Lorcaserin and weight Loss. A double-blind, randomized comparative experiment compared the effect of the drug lorcaserin and a placebo on weight loss in overweight adults. All subjects also underwent diet and exercise counseling. The study reported that, after one year, patients in the lorcaserin group had an average weight loss of 5.8 kilograms (kg), while those on the placebo had an average weight loss of 2.2 kg (P < 0.001).^2 Explain to someone who knows no statistics why these results that there is good reason to think that lorcaserin works. Include an explanation of what P < 0.001 means.

Answer:

= a(7b + 1)

Step-by-step explanation:

7ab + a

a(7b + 1).

Answer:

\(y = -2\, x - 6\).

Step-by-step explanation:

Calculate the gradient of line \(\sf {AB}\). For a non-vertical line that goes through \((x_0,\, y_0)\) and \((x_1,\, y_1)\) where \(x_{0} \ne x_{1}\), the gradient would be:

\(\displaystyle \text{$\frac{y_{1} - y_{0}}{x_{1} - x_{0}}$ given that $x_{0} \ne x_{1}$}\).

For line \(\sf AB\), the two points are \({\sf A}\, (-6,\, -4)\) and \({\sf B}\, (2,\, 0)\). Hence, the gradient of line \(\sf AB\!\) would be:

\(\begin{aligned} \frac{0 - (-4)}{2 - (-6)} = \frac{1}{2}\end{aligned}\).

The perpendicular bisector of a line segment (\(\sf AB\) in this question) is perpendicular to that line segment.

In a cartesian plane, the gradients of two lines perpendicular to one another would be inverse reciprocals. In other words, the product of these two gradients would be \((-1)\).

Hence, if \(m_{1}\) represents the gradient of \({\sf AB}\!\!\), and \(m_{2}\) represents the gradient of the perpendicular bisector, then \(m_{1} \cdot m_{2} = -1\).

Since the gradient of \({\sf AB}\) is \((1/2)\), the gradient of its perpendicular bisector would be:

\(\begin{aligned}m_{2} &= \frac{-1}{m_{1}} \\ &= \frac{-1}{1/2} \\ &= -2\end{aligned}\).

The perpendicular bisector of a line segment (\({\sf AB}\) in this question) goes through the midpoint of that line segment.

Apply the midpoint formula to find the midpoint of segment \({\sf AB}\).

If the endpoints of a line segment are \((x_{0},\, y_{0})\) and \((x_{1},\, y_{1})\), the midpoint of that line segment would be:

\(\begin{aligned} \left(\frac{x_0 + x_1}{2},\, \frac{y_{0} + y_{1}}{2}\right)\end{aligned}\).

The two endpoints of segment \({\sf AB}\) are \({\sf A}\, (-6,\, -4)\) and \({\sf B}\, (2,\, 0)\). The midpoint of segment \({\sf AB}\!\) would be:

\(\begin{aligned} \left(\frac{-6 + 2}{2},\, \frac{-4 + 0}{2}\right) &= (-2,\, -2)\end{aligned}\).

Find the equation of this perpendicular bisector in the point-slope form.

Consider a non-vertical line of slope \(m\). If this line goes through the point \((x_{0},\, y_{0})\), the point-slope form equation of this line would be:

\(y - y_{0} = m\, (x - x_{0})\).

The slope of the perpendicular bisector in this question is \((-2)\). Besides, this line goes through the point \((-2,\, -2)\). Henec, the point-slope equation of this line would be:

\(y - (-2) = (-2)\, (x - (-2))\).

\(y + 2 = -2\, (x + 2)\).

All the choices in this question are in the slope-intercept form. Hence, rewrite the point-slope equation \(y + 2 = -2\, (x + 2)\) to find the corresponding (equivalent) slope-intercept equation of this line:

\(y + 2 = -2\, (x + 2)\).

\(y = -2\, x - 6\).

Answer:

answer is c

Hop it helps you

1. (4a - 5)+(3a + 6) = 7a + 1.
To solve, you simply combine the like terms (4a and 3a) to get 7a, and then combine the constants (-5 and 6) to get 1.

2. (6x + 9)+ (4x^2 - 7) = 4x^2 + 6x + 2.
To solve, you combine the like terms (6x and 4x^2) to get 4x^2 + 6x, and then combine the constants (9 and -7) to get 2.

3. (6xy + 2y + 6x) + (4xy - x) = 10xy + 2y + 6x - x = 10xy + 2y + 5x.
To solve, you combine the like terms (6xy and 4xy) to get 10xy, then combine the constants (2y and -x) to get 2y - x, and finally combine the like terms (6x and 5x) to get 11x. The final answer is 10xy + 2y + 5x.

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shooting percentage: 0.425 (decimal)

To express in percentage:

0.425 x 100 = 42.5%

(3) He made 42.5% of the baskets he attempted.

And also

17/40 = 0.425

(4) He made 17 out of every 40 baskets he attempted

The probability distributions whose graphs can be approximated by bell-shaped curves are commonly known as normal distributions or Gaussian distributions.

These distributions are characterized by their symmetrical shape and the majority of their data falling within a certain range around the mean. The normal distribution is widely used in statistics and is a fundamental concept in many fields of study, including psychology, economics, and engineering. The normal distribution is also known for its many practical applications, such as predicting test scores, stock prices, and medical diagnoses. In summary, the bell-shaped curve is a useful tool in probability theory that can help us understand and make predictions about a wide range of phenomena. The probability distributions whose graphs can be approximated by bell-shaped curves are called Normal Distributions or Gaussian Distributions. They have a symmetrical shape and are characterized by their mean (µ) and standard deviation (σ), which determine the central location and the spread of the distribution, respectively.

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Answer:

multiply the radius by 2 to get the diameter.

Step-by-step explanation:

Answer:

31 : 41

Step-by-step explanation:

Given that Ravi's mass is 31kg . And his brother is 10 kg heavier than him . So his weight will be 31kg + 10kg = 41kg. So we have ,

\(\begin{cases}Mass \ of \ Ravi = 31kg \\ Mass of Brother \ = \ 41kg \end{cases}\)

Therefore their total mass = 31 + 41 = 72 kg .

Therefore the ratio will be 41:72 .

The ratio of Rajar's mass to their total mass is 41:72.

The ratio is defined as the comparison of two quantities of the same units that indicates how much of one quantity is present in the other quantity.

Given that, Ravi's mass is 31 kg.

His brother, Rajar, is 10 kg heavier than Ravi.

Weight of Rajar is 31+10=41 kg

Total weight =31+41

= 72 kg

The ratio of weights = 41:72

Therefore, the ratio of Rajar's mass to their total mass is 41:72.

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i just multiply the 0.31 by the number of days in September to get the average  

The work required to pump all the water over the edge of the tank is approximately 271,433.64 foot-pounds.

To calculate the work required to pump all the water over the edge of the tank, we need to consider the weight of the water in the tank and the height it is lifted.

First, let's find the volume of the water in the tank. The volume of a cone is given by the formula V = (1/3)πr²h, where r is the radius and h is the height. Plugging in the values, we have:

V = (1/3)π(3²)(12)

= (1/3)π(9)(12)

= 36π

Next, we need to find the weight of the water. The weight of an object is given by the formula W = mg, where m is the mass and g is the acceleration due to gravity. The mass of the water can be found by multiplying its volume by the density of water, which is approximately 62.4 pounds per cubic foot:

m = (36π)(62.4)

≈ 22619.47 pounds

Now, we can calculate the work done by multiplying the weight of the water by the height it is lifted. In this case, the height is 12 feet:

W = (22619.47)(12)

≈ 271433.64 foot-pounds

Therefore, the work required to pump all the water over the edge of the tank is approximately 271,433.64 foot-pounds.

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The work required to pump water out of an inverted conical tank involves calculating the pressure-volume work at infinitesimally small volumes within the tank and integrating this over the entire volume of the tank. This provides an interesting application of integral calculus in Physics.

The question requires the concept of work in Physics applied to a fluid, in this case, water lying within an inverted conical tank. Work is done when force is applied over a distance, as stated by work = force x distance. In the fluid analogy, the 'force' link is the pressure exerted on the water and the distance is the change in volume of the fluid. Therefore, work done (W) = Pressure x Change in Volume (ΔV).

In this scenario, you are required to pump out water from an inverted conical tank, hence, the work you do is against the gravitational force pulling the water downwards. To calculate the total work done, you have to consider the work done at each infinitesimally small (hence, constant pressure) strip of volume and integrate over the entire volume of the tank.

The detail of calculation would require the knowledge of integral calculus and the formula for volume of a cone. I recommend considering this as an interesting application of integrals in Physics. Also remember that the volume of a cone = 1/3πr²h, where 'r' is the radius of base and 'h' is the height of cone.

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Answer:

Multiply the bottom equation by -3/2 then add the equations.

Multiply the top equation by-3. then add the equations

Step-by-step explanation:

Given the simultaneous equation

2x- 6y=6  ... 1

6x - 4y = 2 ... 2

To eliminate a variable, we have to make the coefficient of one of the variable to be the same.

Multiply equastion 1 by -3

-6x+18y= -18  

6x - 4y = 2

Add the result:

-6x + 6x + 18y-4y = -18+2

18y-4y = -18+2

14y = -18

y = -9/7

Another way is to Multiply the bottom equation by -3/2 then add the equations.

Multiplying equation 2 by -3/2 will give;

6x(-3/2) - 4y(-3/2) = 2(-3/2)

-9x + 6y = -3

Add to equation 1;

2x- 6y=6

-9x + 2x + 0 = -3+6

-7x = 3

x = -3/7

Hence the correct two options are;

Multiply the bottom equation by -3/2 then add the equations.

Multiply the top equation by-3. then add the equations

The increasing the sample size by a factor of 25 will reduce the width of the confidence interval by a factor of 5.

a. Suppose we have a two-sided confidence interval for the population mean with known variance, given by the equation:

Cl is the confidence interval, x is the sample mean, σ is the population standard deviation, n is the sample size, and zα/2 is the z-score corresponding to the desired level of confidence.

The σ are fixed for a given level of confidence, we can achieve this by increasing n by a factor of 4.

Specifically, if we increase the sample size from n to 4n, then the new confidence interval will have half the width of the original interval.

b. If the sample size is increased by a factor of 25, then the term √n in the denominator of the above equation will be replaced.

Therefore, increasing the sample size by a factor of 25 will reduce the width of the confidence interval by a factor of 5.

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Answer:

Step-by-step explanation:

A measure of central tendency is a summary statistic that represents the center point or typical value of a dataset. These measures indicate where most 

Answer:

I think B.

Step-by-step explanation:

Answer:88.57

Step-by-step explanation: If your doing PEMDAS you do () Wich is 6-3 and gives 3tyen you do Addition which is 4+3=7 Then do Subtraction which is 100-80 so the equation is 100-80/7=88.57 etc.

If the range on an X-bar-R chart suddenly and significantly increases, it indicates an increase in process variation. In this scenario, the mean (X-bar) may or may not be affected.

The mean represents the central tendency or average value of the process, while the range measures the dispersion or variation within the process.

If the mean remains stable and unaffected despite the increase in range, it suggests that the process average is still within control. However, if the range increase is accompanied by a significant shift in the mean, it indicates a potential shift in the process average.

To make a definitive determination, additional analysis and investigation are necessary to identify the underlying cause of the increased range and its impact on the process mean.

This could involve examining individual data points, performing hypothesis testing, or conducting further statistical analysis to assess the process stability and potential issues.

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a) Approximate Doubling Time Dappx = 70/r, where r is the annual interest rate in percentage terms.Assuming that r = 5%, 10%, 20%, 40%, and 60%Doubling time for 5% = 70/5 = 14 yearsDoubling time for 10% = 70/10 = 7 yearsDoubling time for 20% = 70/20 = 3.5 yearsDoubling time for 40% = 70/40 = 1.75 yearsDoubling time for 60% = 70/60 = 1.1667 yearsb) Exact Doubling Time Dexact = ln2/r where r is the annual interest rate in decimal terms.

Assuming that r = 5%, 10%, 20%, 40%, and 60%For 5%: ln2/0.05 ≈ 13.86 yearsFor 10%: ln2/0.1 ≈ 6.93 yearsFor 20%: ln2/0.2 ≈ 3.47 yearsFor 40%: ln2/0.4 ≈ 1.73 yearsFor 60%: ln2/0.6 ≈ 1.16 yearsc)

The percentage error is given by:(Dexact − Dappx)/Dexact × 100%For 5%: (13.86 - 14)/13.86 x 100% ≈ 1.18%For 10%: (6.93 - 7)/6.93 x 100%

≈ 1.15%For 20%: (3.47 - 3.5)/3.47 x 100%

≈ -0.86%For 40%: (1.73 - 1.75)/1.73 x 100%

≈ -1.15%For 60%: (1.16 - 1.1667)/1.16 x 100%

≈ -0.60%Note that the percentage error is small for lower values of interest rates but increases as the interest rate increases.

Also, the percentage error is negative for 20%, 40%, and 60%, which means that the approximate doubling time is actually larger than the exact doubling time.

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The value of K is 2.

Let's denote the scale factor as K. The surface area of a solid after dilation is directly proportional to the square of the scale factor.

We are given that the initial surface area of the solid is 50 units^2, and after dilation, the surface area becomes 200 units^2.

Using the formula for the surface area, we have:

Initial surface area * (scale factor)^2 = Final surface area

50 * K^2 = 200

Dividing both sides of the equation by 50:

K^2 = 200/50

K^2 = 4

Taking the square root of both sides:

K = √4

K = 2

Therefore, the value of K is 2.

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Answer:

3%×300,=900

900÷100=9

Step-by-step explanation:

answer is 9

Answer:

291

Step-by-step explanation:

no prob

The stationing of the end of the excavation is Station 2+080.

The volume of fill by end area method is 1440 cubic meters.

The volume of excavation by end area is 1080 cubic meters.

We have,

The stationing of the end of excavation can be determined by adding the given sections to the starting station.

Assuming the starting station is Station 2+020, the end of excavation would be at Station 2+080 (given as Station 2+020 + 60 meters).

To find the volume of fill by end area method, we need to calculate the area of the end section and multiply it by the distance between the start and end stations.

Given that the base for fill is 12 meters and the sideslope for fill is 2:1, the area of the end section can be calculated as follows:

Area of end section = (Base for fill) * (Sideslope for fill)

Area of end section = 12 * (2/1) = 24 square meters

Now, multiply the area of the end section by the distance between the start and end stations (60 meters) to find the volume of fill:

Volume of fill = Area of end section * Distance

Volume of fill = 24 * 60 = 1440 cubic meters

Therefore, the volume of fill by end area method is 1440 cubic meters.

Similarly, to compute the volume of excavation by end area, we calculate the area of the end section (base for cut * sideslope for cut) and multiply it by the distance between the start and end stations. Given that the base for cut is 12 meters and the sideslope for cut is 1.5:1, the area of the end section can be calculated as follows:

Area of end section = (Base for cut) * (Sideslope for cut)

Area of end section = 12 * (1.5/1) = 18 square meters

Multiply the area of the end section by the distance between the start and end stations (60 meters) to find the volume of excavation:

Volume of excavation = Area of end section * Distance

Volume of excavation = 18 * 60 = 1080 cubic meters

Therefore, the volume of excavation by end area is 1080 cubic meters.

Thus,

The stationing of the end of the excavation is Station 2+080.

The volume of fill by end area method is 1440 cubic meters.

The volume of excavation by end area is 1080 cubic meters.

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The probability of getting the sequence THHT when a fair coin is tossed four times is 1/16.

The probability of getting a specific sequence of tosses (in this case, THHT) with a fair coin can be calculated by multiplying the individual probabilities of each toss.

Since each toss is independent and the coin is fair, the probability of getting a tail (T) is 1/2, and the probability of getting a head (H) is also 1/2.

To find the probability of the sequence THHT, we multiply the probabilities together:

P(THHT) = (1/2) * (1/2) * (1/2) * (1/2)

P(THHT) = 1/16

Therefore, the probability of  the sequence THHT = 1/16.

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The linear function passing through points (0,3) and (4,6) is defined as follows:

f(x) = 0.75x + 3.

The slope-intercept definition of a linear function is given as follows:

y = mx + b.

In which the parameters are given as follows:

The function passes through point (0,3), meaning that when x = 0, y = 3, hence the intercept b is given as follows:

b = 3.

Given two points, the slope is calculated as the change in y divided by the change in x, hence:

m = (6 - 3)/(4 - 0) = 3/4 = 0.75.

Thus the linear function f(x) is defined as follows:

f(x) = 0.75x + 3.

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The correct sequence of project phases is Definition - Planning  - Performance. The correct answer is B.

This is the typical order of project phases in a traditional project management approach. It starts with the definition phase, where the project's goals, the scope, and the requirements are established.

Then comes the planning phase, where the project plan is developed, including the allocation of resources, scheduling, and budgeting.

Finally, the performance phase begins, during which the project activities are executed, monitored, and controlled to ensure the successful project completion.

Therefore the sequence of project phase is  Definition - Planning  - Performance The correct answer is B.

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Given:

The equation is:

\(2y=3x-6\)

To find:

The x-intercept and y-intercept of the graph of the given equation.

Solution:

We have,

\(2y=3x-6\)

Substitute \(y=0\) to find the x-intercept.

\(2(0)=3x-6\)

\(0+6=3x\)

\(\dfrac{6}{3}=x\)

\(2=x\)

So, the x-intercept of the graph of the given equation is 2.

Substitute \(x=0\) to find the y-intercept.

\(2y=3(0)-6\)

\(2y=-6\)

\(y=\dfrac{-6}{2}\)

\(y=-3\)

So, the y-intercept of the graph of the given equation is -3.

Therefore, the x-intercept is 2 and the y-intercept is -3.

Answer:

The 1st term is -67

Step-by-step explanation:

I. When Sn = \(\frac{n}{2}\)(\(a_{1} + a_{n}\))

   If  Sn = 124, n = 8 and \(a_{n}\) = 98

                124 = 8/2(a + 98)

                124 = 4(a + 98)

                124 = 4a + 392

                124 - 392 = 4a

                -268/4 = a

                -67 = \(a_{1}\)

Hope that help :D

The most specific name for the figure is an isosceles trapezoid

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

The figure

The properties of the given figure are

Using the above as a guide, we have the following:

The figure is a trapezoid

Because the nonparallel sides are congruent, then the most specific name for the figure is an isosceles trapezoid

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(a) The expected profit of this policy to the insurance company is $520.10.

(b) The interpretation value of expected profit in (a) in the context of the problem is small probability of the policyholder's death.

(a) The expected profit to the insurance company can be calculated as follows:

Expected Profit = (Probability of Survival * Premium) - (Probability of Death * Death Benefit)

Probability of Survival = 0.997

Probability of Death = 1 - Probability of Survival = 1 - 0.997 = 0.003

Premium = $780

Death Benefit = $250,000

Expected Profit = (0.997 * $780) - (0.003 * $250,000) = $520.10

Therefore, the expected profit to the insurance company from selling this policy is $520.10.

(b) The expected profit of $520.10 represents the average profit that the insurance company can expect to earn from selling this type of policy to similar customers over time.

In the context of this problem, it means that if the insurance company sells many policies like this one, they can expect to make a profit of about $520.10 per policy, on average, after accounting for the small probability of the policyholder's death.

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