What is the absolute value of 3/4 on the number line

The probability that a student scores between 450 and 600 on the SAT math section is approximately 0.4147 or 41.47%.

To find the probability that a student scores between 450 and 600 on the SAT math section, we need to use the properties of the normal distribution. We know that the mean is 511 and the standard deviation is 110.

First, we need to standardize the values of 450 and 600 using the formula:

z = (x - μ) / σ

where x is the value we want to standardize, μ is the mean, and σ is the standard deviation.

For 450:

z = (450 - 511) / 110 = -0.55

For 600:

z = (600 - 511) / 110 = 0.81

Next, we need to find the area under the normal curve between these two standardized values. We can use a table or a calculator to find that the area between z = -0.55 and z = 0.81 is approximately 0.4147.

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The solution to the equation 4n^2 = 64 is n = 4.

The values for the variable in 2x^2 + 5x -3 are 1/2 and -3.

The equation 4n^2 = 64 can be solved by dividing the two sides by four. As a result, we get n^2 = 16. After taking the square root of both sides, it becomes apparent that n=±4, thereby providing us with solutions for n as being equal to both 4 and -4.

For equations such as 2x^2 + 5x -3 = (2x - 1)(x+3) = 0, setting each factor equal to zero is necessary to solve for x with either simultaneous or individual approaches. If done concurrently, two separate sets of equations come up: 2x-1=0 or x+3=0. Alternatively, solving these separately yields values equaling x=1/2 and x=-3.

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The volume of the solid is  17124 cubic units. To find the volume of the solid, we need to integrate the area of each cross-section perpendicular to the base and parallel to the diameter.

Since these cross-sections are squares. We can find their area by squaring the side length.

Let's call the side length of each square "x". We know that the diameter of the semicircle is 32 (twice the radius of 16), so the side length of each square is also 32.

Now we need to express the volume of the solid as an integral. We can do this by summing the areas of all the infinitesimal squares as we slice the solid perpendicular to the base.

The infinitesimal thickness of each slice is dx. The width of each slice is also dx, since the squares are perpendicular to the base. The height of each slice is the length of the chord of the semicircle that corresponds to the x-coordinate of the slice.

We can use the Pythagorean theorem to find this height. The chord has length 2(sqrt(16^2 - x^2)), so the height is sqrt(16^2 - x^2).

Therefore, the volume of the solid is given by the integral:

V = ∫(0 to 16) of [x^2 * (2sqrt(16^2 - x^2))] dx

Evaluating this integral, we get:

V = (2/3)(16^3)(pi) - (2/3)(16^3)

So the volume of the solid is approximately 17124 cubic units.

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The drive-thru is able to estimate their wait time more consistently will be;

⇒ A. Burger Quick, because it has a smaller IQR.

Since, The range of values in the middle of the scores is known as the interquartile range, or IQR.

The appropriate measure of variability is the interquartile range when a distribution is skewed and the median is used instead of the mean to show a central tendency.

Since, IQR is the difference between the third quartile and the first quartile, which is represented by the box in the box plot.

Now, In this case, the IQR for Burger Quick is,

15.5 - 8.5 = 7.0,

while the IQR for Fast Chicken is,

14.5 - 10 = 4.5.

Hence, A smaller IQR indicates that the data is more consistent and less spread out.

Thus, The correct option here is Burger Quick, because it has a smaller IQR (Interquartile Range).

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The graph that shows the image of triangle LMN after a dilation with scale factor of 3, followed by a translation 4 units down and 2 units left is given as follows:

Graph D.

The vertices of the original figure are given as follows:

L(1,0), M(0,3), N(2,2).

The dilation by a scale factor of 3 means that each coordinate of each vertex is multiplied by 3, hence the vertices are given as follows:

L'(3,0), M'(0,9) and N'(6,6).

The rule for the translation 4 units down and 2 units left is given as follows:

(x,y) -> (x - 2, y - 4).

Hence the vertices of the image are given as follows:

L''(1,-4), M''(-2, 5) and N''(4, 2).

Which are shown on Graph D.

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The annual payment is $7,351.98 if the APR rate is 6.75% over a 20-year period.

Area of land =  275 acres

Price = $4,300/acre

Time = 20-year

APR rate =  6.75%

down payment  = 25% of the total cost

Total cost = 275 acres x $4,300/acre

Total cost  = $1,182,500

Down payment = 0.25 x $1,182,500

Down payment = $295,625

The remaining amount = Total cost - Down payment

The remaining amount  = $1,182,500 - $295,625

The remaining amount = $886,875

The present value of an annuity,

PMT = [tex](r * PV) / (1 - (1 + r)^{n} )[/tex]

The interest rate =  6.75% / 12 = 0.5625% per month

The total number of periods = 20 years x 12 months/year = 240 months.

PMT = [tex](0.005625 * $886875) / (1 - (1 +0.005625)^{240} )[/tex]

PMT  = $7,351.98

Therefore, we can conclude that the annual payment is approximately $7,351.98

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False. If Determinant  A is zero, it does not necessarily imply that two columns of A must be the same or that all elements in a row or column of A are zero.

If det A is zero, it does not necessarily imply that two columns of A must be the same or that all elements in a row or column of A are zero.

For example, consider the following 2x2 matrix:

A = [1 2]
   [2 4]

The determinant of A is det(A) = (1*4 - 2*2) = 0, but the columns of A are not the same, and not all elements in a row or column are zero.

However, it is true that if two columns of A are the same or all of the elements in a row or column of A are zero, then det A must be zero.

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If there were only twenty-two students instead of fifty-two who contributed their moms' heights and we wanted to use a one-sample confidence interval procedure (whether it would be z or t), we would have had to assume that the population of mom-heights was normally distributed.

The one-sample confidence interval is used to estimate the true population mean based on a sample mean and standard deviation. The procedure assumes that the sample is a random sample from a normally distributed population. With a sample size of at least 30, the central limit theorem allows for the use of a z-test to construct the confidence interval. With a smaller sample size, a t-test would be more appropriate. However, the assumption of normality still needs to hold for the validity of the confidence interval.

If the sample size is smaller than 30, the assumption of normality can be replaced by the assumption of approximately normal data. This means that the data follows a distribution that is symmetric and bell-shaped, even if it is not exactly normal.

However, the validity of the confidence interval can be compromised if the data is highly skewed or contains outliers. In such cases, it may be necessary to use non-parametric methods to construct a confidence interval.

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Based on fractional values, if Susan originally has 7 yards of fabric, after making 3 aprons consuming 5⁵/₈ yards, the quantity of fabric left is 1³/₈ yards.

Fractional values are the results of fractional computations.

Fractions may be proper, improper, and complex fractions, depending on the values of the denominators and the numerators.

Algebraic expressions that have fractions are stated as fractional values.

The original quantity of fabric that Susan has = 7 yards

The quantity of fabric used for the front of each apron = 1¹/₄ yards

The quantity of fabric used for the tie of each apron = ⁵/₈ yards

The total quantity of fabric used for each apron = 1⁷/₈ yards (1¹/₄ + ⁵/₈)

The total quantity of fabric used for the 3 aprons made = 5⁵/₈ yards (1⁷/₈ x 3)

Therefore, the remaining quantity of fabric that Susan has after making the 3 aprons = 1³/₈ yards (7 - 5⁵/₈).

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Susan first made 3 aprons using 1¹/₄ yards for the front and ⁵/₈ yards for the tie.

Rounding up to the nearest whole number, we get a sample size of 314 students. Therefore, if we randomly select 314 students who are classified as high on their need for closure. we can be 99% confident that the sample mean score is within 1.5 points of the population mean score.

To determine the sample size needed, we can use the formula:

n = (z * σ / E)^2

Where:
z = the z-score corresponding to the desired level of confidence (in this case, 2.576 for 99% confidence)
σ = the population standard deviation (8.35)
E = the maximum allowable error (1.5)

Plugging in these values, we get:

n = (2.576 * 8.35 / 1.5)^2
n = 313.15

To determine the required sample size for a 99% confidence interval within 1.5 points of the population mean score, follow these steps:

1. Identify the given information:
- Population standard deviation (σ) = 8.35
- Desired margin of error (E) = 1.5
- Confidence level (z-score) = 2.576 (for 99% confidence interval)

2. Use the formula for sample size calculation:
n = (Z * σ / E)^2

3. Plug in the values:
n = (2.576 * 8.35 / 1.5)^2

4. Calculate the result:
n ≈ 121.22

5. Round up to the nearest whole number:
n = 122 students

So, a sample size of 122 students is needed to be 99% confident that the sample mean score is within 1.5 points of the population mean score for students who are high on the need for closure.

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The difference in the beatles length is 1/10 cm.

Given that Serena is measuring the length of beetles for a science project

Beetle measures 4/5 cm and another measure 7/10 cm.

We have to find the  difference in the beatles length

Let us convert one fraction  4/5  to denominator 10 by multiplying numerator and denominator by 2

4/5=8/10

Now difference is 8/10 - 7/10

Which is 1/10

Hence, the difference in the beatles length is 1/10 cm.

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We need a sample size of at least 665 drivers to estimate the proportion of all drivers exceeding the legal speed limit on the certain stretch of road between Los Angeles and Bakersfield with a margin of error of 0.04 and a 99% confidence level.

To estimate the proportion of all drivers exceeding the legal speed limit on a certain stretch of road between Los Angeles and Bakersfield with a margin of error of 0.04 and a 99% confidence level, we need to use the following formula:

[tex]n = (Z^2 * p * (1 - p)) / E^2[/tex]

where n is the sample size, Z is the Z-score for the desired confidence level (2.576 for 99% confidence level), p is the estimated proportion of drivers exceeding the speed limit (we don't have an estimate, so we'll use 0.5 for maximum variability), and E is the margin of error we want (0.04).

Plugging in the values, we get:

[tex]n = (2.576^2 * 0.5 * (1 - 0.5)) / 0.04^2[/tex]

n = 664.92

Therefore, we need a sample size of at least 665 drivers to estimate the proportion of all drivers exceeding the legal speed limit on the certain stretch of road between Los Angeles and Bakersfield with a margin of error of 0.04 and a 99% confidence level.

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The p-value is 0.05 or less if the chi-square statistic is 3.84 or more.

The p-value is 0.05 or less if the chi-square statistic is 3.84 or more. This indicates that there is a statistically significant difference between the observed and expected frequencies, and we reject the null hypothesis at a 5% significance level.

A statistical technique called the chi-squared test is used to assess whether there is a significant discrepancy between observed and predicted data. The test is frequently employed in disciplines like biology, sociology, and psychology to analyse categorical data.

The chi-squared test's fundamental premise is to contrast observed frequencies of a collection of categorical data with expected frequencies that would be anticipated if the data were distributed in a particular way. The test determines whether there is a statistically significant difference between the observed and anticipated frequencies.

The goodness-of-fit test and the test of independence are the two primary categories of chi-squared tests.

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In a double-elimination tournament with n players, the minimum number of matches that can be played is 2n - 3, and the maximum number of matches is 3n - 3.

In a double-elimination tournament with n players, we need to determine the minimum and maximum number of matches that can be played. Here's the step-by-step explanation:

1. the Minimum number of matches:
In a double-elimination tournament, each player is eliminated after their second loss. The minimum number of matches occurs when all players except the eventual winner lose twice in succession. In this case, there will be (n-1) matches in the winner's bracket and (n-2) matches in the loser's bracket.

Minimum number of matches = (n-1) + (n-2)
                                     = 2n - 3

2.The maximum number of matches:
In the maximum case scenario, each player has to be defeated twice except the eventual winner who will only have one defeat. This means there will be (n-1) matches in the winner's bracket and 2(n-1) matches in the loser's bracket.

Maximum number of matches = (n-1) + 2(n-1)
                                      = 3n - 3

So, in a double-elimination tournament with n players, the minimum number of matches that can be played is 2n - 3, and the maximum number of matches is 3n - 3.

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The calclated length of segment OG is 8 units

From the question, we have the following parameters that can be used in our computation:

DG = 4√5

DO = 4.

The length OG is calculated as

OG^2 = DG^2 - DO^2

substitute the known values in the above equation, so, we have the following representation

OG^2 = (4√5)^2 - 4^2

Evaluate

OG^2 = 64

So, we have

OG = 8

Hence, the solution is 8

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Answer:

Step-by-step explanation:

dad hates me sorry bye

Checking for approximate normality in the population is essential for constructing a valid confidence interval, particularly when dealing with small sample sizes. This ensures the accuracy and reliability of the interval in estimating the true population parameter.

It's important to check whether the population is approximately normal before constructing a confidence interval because the accuracy and validity of the interval depend on the underlying distribution of the population. Here's a step-by-step explanation:

1. A confidence interval is a range of values within which the true population parameter (e.g., mean or proportion) is likely to fall, with a certain level of confidence (e.g., 95% or 99%).

2. The process of constructing a confidence interval relies on the Central Limit Theorem, which states that, for large sample sizes, the sampling distribution of the sample mean will be approximately normal, regardless of the population distribution.

3. However, for small sample sizes, the distribution of the population needs to be approximately normal in order to obtain an accurate confidence interval. This is because the normality assumption is crucial for the proper interpretation of the interval.

4. If the population is not approximately normal, the confidence interval may not provide a reliable estimate of the true population parameter, leading to incorrect conclusions and potentially invalid results.

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The time taken by them to complete the work is 18 minutes.

Time taken by Sophie to peel all the potatoes = 45 minutes

Time taken by Simon to peel all the potatoes = 30 minutes

Amount of work done by Sophie in one minute = 1/45

Amount of work done by Simon in one minute = 1/30

Let the time taken by both of them to complete the work together be x.

So, the time taken by them to complete the work,

1/x = (1/45) + (1/30)

x = 1350/75

x = 18 minutes

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Therefore, the linearization of f(x) at a = -1 is L(x) = -10x - 6.

To find the linearization L(x) of the function f(x) = x⁴ + 3x² at a = -1, we need to use the formula:

L(x) = f(a) + f'(a)(x-a)

where f'(x) is the derivative of f(x) with respect to x.

First, we need to find f(-1) and f'(-1).

f(-1) = (-1)⁴ + 3(-1)²

= 1 + 3

= 4

f'(x) = 4x³ + 6x

f'(-1) = 4(-1)³ + 6(-1)

= -4 - 6

= -10

Now we can substitute these values into the linearization formula:

L(x) = f(-1) + f'(-1)(x - (-1))

L(x) = 4 - 10(x + 1)

L(x) = -10x - 6

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the height of the building is approximately 732.7 feet.

The tangent function is a trigonometric function that relates the ratio of the lengths of the opposite and adjacent sides of a right triangle to the angle between them. Specifically, for an acute angle theta in a right triangle, the tangent of theta is defined as:

tan(theta) = opposite/adjacent

We can use the tangent function to find the height of the building.

Let h be the height of the building. Then, we have:

tan(8 degrees) = h/1 mile

We can solve for h by multiplying both sides by 1 mile:

h = 1 mile * tan(8 degrees)

Using a calculator, we get:

h ≈ 0.139 miles

To convert this to feet, we can multiply by the conversion factor 5280 feet/mile:

h ≈ 0.139 * 5280 feet

h ≈ 732.72 feet

Therefore, the height of the building is approximately 732.7 feet.

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The length of side x based on the triangle given will be 26.2cm.

It should be noted that the image of the triangle is missing, so i have attached it.

In this case, to find the value of x, we will use cosine rule;

x² = 15² + 18² - 2(15 × 18)cos105

x² = 225 + 324 - 540(-0.2558)

x² = 549 + 138.132

x² = 687.132

x ≈ 26.2 cm

Therefore, length of side x based on the triangle given will be 26.2cm.

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This is greater than 4000 hours, so

a) First, we need to determine the travel time for each route at the current peak-hour demand of 10,000 trips/hour. For Route 1, we have:

ti = 0.33(1 + x1/ci) = 0.33(1 + 10,000/4,400) = 0.33(3.27) = 1.08 hours

For Route 2, we have:

t2 = 0.75 hours

Next, we need to determine the user-optimized assignment (UOA) of trips to these routes. The UOA is the assignment of trips that minimizes the total travel time for all travelers. In this case, since we only have two routes, the UOA is to assign all trips to the route with the lower travel time, which is Route 2. Therefore, all 10,000 trips/hour will be assigned to Route 2, and the total travel time will be:

Total travel time = 10,000 * 0.75 = 7,500 hours

b) With the expansion of the highway route to 6600 vehicles/hour, the travel time on Route 1 will change. We can calculate the new travel time using the same formula as before, but with the new capacity:

ti = 0.33(1 + x1/ci) = 0.33(1 + 10,000/6,600) = 0.33(1.51) = 0.50 hours

Now we need to determine the UOA with the new capacity. Again, the UOA is to assign all trips to the route with the lower travel time, which is now Route 1. Therefore, all 10,000 trips/hour will be assigned to Route 1, and the total travel time will be:

Total travel time = 10,000 * 0.50 = 5,000 hours

The percent improvement of travel time on the highway route is:

Percent improvement = (1.08 - 0.50)/1.08 * 100% = 53.7%

c) To limit the highway travel to 4000 vehicle-hours during the peak hour, we need to discourage some travelers from using Route 1. We can do this by imposing a congestion toll on Route 1. The toll should be set at a level that makes the UOA consistent with the limit of 4000 vehicle-hours. We can use the following steps to find the appropriate toll:

Assume a toll amount and calculate the resulting travel time on Route 1.

Use the UOA to calculate the total travel time.

If the total travel time is less than or equal to 4000 hours, the toll is appropriate. If the total travel time is greater than 4000 hours, adjust the toll amount and repeat steps 1-3 until the total travel time is less than or equal to 4000 hours.

Let's start with an initial toll of $0. Using the formula for travel time on Route 1, we can calculate the travel time as:

ti = 0.33(1 + x1/ci + t), where t is the toll amount in dollars

ti = 0.33(1 + 10,000/4,400 + 0) = 1.08 hours

The UOA is to assign all trips to Route 2, since the travel time on Route 2 is still lower than the travel time on Route 1. Therefore, the total travel time is:

Total travel time = 10,000 * 0.75 = 7,500 hours

This is greater than 4000 hours, so

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The probability of getting the majority of the questions right is P(X>=3) = P(X=3) + P(X=4) + P(X=5) = 15/128 + 6/1024 = 0.2266, rounded to four decimal places.

(a) The probability of guessing the correct answer for any single question is 1/4. Since there are no dependencies between questions, the probability of guessing the third question correctly is also 1/4. The probability of getting the first two questions wrong is (3/4)^2 = 9/16. Therefore, the probability that the first question Robin gets right is the third question is the product of these probabilities, which is (1/4)*(9/16) = 9/64. Rounded to four decimal places, this is 0.1406.

(b) To find the probability that Robin gets exactly 3 or exactly 4 questions right, we can use the binomial distribution. Let X be the number of questions Robin gets right. Then X follows a binomial distribution with n=5 and p=1/4, since each question has a probability of 1/4 of being answered correctly, and there are 5 questions in total.

The probability of getting exactly 3 questions right is P(X=3) = (5 choose 3) * (1/4)^3 * (3/4)^2 = 15/128. Similarly, the probability of getting exactly 4 questions right is P(X=4) = (5 choose 4) * (1/4)^4 * (3/4)^1 = 5/1024. The probability of getting both exactly 3 and exactly 4 questions right is 0, since they are mutually exclusive events.

Therefore, the probability of getting exactly 3 or exactly 4 questions right is P(X=3) + P(X=4) = 15/128 + 5/1024 = 0.1719, rounded to four decimal places.

(c) The majority of the questions means getting at least 3 questions right. We can calculate this probability using the binomial distribution again. The probability of getting 3 questions right is P(X=3) = 15/128, as calculated above. The probability of getting 4 or 5 questions right is P(X=4) + P(X=5) = (5 choose 4) * (1/4)^4 * (3/4)^1 + (5 choose 5) * (1/4)^5 * (3/4)^0 = 5/1024 + 1/1024 = 6/1024. Therefore, the probability of getting the majority of the questions right is P(X>=3) = P(X=3) + P(X=4) + P(X=5) = 15/128 + 6/1024 = 0.2266, rounded to four decimal places.

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The probability that a 0 is received can be found using conditional probability. Let's denote the event that a 0 is sent as S0, and the event that a 0 is received as R0. We want to find P(R0), the probability that a 0 is received.

Using the law of total probability, we can express P(R0) as the sum of the probabilities of receiving a 0 given that a 0 or a 1 was sent, weighted by the probabilities of sending a 0 or a 1:

[tex]P(R0) = P(R0|S0)P(S0) + P(R0|S1)P(S1)[/tex]

We are given that the probe sends a 1 one-third of the time and a 0 two-thirds of the time, so we have:

P(S0) = 2/3
P(S1) = 1/3

We are also given the probabilities of receiving a 0 or a 1 correctly or incorrectly, so we have:

P(R0|S0) = 0.6
P(R1|S0) = 0.4
P(R0|S1) = 0.2
P(R1|S1) = 0.8

Plugging these values into the formula for P(R0), we get:

P(R0) = (0.6)(2/3) + (0.8)(1/3)
     = 1/2

Therefore, the probability that a 0 is received is 1/2, or 50%.

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The area of the shape is 843.76mm²

The area of a shape is the space occupied by the boundary of a plane figures like circles, rectangles, and triangles.

The shape can be divided into a trapezoid , square and a semi circle.

The area of the trapezoid = 1/2(a+b) h

= 1/2( 55+12.5) ×12.5

= 1/2 × 67.5 × 12.5

= 421.88mm²

Area of square = 12.5 × 12.5

= 156.25mm²

Area of semi circle = 1/2πr²

= 1/2 × 3.14 × 12.5²

= 265.63mm²

Area of the shape = 265.63 + 421.88 + 156.25 = 843.76mm²

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Answer:

The greater fraction is 8/9.  The difference is 4/45

Step-by-step explanation:

To compare we need a common denominator. That number is 45

[tex]\frac{4}{5}[/tex]·[tex]\frac{9}{9}[/tex] = [tex]\frac{36}{45}[/tex]

[tex]\frac{8}{9}[/tex]·[tex]\frac{5}{5}[/tex] = [tex]\frac{40}{45}[/tex]  This has the largest value.

[tex]\frac{40}{45}[/tex] - [tex]\frac{36}{45}[/tex] = [tex]\frac{4}{45}[/tex]

Helping in the name of Jesus.

the time it would take for person 1 and person 2 to complete the task together is 2xy / (x + y) hours

If person 1 can do a task in x hours and person 2 can do the same task in y hours, then the combined rate at which they can complete the task is:

rate = 1/x + 1/y

This is because each person's rate of completing the task is the reciprocal of their time to complete the task, and their combined rate is the sum of their individual rates.

To find the time it would take for both persons to complete the task working together, we can use the formula:

time = 1 / rate

Substituting the expression for the rate above, we get:

time = 1 / (1/x + 1/y)

Simplifying this expression, we can use the formula for the harmonic mean of two numbers:

time = 2xy / (x + y)

Therefore, the time it would take for person 1 and person 2 to complete the task together is 2xy / (x + y) hours

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a. No, this model should not be used to predict the mean value of Y when X equals 7.
b. No, this model should not be used to predict the mean value of Y when X equals -7
c. Yes, this model can be used to predict the mean value of Y when X equals 0.
d. Yes, this model can be used to predict the mean value of Y when X equals 32.

a. Yes, this model can be used to predict the mean value of Y when X equals 7 because 7 is within the range of X values (6 to 33) used to find the prediction line in other words 7 is outside the range of values used to find the prediction line.

b. No, this model should not be used to predict the mean value of Y when X equals -7 because -7 is outside the range of X values (6 to 33) used to find the prediction line. In other words, the values of X used to find the prediction line are all positive.

c. No, this model should not be used to predict the mean value of Y when X equals 0 because 0 is outside the range of X values (6 to 33) used to find the prediction line. In other words, because it falls within the range of values used to find the prediction line.

d. Yes, this model can be used to predict the mean value of Y when X equals 32 because 32 is within the range of X values (6 to 33) used to find the prediction line. In other words, because it falls within the range of values used to find the prediction line.

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The Domain of the function is Domain = {-1, -1, -2, -2, -3, -3, -4, -4}.

Yes the relation a function.

We have the set,

{(-1,1), (-1,-1),(-2,2), (-2,-2),(-3,3),(-3,-3),(-4,4), (-4,-4)}

We know that the domain is the input value or the x value.

So, Domain = {-1, -1, -2, -2, -3, -3, -4, -4}

and, the range is the output value or y value

So, Range = {1, -1, 2, -2, 3, -3, 4, -4}

As, each input value have distinct output s then the relation a function.

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The domain of the original function is all real numbers, and the range is also all real numbers. The domain of the inverse function is also all real numbers, and the range is also all real numbers.

The domain of the original function is all real numbers, and the range is also all real numbers. The domain of the inverse function is also all real numbers, and the range is also all real numbers.

For the third number, the domain of the original function is all real numbers, and the range is also all real numbers. The domain of the inverse function is also all real numbers, and the range is also all real numbers.

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The best point estimate for the average number of roommates per semester for all students at the local college is 0.7, based on the study conducted with a sample of 133 students.

The best point estimate for the average number of roommates per semester for all students at the local college is 0.7, which is the average found in the study of the sample of 133 students. Since the sample is representative of the population, we can use the sample mean as a point estimate for the population mean. Therefore, we can estimate that the average number of roommates per semester for all students at the local college is 0.7.

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