back into the product rule formula:

g'(x) = f'(x) * h(x) + f(x) * h'(x)

       = 10(x² - x + 1)² * (2x - 1) * (tan(x))³ + 3(x² - x + 1\()^1^0\)* (tan(x))² * sec²(x).

In summary, the derivative of the function g(x) = (x² - x + 1\()^1^0\) * (tan(x))³ is g'(x) = 10(x² - x + 1)⁹  * (2x - 1) * (tan(x))³ + 3(x² - x + 1\()^1^0\) * (tan(x))² * sec²(x).

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Answer:

5

Step-by-step explanation:

Answer:

5

hope it helped you out

The statement "The Taylor series centered at 0 (aka MacLaurin Series) for f(x) = \(x^{11}\) sin x has only even powers of x" is false.

The Taylor series centered at 0 (or MacLaurin series) for the function f(x) = \(x^{11}\) sin(x) includes terms with both even and odd powers of x. The general form of the Taylor series for this function is:

f(x) = f(0) + f'(0)x + f''(0)\(x^{2}\)/2! + f'''(0)\(x^{3}\)/3! + ...

The derivative of f(x) = \(x^{11}\) sin(x) involves both the derivative of \(x^{11}\) (which has only even powers) and the derivative of sin(x) (which has both even and odd powers). Therefore, when the Taylor series is expanded, terms with both even and odd powers of x will be present.

Hence, the statement is false.

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