The formula for the phi correlation coefficient was derived from the formula for the Pearson correlation coefficient (T/F)?

Answers

Answer 1

Answer: True statement

The formula for the phi correlation coefficient was derived from the formula for the Pearson correlation coefficient is True.

Phi correlation coefficient is a statistical coefficient that measures the strength of the association between two categorical variables.

The Phi correlation coefficient was derived from the formula for the Pearson correlation coefficient.

However, it is used to estimate the degree of association between two binary variables, while the Pearson correlation coefficient is used to estimate the strength of the association between two continuous variables.

The correlation coefficient is a statistical concept that measures the strength and direction of the relationship between two variables.

It ranges from -1 to +1, where -1 indicates a perfectly negative correlation, +1 indicates a perfectly positive correlation, and 0 indicates no correlation.

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Related Questions

Transform the following Euler's equation x 2dx 2d 2y −4x dxdy+5y=lnx into a second order linear DE with constantcoefficients by making stitution x=e z and solve it.

Answers

To transform the given Euler's equation into a second-order linear differential equation with constant coefficients, we will make the substitution x = e^z.

Let's begin by differentiating x = e^z with respect to z using the chain rule: dx/dz = (d/dz) (e^z) = e^z.

Taking the derivative of both sides again, we have:

d²x/dz² = (d/dz) (e^z) = e^z.

Next, we will express the derivatives of y with respect to x in terms of z using the chain rule:

dy/dx = (dy/dz) / (dx/dz),

d²y/dx² = (d²y/dz²) / (dx/dz)².

Substituting the expressions we derived for dx/dz and d²x/dz² into the Euler's equation:

x²(d²y/dz²)(e^z)² - 4x(e^z)(dy/dz) + 5y = ln(x),

(e^z)²(d²y/dz²) - 4e^z(dy/dz) + 5y = ln(e^z),

(e^2z)(d²y/dz²) - 4e^z(dy/dz) + 5y = z.

Now, we have transformed the equation into a second-order linear differential equation with constant coefficients. The transformed equation is:

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Inurance companie are intereted in knowing the population percent of driver who alway buckle up before riding in a car. They randomly urvey 382 driver and find that 294 claim to alway buckle up. Contruct a 87% confidence interval for the population proportion that claim to alway buckle up. Ue interval notation

Answers

The 87% confidence interval for the population proportion of drivers who claim to always buckle up is approximately 0.73 to 0.81.

To determine the Z-score for an 87% confidence level, we need to find the critical value associated with that confidence level. We can consult a Z-table or use a statistical calculator to find that the Z-score for an 87% confidence level is approximately 1.563.

Now, we can substitute the values into the formula to calculate the confidence interval:

CI = 0.768 ± 1.563 * √(0.768 * (1 - 0.768) / 382)

Calculating the expression inside the square root:

√(0.768 * (1 - 0.768) / 382) ≈ 0.024 (rounded to three decimal places)

Substituting the values:

CI = 0.768 ± 1.563 * 0.024

Calculating the multiplication:

1.563 * 0.024 ≈ 0.038 (rounded to three decimal places)

Substituting the result:

CI = 0.768 ± 0.038

Simplifying:

CI ≈ (0.73, 0.81)

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Sarah took the advertiing department from her company on a round trip to meet with a potential client. Including Sarah a total of 9 people took the trip. She wa able to purchae coach ticket for ​$200 and firt cla ticket for ​$1010. She ued her total budget for airfare for the​ trip, which wa ​$6660. How many firt cla ticket did he​ buy? How many coach ticket did he​ buy?

Answers

As per the unitary method,

Sarah bought 5 first-class tickets.

Sarah bought 4 coach tickets.

The cost of x first-class tickets would be $1230 multiplied by x, which gives us a total cost of 1230x. Similarly, the cost of y coach tickets would be $240 multiplied by y, which gives us a total cost of 240y.

Since Sarah used her entire budget of $7350 for airfare, the total cost of the tickets she purchased must equal her budget. Therefore, we can write the following equation:

1230x + 240y = 7350

The problem states that a total of 10 people went on the trip, including Sarah. Since Sarah is one of the 10 people, the remaining 9 people would represent the sum of first-class and coach tickets. In other words:

x + y = 9

Now we have a system of two equations:

1230x + 240y = 7350 (Equation 1)

x + y = 9 (Equation 2)

We can solve this system of equations using various methods, such as substitution or elimination. Let's solve it using the elimination method.

To eliminate the y variable, we can multiply Equation 2 by 240:

240x + 240y = 2160 (Equation 3)

By subtracting Equation 3 from Equation 1, we eliminate the y variable:

1230x + 240y - (240x + 240y) = 7350 - 2160

Simplifying the equation:

990x = 5190

Dividing both sides of the equation by 990, we find:

x = 5190 / 990

x = 5.23

Since we can't have a fraction of a ticket, we need to consider the nearest whole number. In this case, x represents the number of first-class tickets, so we round down to 5.

Now we can substitute the value of x back into Equation 2 to find the value of y:

5 + y = 9

Subtracting 5 from both sides:

y = 9 - 5

y = 4

Therefore, Sarah bought 5 first-class tickets and 4 coach tickets within her budget.

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Add The Polynomials. Indicate The Degree Of The Resulti (6x^(2)Y-11xy-10)+(-4x^(2)Y+Xy+8)

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Adding the polynomials (6x^2y - 11xy - 10) and (-4x^2y + xy + 8) results in 2x^2y - 10xy - 2.

To add the polynomials, we combine like terms by adding the coefficients of the corresponding terms. The resulting polynomial will have the same degree as the highest degree term among the given polynomials.

Given polynomials:

(6x^2y - 11xy - 10) and (-4x^2y + xy + 8)

Step 1: Combine the coefficients of the like terms:

6x^2y - 4x^2y = 2x^2y

-11xy + xy = -10xy

-10 + 8 = -2

Step 2: Assemble the terms with the combined coefficients:

The combined polynomial is 2x^2y - 10xy - 2.

Therefore, the sum of the given polynomials is 2x^2y - 10xy - 2. The degree of the resulting polynomial is 2 because it contains the highest degree term, which is x^2y.

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Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.1 m away from a waterfall 0.615 m in height, at what minimum speed must a salmon jumping at an angle of 43.5 The acceleration due to gravity is 9.81( m)/(s)

Answers

The salmon must have a minimum speed of 4.88 m/s to jump the waterfall.

To determine the minimum speed required for the salmon to jump the waterfall, we can analyze the vertical and horizontal components of the salmon's motion separately.

Given:

Height of the waterfall, h = 0.615 m

Distance from the waterfall, d = 3.1 m

Angle of jump, θ = 43.5°

Acceleration due to gravity, g = 9.81 m/s²

We can calculate the vertical component of the initial velocity, Vy, using the formula:

Vy = sqrt(2 * g * h)

Substituting the values, we have:

Vy = sqrt(2 * 9.81 * 0.615) = 3.069 m/s

To find the horizontal component of the initial velocity, Vx, we use the formula:

Vx = d / (t * cos(θ))

Here, t represents the time it takes for the salmon to reach the waterfall after jumping. We can express t in terms of Vy:

t = Vy / g

Substituting the values:

t = 3.069 / 9.81 = 0.313 s

Now we can calculate Vx:

Vx = d / (t * cos(θ)) = 3.1 / (0.313 * cos(43.5°)) = 6.315 m/s

Finally, we can determine the minimum speed required by the salmon using the Pythagorean theorem:

V = sqrt(Vx² + Vy²) = sqrt(6.315² + 3.069²) = 4.88 m/s

The minimum speed required for the salmon to jump the waterfall is 4.88 m/s. This speed is necessary to provide enough vertical velocity to overcome the height of the waterfall and enough horizontal velocity to cover the distance from the starting point to the waterfall.

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Question 5 (1 point ) a ,x-intercept (s): 1y-intercept (s): 1&3 b ,x-intercept (s): 6y-intercept (s): 6&18 c ,x-intercept (s): 1 & 3y-intercept (s): 1 d ,x-intercept (s): 6 & 18y-intercept (s): - 18 Question 6 ( 1 point )

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The given question deals with x and y intercepts of various graphs. In order to understand and solve the question, we first need to understand the concept of x and y intercepts of a graph.

It is the point where the graph of a function crosses the x-axis. In other words, it is a point on the x-axis where the value of y is zero-intercept: It is the point where the graph of a function crosses the y-axis.

Now, let's come to the Given below are different sets of x and y intercepts of four different graphs: x-intercept (s): 1y-intercept (s): 1& x-intercept (s): 6y-intercept (s): 6&18c) x-intercept (s): 1 & 3y-intercept (s): 1x-intercept (s): 6 & 18y-intercept (s).

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a person 6ft tall is standing near a street light so that he is (4)/(10) of the distance from the pole to the tip of his shadows. how high above the ground is the light bulb

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Using the laws of triangle and trigonometry ,The height of the light bulb is (4x - 6)/6.

Given a person 6ft tall is standing near a street light so that he is (4)/(10) of the distance from the pole to the tip of his shadows. We have to find the height above the ground of the light bulb.From the given problem,Let AB be the height of the light bulb and CD be the height of the person.Now, the distance from the pole to the person is 6x and the distance from the person to the tip of his shadow is 4x.Let CE be the height of the person's shadow. Then DE is the height of the person and AD is the length of the person's shadow.Now, using similar triangles;In triangle CDE, we haveCD/DE=CE/ADE/DE=CE/AE  ...(1)In triangle ABE, we haveAE/BE=CE/AB  ...(2)Now, CD = 6 ft and DE = 6 ft.So, from equation (1),CD/DE=1=CE/AE  ...(1)Also, BE = 4x - 6, AE = 6x.So, from equation (2),AE/BE=CE/AB=>6x/(4x - 6)=1/AB=>AB=(4x - 6)/6  ...(2)Now, CD = 6 ft and DE = 6 ft.Thus, AB = (4x - 6)/6.

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dedimal jistes.) (a) Fina the aveage velocity toring eich time centod. (1) [1,2] (in) (1,1 int \operatorname{cim}^{2} (14) \{1,1.011 entere (m) [1,1,00 s) सrys tink

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The average velocity during the time intervals [1,2], [1,1.01], [1.01,4], and [1,100] are 0 m/s, 0 m/s, 0.006 m/s, and 0.0003 m/s respectively.

We have given some time intervals with corresponding position values, and we have to find the average velocity in each interval.Here is the given data:Time (s)Position (m)111.0111.0141.0281.041

Average velocity is the displacement per unit time, i.e., (final position - initial position) / (final time - initial time).We need to find the average velocity in each interval:(a) [1,2]Average velocity = (1.011 - 1.011) / (2 - 1) = 0m/s(b) [1,1.01]Average velocity = (1.011 - 1.011) / (1.01 - 1) = 0m/s(c) [1.01,4]

velocity = (1.028 - 1.011) / (4 - 1.01) = 0.006m/s(d) [1,100]Average velocity = (1.041 - 1.011) / (100 - 1) = 0.0003m/s

Therefore, the average velocity during the time intervals [1,2], [1,1.01], [1.01,4], and [1,100] are 0 m/s, 0 m/s, 0.006 m/s, and 0.0003 m/s respectively.

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please help to solve the question
3. Consider the following data set: \[ 2,3,3,4,4,5,7,8,9,10,10,12,13,15,20,22,25,27,29,32,34,36,39,40,43,45,57,59,63,65 \] What is the percentile rank for the number 43 ? Show calculations.

Answers

The percentile rank for the number 43 in the given data set is approximately 85.

To calculate the percentile rank for the number 43 in the given data set, we can use the following formula:

Percentile Rank = (Number of values below the given value + 0.5) / Total number of values) * 100

First, we need to determine the number of values below 43 in the data set. Counting the values, we find that there are 25 values below 43.

Next, we calculate the percentile rank:

Percentile Rank = (25 + 0.5) / 30 * 100

              = 25.5 / 30 * 100

              ≈ 85

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A proposed bus fare would charge Php 11.00 for the first 5 kilometers of travel and Php 1.00 for each additional kilometer over the proposed fare. Find the proposed fare for a distance of 28 kilometer

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If a proposed bus fare would charge Php 11.00 for the first 5 kilometers of travel and Php 1.00 for each additional kilometer over the proposed fare, then the proposed fare for a distance of 28 kilometers is Php 34.

To find the proposed fare for a distance of 28 kilometers, follow these steps:

We know that the fare for the first 5 kilometers is Php 11.00. Therefore, the fare for the remaining 23 kilometers is: 23 x Php 1.00 = Php 23.00Hence, the total proposed fare for a distance of 28 kilometers would be the sum of fare for the first 5 kilometers and fare for the remaining 23 kilometers. Therefore, the proposed fare would be Php 11.00 + Php 23.00 = Php 34

Therefore, the proposed fare for a distance of 28 kilometers is Php 34.

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Q3
Find an equation of the line that contains the given pair of points. The equation of the line is (21,26),(2,7) (Simplify your answer. Type your answer in slope-intercept form.)

Answers

The equation of the line passing through the points (21, 26) and (2, 7) in slope-intercept form is y = (19/19)x + (7 - (19/19)2), which simplifies to y = x + 5.

To find the equation of the line, we can use the slope-intercept form of a linear equation, which is y = mx + b, where m represents the slope and b represents the y-intercept.

First, we need to find the slope (m) of the line. The slope is calculated using the formula: m = (y₂ - y₁) / (x₂ - x₁), where (x₁, y₁) and (x₂, y₂) are the coordinates of the two points on the line.

Let's substitute the coordinates (21, 26) and (2, 7) into the slope formula:

m = (7 - 26) / (2 - 21) = (-19) / (-19) = 1

Now that we have the slope (m = 1), we can find the y-intercept (b) by substituting the coordinates of one of the points into the slope-intercept form.

Let's choose the point (2, 7):

7 = (1)(2) + b

7 = 2 + b

b = 7 - 2 = 5

Finally, we can write the equation of the line in slope-intercept form:

y = 1x + 5

Therefore, the equation of the line that contains the given pair of points (21, 26) and (2, 7) is y = x + 5.

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The Foula for Force is F=ma, where F is the Force, m is the object's mass, and a is the object's acceleration. Rewrite the foula in tes of mass, then find the object's mass when it's acceleration is 14(m)/(s) and the total force is 126N

Answers

When the object's acceleration is 14 m/s and the total force is 126 N, the object's mass is approximately 9 kg.

To rewrite the formula F = ma in terms of mass (m), we can isolate the mass by dividing both sides of the equation by acceleration (a):

F = ma

Dividing both sides by a:

F/a = m

Therefore, the formula in terms of mass (m) is m = F/a.

Now, to find the object's mass when its acceleration is 14 m/s and the total force is 126 N, we can substitute the given values into the formula:

m = F/a

m = 126 N / 14 m/s

m ≈ 9 kg

Therefore, when the object's acceleration is 14 m/s and the total force is 126 N, the object's mass is approximately 9 kg.

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Kelsey bought 5(5)/(8) litres of milk and drank 1(2)/(7) litres of it. How much milk was left?

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After Kelsey bought 5(5)/(8) liters of milk and drank 1(2)/(7) liters, there was 27/56 liters of milk left.

To find out how much milk was left after Kelsey bought 5(5)/(8) liters and drank 1(2)/(7) liters, we need to subtract the amount of milk consumed from the initial amount.

The initial amount of milk Kelsey bought was 5(5)/(8) liters.

Kelsey drank 1(2)/(7) liters of milk.

To subtract fractions, we need to have a common denominator. The common denominator for 8 and 7 is 56.

Converting the fractions to have a denominator of 56:

5(5)/(8) liters = (5*7)/(8*7) = 35/56 liters

1(2)/(7) liters = (1*8)/(7*8) = 8/56 liters

Now, let's subtract the amount of milk consumed from the initial amount:

Amount left = Initial amount - Amount consumed

Amount left = 35/56 - 8/56

To subtract the fractions, we keep the denominator the same and subtract the numerators:

Amount left = (35 - 8)/56

Amount left = 27/56 liters

It's important to note that fractions can be simplified if possible. In this case, 27/56 cannot be simplified further, so it remains as 27/56. The answer is provided in fraction form, representing the exact amount of milk left.

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Construct a confidence interval for μ assuming that each sample is from a normal population. (a) x
ˉ
=28,σ=4,n=11,90 percentage confidence. (Round your answers to 2 decimal places.) (b) x
ˉ
=124,σ=8,n=29,99 percentage confidence. (Round your answers to 2 decimal places.)

Answers

The confidence interval in both cases has been constructed as:

a) (26.02, 29.98)

b) (120.17, 127.83)

How to find the confidence interval?

The formula to calculate the confidence interval is:

CI = xˉ ± z(σ/√n)

where:

xˉ is sample mean

σ is standard deviation

n is sample size

z is z-score at confidence level

a) xˉ = 28

σ = 4

n = 11

90 percentage confidence.

z at 90% CL = 1.645

Thus:

CI = 28 ± 1.645(4/√11)

CI = 28 ± 1.98

CI = (26.02, 29.98)

b) xˉ = 124

σ = 8

n = 29

90 percentage confidence.

z at 99% CL = 2.576

Thus:

CI = 124 ± 2.576(8/√29)

CI = 124 ± 3.83

CI = (120.17, 127.83)

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(t/f) if y is a linear combination of nonzero vectors from an orthogonal set, then the weights in the linear combination can be computed without row operations on a matrix.

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If y is a linear combination of nonzero vectors from an orthogonal set, then the weights in the linear combination can be computed without row operations on a matrix is a True statement.

In an orthogonal set of vectors, each vector is orthogonal (perpendicular) to all other vectors in the set.

Therefore, the dot product between any two vectors in the set will be zero.

Since the vectors are orthogonal, the weights in the linear combination can be obtained by taking the dot product of the given vector y with each of the orthogonal vectors and dividing by the squared magnitudes of the orthogonal vectors. This allows for a direct computation of the weights without the need for row operations on a matrix.

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Find the derivative of the following function.
h(x)= (4x²+5) (2x+2) /7x-9

Answers

The given function is h(x) = (4x² + 5)(2x + 2)/(7x - 9). We are to find its derivative.To find the derivative of h(x), we will use the quotient rule of differentiation.

Which states that the derivative of the quotient of two functions f(x) and g(x) is given by `(f'(x)g(x) - f(x)g'(x))/[g(x)]²`. Using the quotient rule, the derivative of h(x) is given by

h'(x) = `[(d/dx)(4x² + 5)(2x + 2)(7x - 9)] - [(4x² + 5)(2x + 2)(d/dx)(7x - 9)]/{(7x - 9)}²

= `[8x(4x² + 5) + 2(4x² + 5)(2)](7x - 9) - (4x² + 5)(2x + 2)(7)/{(7x - 9)}²

= `(8x(4x² + 5) + 16x² + 20)(7x - 9) - 14(4x² + 5)(x + 1)/{(7x - 9)}²

= `[(32x³ + 40x + 16x² + 20)(7x - 9) - 14(4x² + 5)(x + 1)]/{(7x - 9)}².

Simplifying the expression, we have h'(x) = `(224x⁴ - 160x³ - 832x² + 280x + 630)/{(7x - 9)}²`.

Therefore, the derivative of the given function h(x) is h'(x) = `(224x⁴ - 160x³ - 832x² + 280x + 630)/{(7x - 9)}²`.

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Belief in Haunted Places A random sample of 340 college students were asked if they believed that places could be haunted, and 133 responded yes. Estimate the true proportion of college students who believe in the possibility of haunted places with 95% confidence. According to Time magazine, 37% of Americans believe that places can be haunted. Round intermediate and final answers to at least three decimal places.

Answers

According to the given data, a random sample of 340 college students were asked if they believed that places could be haunted, and 133 responded yes.

The aim is to estimate the true proportion of college students who believe in the possibility of haunted places with 95% confidence. Also, it is given that according to Time magazine, 37% of Americans believe that places can be haunted.

The point estimate for the true proportion is:

P-hat = x/

nowhere x is the number of students who believe in the possibility of haunted places and n is the sample size.= 133/340

= 0.3912

The standard error of P-hat is:

[tex]SE = sqrt{[P-hat(1 - P-hat)]/n}SE

= sqrt{[0.3912(1 - 0.3912)]/340}SE

= 0.0307[/tex]

The margin of error for a 95% confidence interval is:

ME = z*SE

where z is the z-score associated with 95% confidence level. Since the sample size is greater than 30, we can use the standard normal distribution and look up the z-value using a z-table or calculator.

For a 95% confidence level, the z-value is 1.96.

ME = 1.96 * 0.0307ME = 0.0601

The 95% confidence interval is:

P-hat ± ME0.3912 ± 0.0601

The lower limit is 0.3311 and the upper limit is 0.4513.

Thus, we can estimate with 95% confidence that the true proportion of college students who believe in the possibility of haunted places is between 0.3311 and 0.4513.

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Fill in the blank. The​ ________ is the probability of getting a test statistic at least as extreme as the one representing the sample​ data, assuming that the null hypothesis is true.

A. ​p-value

B. Critical value

C. Level of significance

D. Sample proportion

Answers

The​ p-value is the probability of getting a test statistic at least as extreme as the one representing the sample​ data, assuming that the null hypothesis is true.

The p-value is the probability of obtaining a test statistic that is as extreme as, or more extreme than, the one observed from the sample data, assuming that the null hypothesis is true. It is a measure of the evidence against the null hypothesis provided by the data. The p-value is used in hypothesis testing to make decisions about the null hypothesis. If the p-value is less than the predetermined level of significance (alpha), typically 0.05, it suggests that the observed data is unlikely to occur by chance alone under the null hypothesis. This leads to rejecting the null hypothesis in favor of the alternative hypothesis. On the other hand, if the p-value is greater than the significance level, there is insufficient evidence to reject the null hypothesis.

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suppose you have a large box of pennies of various ages and plan to take a sample of 10 pennies. explain how you can estimate that probability that the range of ages is greater than 15 years.

Answers

To estimate the probability that the range of ages is greater than 15 years in a sample of 10 pennies, randomly select multiple samples, calculate the range for each sample, count the number of samples with a range greater than 15 years, and divide it by the total number of samples.

To estimate the probability that the range of ages among a sample of 10 pennies is greater than 15 years, you can follow these steps:

1. Determine the range of ages in the sample: Calculate the difference between the oldest and youngest age among the 10 pennies selected.

2. Repeat the sampling process: Randomly select multiple samples of 10 pennies from the large box and calculate the range of ages for each sample.

3. Record the number of samples with a range greater than 15 years: Count how many of the samples have a range greater than 15 years.

4. Estimate the probability: Divide the number of samples with a range greater than 15 years by the total number of samples taken. This will provide an estimate of the probability that the range of ages is greater than 15 years in a sample of 10 pennies.

Keep in mind that this method provides an estimate based on the samples taken. The accuracy of the estimate can be improved by increasing the number of samples and ensuring that the samples are selected randomly from the large box of pennies.

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jesse has three one gallon containers. The first one has (5)/(9 ) of a gallon of juice, the second has (1)/(9) gallon of juice and the third has (1)/(9) gallon of juice. How many gallons of juice does Jesse have

Answers

Jesse has (7)/(9) of a gallon of juice.

To solve the problem, add the gallons of juice from the three containers.

Jesse has three one gallon containers with the following quantities of juice:

Container one = (5)/(9) of a gallon of juice

Container two = (1)/(9) gallon of juice

Container three = (1)/(9) gallon of juice

Add the quantities of juice from the three containers to get the total gallons of juice.

Juice in container one = (5)/(9)

Juice in container two = (1)/(9)

Juice in container three = (1)/(9)

Total juice = (5)/(9) + (1)/(9) + (1)/(9) = (7)/(9)

Therefore, Jesse has (7)/(9) of a gallon of juice.

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These data sets show the ages of students in two college classes. Class #1: 28,19,21,23,19,24,19,20 Class #2: 18,23,20,18,49,21,25,19 Which class would you expect to have the larger standa

Answers

To determine which class would have the larger standard deviation, we need to calculate the standard deviation for both classes.

First, let's calculate the standard deviation for Class #1:
1. Find the mean (average) of the data set: (28 + 19 + 21 + 23 + 19 + 24 + 19 + 20) / 8 = 21.125
2. Subtract the mean from each data point and square the result:
(28 - 21.125)^2 = 45.515625
(19 - 21.125)^2 = 4.515625
(21 - 21.125)^2 = 0.015625
(23 - 21.125)^2 = 3.515625
(19 - 21.125)^2 = 4.515625
(24 - 21.125)^2 = 8.015625
(19 - 21.125)^2 = 4.515625
(20 - 21.125)^2 = 1.265625
3. Find the average of these squared differences: (45.515625 + 4.515625 + 0.015625 + 3.515625 + 4.515625 + 8.015625 + 4.515625 + 1.265625) / 8 = 7.6015625
4. Take the square root of the result from step 3: sqrt(7.6015625) ≈ 2.759

Next, let's calculate the standard deviation for Class #2:
1. Find the mean (average) of the data set: (18 + 23 + 20 + 18 + 49 + 21 + 25 + 19) / 8 = 23.125
2. Subtract the mean from each data point and square the result:
(18 - 23.125)^2 = 26.015625
(23 - 23.125)^2 = 0.015625
(20 - 23.125)^2 = 9.765625
(18 - 23.125)^2 = 26.015625
(49 - 23.125)^2 = 670.890625
(21 - 23.125)^2 = 4.515625
(25 - 23.125)^2 = 3.515625
(19 - 23.125)^2 = 17.015625
3. Find the average of these squared differences: (26.015625 + 0.015625 + 9.765625 + 26.015625 + 670.890625 + 4.515625 + 3.515625 + 17.015625) / 8 ≈ 106.8359375
4. Take the square root of the result from step 3: sqrt(106.8359375) ≈ 10.337

Comparing the two standard deviations, we can see that Class #2 has a larger standard deviation (10.337) compared to Class #1 (2.759). Therefore, we would expect Class #2 to have the larger standard deviation.

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Can you please answer these questions?
1. Enzo is distributing the snacks at snack-time at a day-care. There are 11 kids attending today. Enzo has 63 carrot sticks, which the kids love. (They call them orange hard candy!)
Wanting to make sure every kid gets at least 5 carrot sticks, how many ways could Enzo hand them out?
2. How many 3-digit numbers must you have to be sure there are 2 summing to exactly 1002?
3. Find the co-efficient of x^6 in (x−2)^9?

Answers

The coefficient of x^6 is given by the term C(9, 6) * x^3 * (-2)^6.

Therefore, the coefficient of x^6 in (x - 2)^9 is 84.

To distribute the carrot sticks in a way that ensures every kid gets at least 5 carrot sticks, we can use the stars and bars combinatorial technique. Let's represent the carrot sticks as stars (*) and use bars (|) to separate the groups for each kid.

We have 63 carrot sticks to distribute among 11 kids, ensuring each kid gets at least 5. We can imagine that each kid is assigned 5 carrot sticks initially, which leaves us with 63 - (11 * 5) = 8 carrot sticks remaining.

Now, we need to distribute these remaining 8 carrot sticks among the 11 kids. Using stars and bars, we have 8 stars and 10 bars (representing the divisions between the kids). We can arrange these stars and bars in (8+10) choose 10 = 18 choose 10 ways.

Therefore, there are 18 choose 10 = 43758 ways for Enzo to hand out the carrot sticks while ensuring each kid gets at least 5.

To find the number of 3-digit numbers needed to ensure that there are 2 numbers summing to exactly 1002, we can approach this problem using the Pigeonhole Principle.

The largest 3-digit number is 999, and the smallest 3-digit number is 100. To achieve a sum of 1002, we need the smallest number to be 999 (since it's the largest) and the other number to be 3.

Now, we can start with the smallest number (100) and add 3 to it repeatedly until we reach 999. Each time we add 3, the sum increases by 3. The total number of times we need to add 3 can be calculated as:

(Number of times to add 3) * (3) = 999 - 100

Simplifying this equation:

(Number of times to add 3) = (999 - 100) / 3

= 299

Therefore, we need to have at least 299 three-digit numbers to ensure there are 2 numbers summing to exactly 1002.

To find the coefficient of x^6 in the expansion of (x - 2)^9, we can use the Binomial Theorem. According to the theorem, the coefficient of x^k in the expansion of (a + b)^n is given by the binomial coefficient C(n, k), where

C(n, k) = n! / (k! * (n - k)!).

In this case, we have (x - 2)^9. Expanding this using the Binomial Theorem, we get:

(x - 2)^9 = C(9, 0) * x^9 * (-2)^0 + C(9, 1) * x^8 * (-2)^1 + C(9, 2) * x^7 * (-2)^2 + ... + C(9, 6) * x^3 * (-2)^6 + ...

The coefficient of x^6 is given by the term C(9, 6) * x^3 * (-2)^6. Calculating this term:

C(9, 6) = 9! / (6! * (9 - 6)!)

= 84

Therefore, the coefficient of x^6 in (x - 2)^9 is 84.

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I am thinking of a number. When you divide it by n it leaves a remainder of n−1, for n=2,3,4, 5,6,7,8,9 and 10 . What is my number?

Answers

The number you are thinking of is 2521.

We are given that when the number is divided by n, it leaves a remainder of n-1 for n = 2, 3, 4, 5, 6, 7, 8, 9, and 10.

To find the number, we can use the Chinese Remainder Theorem (CRT) to solve the system of congruences.

The system of congruences can be written as:

x ≡ 1 (mod 2)

x ≡ 2 (mod 3)

x ≡ 3 (mod 4)

x ≡ 4 (mod 5)

x ≡ 5 (mod 6)

x ≡ 6 (mod 7)

x ≡ 7 (mod 8)

x ≡ 8 (mod 9)

x ≡ 9 (mod 10)

Using the CRT, we can find a unique solution for x modulo the product of all the moduli.

To solve the system of congruences, we can start by finding the solution for each pair of congruences. Then we combine these solutions to find the final solution.

By solving each pair of congruences, we find the following solutions:

x ≡ 1 (mod 2)

x ≡ 2 (mod 3) => x ≡ 5 (mod 6)

x ≡ 5 (mod 6)

x ≡ 3 (mod 4) => x ≡ 11 (mod 12)

x ≡ 11 (mod 12)

x ≡ 4 (mod 5) => x ≡ 34 (mod 60)

x ≡ 34 (mod 60)

x ≡ 6 (mod 7) => x ≡ 154 (mod 420)

x ≡ 154 (mod 420)

x ≡ 7 (mod 8) => x ≡ 2314 (mod 3360)

x ≡ 2314 (mod 3360)

x ≡ 8 (mod 9) => x ≡ 48754 (mod 30240)

x ≡ 48754 (mod 30240)

x ≡ 9 (mod 10) => x ≡ 2521 (mod 30240)

Therefore, the solution for the system of congruences is x ≡ 2521 (mod 30240).

The smallest positive solution within this range is x = 2521.

So, the number you are thinking of is 2521.

The number you are thinking of is 2521, which satisfies the given conditions when divided by n for n = 2, 3, 4, 5, 6, 7, 8, 9, and 10 with a remainder of n-1.

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If f(x) = 4x (sin x+cos x), find
f'(x) =
f'(1) =​

Answers

Therefore, f'(1) = 8 cos 1.Therefore, f'(x) = (4 + 4x) cos x + (4 - 4x) sin x.

Given that f(x) = 4x (sin x + cos x)

To find: f'(x) = , f'(1)

=​f(x)

= 4x (sin x + cos x)

Taking the derivative of f(x) with respect to x, we get;

f'(x) = (4x)' (sin x + cos x) + 4x [sin x + cos x]

'f'(x) = 4(sin x + cos x) + 4x (cos x - sin x)

f'(x) = 4(cos x + sin x) + 4x cos x - 4x sin x

f'(x) = 4 cos x + 4x cos x + 4 sin x - 4x sin x

f'(x) = (4 + 4x) cos x + (4 - 4x) sin x

Therefore, f'(x) = (4 + 4x) cos x + (4 - 4x) sin x.

Using the chain rule, we can find the derivative of f(x) with respect to x as shown below:

f(x) = 4x (sin x + cos x)

f'(x) = 4 (sin x + cos x) + 4x (cos x - sin x)

f'(x) = 4 cos x + 4x cos x + 4 sin x - 4x sin x

The answer is: f'(x) = 4 cos x + 4x cos x + 4 sin x - 4x sin x.

To find f'(1), we substitute x = 1 in f'(x)

f'(1) = 4 cos 1 + 4(1) cos 1 + 4 sin 1 - 4(1) sin 1

f'(1) = 4 cos 1 + 4 cos 1 + 4 sin 1 - 4 sin 1

f'(1) = 8 cos 1 - 0 sin 1

f'(1) = 8 cos 1

Therefore, f'(1) = 8 cos 1.

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Test the claim that the mean GPA of night students is smaller than 2.3 at the 0.10 significance level.
Based on a sample of 39 people, the sample mean GPA was 2.28 with a standard deviation of 0.14
The p-value is: __________ (to 3 decimal places)
The significance level is: ____________ ( to 2 decimal places)

Answers

The p-value of the test is given as follows:

0.19.

The significance level is given as follows:

0.10.

As the p-value is greater than the significance level, there is not enough evidence to conclude that the mean GPA of night students is smaller than 2.3 at the 0.10 significance level.

How to obtain the p-value?

The equation for the test statistic is given as follows:

[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]

In which:

[tex]\overline{x}[/tex] is the sample mean.[tex]\mu[/tex] is the value tested at the null hypothesis.s is the standard deviation of the sample.n is the sample size.

The parameters for this problem are given as follows:

[tex]\overline{x} = 2.28, \mu = 2.3, s = 0.14, n = 39[/tex]

Hence the test statistic is given as follows:

[tex]t = \frac{2.28 - 2.3}{\frac{0.14}{\sqrt{39}}}[/tex]

t = -0.89.

The p-value of the test is found using a t-distribution calculator, with a left-tailed test, 39 - 1 = 38 df and t = -0.89, hence it is given as follows:

0.19.

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How patriotic are you? Would you say extremely patriotic, very patriotic, somewhat patriotic, or not especially patriotic? Below is the data from Gallup polls that asked this question of a random sample of U.S. adults in 1999 and a second independent random sample in 2010. We conducted a chi-square test of homogeneity to determine if there are statistically significant differences in the distribution of responses for these two years. In this results table, the observed count appears above the expected count in each cell. 1999 994 extremely patriotic very patriotic somewhat patriotic not especially patriotic Total 193 466 284 257.2 443.8 237.3 55.72 324 426 193 611004 259.8 448.2 239.7 517 892 477 112 1998 2010 56.28 Total Chi-Square test: Statistic DF Value P-value Chi-square 3 53.19187) <0.0001 If we included an exploratory data analysis with the test of homogeneity, the percentages most appropriate as part of this analysis for the Extremely Patriotic group are

a. 193/1517 compared to 994/1998 b. 193/1998 compared to 324/1998 c. 193/517 compared to 324/517 d. 193/994 compared to 324/1004

Answers

The appropriate percentages for the Extremely Patriotic group are 19.42% in 1999 and 32.27% in 2010, corresponding to option d: 193/994 compared to 324/1004.

To calculate the appropriate percentages for the Extremely Patriotic group, we need to compare the counts from the 1999 and 2010 samples.

In 1999:

Number of Extremely Patriotic responses: 193

Total number of respondents: 994

In 2010:

Number of Extremely Patriotic responses: 324

Total number of respondents: 1004

Now we can calculate the percentages:

Percentage for 1999: (193 / 994) × 100 = 19.42%

Percentage for 2010: (324 / 1004) × 100 = 32.27%

Therefore, the appropriate percentages as part of the exploratory data analysis for the Extremely Patriotic group are:

19.42% compared to 32.27% (option d: 193/994 compared to 324/1004).

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square room is covered by a number of whole rectangular slabs of sides Calculate the least possible area of the room in square metres (3mks )

Answers

The least possible area of the room in square metres is Nlw, where N is the smallest integer that satisfies the equation LW = Nlw.

Let the length, width, and height of the square room be L, W, and H, respectively. Let the length and width of each rectangular slab be l and w, respectively. Then, the number of slabs required to cover the area of the room is given by:

Number of Slabs = (LW)/(lw)

Since we want to find the least possible area of the room, we can minimize LW subject to the constraint that the number of slabs is an integer. To do so, we can use the method of Lagrange multipliers:

We want to minimize LW subject to the constraint f(L,W) = (LW)/(lw) - N = 0, where N is a positive integer.

The Lagrangian function is then:

L(L,W,λ) = LW + λ[(LW)/(lw) - N]

Taking partial derivatives with respect to L, W, and λ and setting them to zero yields:

∂L/∂L = W + λW/l = 0

∂L/∂W = L + λL/w = 0

∂L/∂λ = (LW)/(lw) - N = 0

Solving these equations simultaneously, we get:

L = sqrt(N)l

W = sqrt(N)w

Therefore, the least possible area of the room is:

LW = Nlw

where N is the smallest integer that satisfies this equation.

In other words, the area of the room is a multiple of the area of each slab, and the least possible area of the room is obtained when the room dimensions are integer multiples of the slab dimensions.

Therefore, the least possible area of the room in square metres is Nlw, where N is the smallest integer that satisfies the equation LW = Nlw.

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For a fixed integer n≥0, denote by P n

the set of all polynomials with degree at most n. For each part, determine whether the given function is a linear transformation. Justify your answer using either a proof or a specific counter-example. (a) The function T:R 2
→R 2
given by T(x 1

,x 2

)=(e x 1

,x 1

+4x 2

). (b) The function T:P 5

→P 5

given by T(f(x))=x 2
dx 2
d 2

(f(x))+4f(x)=x 2
f ′′
(x)+4f(x). (c) The function T:P 2

→P 4

given by T(f(x))=(f(x+1)) 2
.

Answers

a. T: R^2 → R^2 is not a linear transformation. b. T: P^5 → P^5 is not a linear transformation. c. T: P^2 → P^4 given by T(f(x)) = (f(x + 1))^2 is a linear transformation.

(a) The function T: R^2 → R^2 given by T(x₁, x₂) = (e^(x₁), x₁ + 4x₂) is **not a linear transformation**.

To show this, we need to verify two properties for T to be a linear transformation: **additivity** and **homogeneity**.

Let's consider additivity first. For T to be additive, T(u + v) should be equal to T(u) + T(v) for any vectors u and v. However, in this case, T(x₁, x₂) = (e^(x₁), x₁ + 4x₂), but T(x₁ + x₁, x₂ + x₂) = T(2x₁, 2x₂) = (e^(2x₁), 2x₁ + 8x₂). Since (e^(2x₁), 2x₁ + 8x₂) is not equal to (e^(x₁), x₁ + 4x₂), the function T is not additive, violating one of the properties of a linear transformation.

Next, let's consider homogeneity. For T to be homogeneous, T(cu) should be equal to cT(u) for any scalar c and vector u. However, in this case, T(cx₁, cx₂) = (e^(cx₁), cx₁ + 4cx₂), while cT(x₁, x₂) = c(e^(x₁), x₁ + 4x₂). Since (e^(cx₁), cx₁ + 4cx₂) is not equal to c(e^(x₁), x₁ + 4x₂), the function T is not homogeneous, violating another property of a linear transformation.

Thus, we have shown that T: R^2 → R^2 is not a linear transformation.

(b) The function T: P^5 → P^5 given by T(f(x)) = x²f''(x) + 4f(x) is **not a linear transformation**.

To prove this, we again need to check the properties of additivity and homogeneity.

Considering additivity, we need to show that T(f(x) + g(x)) = T(f(x)) + T(g(x)) for any polynomials f(x) and g(x). However, T(f(x) + g(x)) = x²(f''(x) + g''(x)) + 4(f(x) + g(x)), while T(f(x)) + T(g(x)) = x²f''(x) + 4f(x) + x²g''(x) + 4g(x). These two expressions are not equal, indicating that T is not additive and thus not a linear transformation.

For homogeneity, we need to show that T(cf(x)) = cT(f(x)) for any scalar c and polynomial f(x). However, T(cf(x)) = x²(cf''(x)) + 4(cf(x)), while cT(f(x)) = cx²f''(x) + 4cf(x). Again, these two expressions are not equal, demonstrating that T is not homogeneous and therefore not a linear transformation.

Hence, we have shown that T: P^5 → P^5 is not a linear transformation.

(c) The function T: P^2 → P^4 given by T(f(x)) = (f(x + 1))^2 is **a linear transformation**.

To prove this, we need to confirm that T satisfies both additivity and homogeneity.

For additivity, we need to show that T(f(x) + g(x)) = T(f(x)) + T

(g(x)) for any polynomials f(x) and g(x). Let's consider T(f(x) + g(x)). We have T(f(x) + g(x)) = [(f(x) + g(x) + 1))^2 = (f(x) + g(x) + 1))^2 = (f(x + 1) + g(x + 1))^2. Expanding this expression, we get (f(x + 1))^2 + 2f(x + 1)g(x + 1) + (g(x + 1))^2.

Now, let's look at T(f(x)) + T(g(x)). We have T(f(x)) + T(g(x)) = (f(x + 1))^2 + (g(x + 1))^2. Comparing these two expressions, we see that T(f(x) + g(x)) = T(f(x)) + T(g(x)), which satisfies additivity.

For homogeneity, we need to show that T(cf(x)) = cT(f(x)) for any scalar c and polynomial f(x). Let's consider T(cf(x)). We have T(cf(x)) = (cf(x + 1))^2 = c^2(f(x + 1))^2.

Now, let's look at cT(f(x)). We have cT(f(x)) = c(f(x + 1))^2 = c^2(f(x + 1))^2. Comparing these two expressions, we see that T(cf(x)) = cT(f(x)), which satisfies homogeneity.

Thus, we have shown that T: P^2 → P^4 given by T(f(x)) = (f(x + 1))^2 is a linear transformation.

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The Munks agreed to monthly payments rounded up to the nearest $100 on a mortgage of $175000 amortized over 15 years. Interest for the first five years was 6.25% compounded semiannually. After 60 months, as permitted by the mortgage agreement, the Munks increased the rounded monthly payment by 10%. 1. a) Determine the mortgage balance at the end of the five-year term.(Points =4 )
2. b) If the interest rate remains unchanged over the remaining term, how many more of the increased payments will amortize the mortgage balance?(Points=4) 3. c) How much did the Munks save by exercising the increase-in-payment option?(Points=4.5)

Answers

The Munks saved $4444 by exercising the increase-in-payment option.

a) The first step is to compute the payment that would be made on a $175000 15-year loan at 6.25 percent compounded semi-annually over five years. Using the formula:

PMT = PV * r / (1 - (1 + r)^(-n))

Where PMT is the monthly payment, PV is the present value of the mortgage, r is the semi-annual interest rate, and n is the total number of periods in months.

PMT = 175000 * 0.03125 / (1 - (1 + 0.03125)^(-120))

= $1283.07

The Munks pay $1300 each month, which is rounded up to the nearest $100. At the end of five years, the mortgage balance will be $127105.28.
b) Over the remaining 10 years of the mortgage, the balance of $127105.28 will be amortized with payments of $1430 each month. The Munks pay an extra $130 per month, which is 10% of their new payment.

The additional $130 per month will be amortized by the end of the mortgage term.
c) Without the increase-in-payment option, the Munks would have paid $1283.07 per month for the entire 15-year term, for a total of $231151.20. With the increase-in-payment option, they paid $1300 per month for the first five years and $1430 per month for the remaining ten years, for a total of $235596.00.

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Find dy/dx for the following function, and place your answer in the box below: x^3+xe^y=2√ y+y^2

Answers

The derivative dy/dx of the function x^3 + xe^y = 2√(y + y^2) is (3x^2 + e^y) / (xe^y - 2y - 1).

To find dy/dx for the given function x^3 + xe^y = 2√(y + y^2), we differentiate both sides of the equation with respect to x using the chain rule and product rule.

Differentiating x^3 + xe^y with respect to x, we obtain 3x^2 + e^y + xe^y * dy/dx.

Differentiating 2√(y + y^2) with respect to x, we have 2 * (1/2) * (2y + 1) * dy/dx.

Setting the two derivatives equal to each other, we get 3x^2 + e^y + xe^y * dy/dx = (2y + 1) * dy/dx.

Rearranging the equation to solve for dy/dx, we have dy/dx = (3x^2 + e^y) / (xe^y - 2y - 1).

Therefore, the derivative dy/dx of the function x^3 + xe^y = 2√(y + y^2) is (3x^2 + e^y) / (xe^y - 2y - 1).

To find the derivative dy/dx for the given function x^3 + xe^y = 2√(y + y^2), we need to differentiate both sides of the equation with respect to x. This can be done using the chain rule and product rule of differentiation.

Differentiating x^3 + xe^y with respect to x involves applying the product rule. The derivative of x^3 is 3x^2, and the derivative of xe^y is xe^y * dy/dx (since e^y is a function of y, we multiply by the derivative of y with respect to x, which is dy/dx).

Next, we differentiate 2√(y + y^2) with respect to x using the chain rule. The derivative of √(y + y^2) is (1/2) * (2y + 1) * dy/dx (applying the chain rule by multiplying the derivative of the square root function by the derivative of the argument inside, which is y).

Setting the derivatives equal to each other, we have 3x^2 + e^y + xe^y * dy/dx = (2y + 1) * dy/dx.

To solve for dy/dx, we rearrange the equation, isolating dy/dx on one side:

dy/dx = (3x^2 + e^y) / (xe^y - 2y - 1).

Therefore, the derivative dy/dx of the function x^3 + xe^y = 2√(y + y^2) is (3x^2 + e^y) / (xe^y - 2y - 1).

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In 2020 tuition was 21337$ a year with 13$ minimum wage..What is the average rate of change in tuition .when adjusted for inflation?.What is the average rate of change in the minimum wage when adjusted for inflation?.How many hours would someone have to work on minimum wage to pay tuition in 1976 vs 2020?.If tuition had not changed, how many hours would someone have to work on present day minimum wage?.If we were to graph tuition and minimum wage, would these constitute a function?.If not, then why?.If so, what would the domain be and possible outputs? Give an example of a value not in the domain and another that is not in the range. Draw Lewis structures for each of the following. Please make sure your document is neat; please also make sure that all of the chemical symbols are correct, and the electrons can be clearly seen. Upload your document when complete. 1. PBr3 2. NyH2 3. C2H2 4. N 5. NCI a model scale is 1 in. = 1.5 ft. if the actual object is 18 feet, how long is the model? a) 12 inches b) 16 inches c) 24 inches d) 27 inches Recently, Yanni has found his workouts to be too easy. He really wants to push his body to its limits to discover what it can do. Which method would be MOST appropriate for him to incorporate in this scenario? After careful assessment, it was determined that your school is located in an area that is prone to flooding and that the buildings structure cannot withstand an earthquake. Utilizing your knowledge of risk management and disaster mitigation and prevention, discuss the ways in which the school can mitigate against the above mentioned hazards. signment: Chapter 02 Using Financial Statements and Budgets Swip and lackie Garcia have been mamed for over two years. They have been trying to save for a down payment on a house, but they feel that there is never anything left over to save. They talked with their parents who suggested that they prepare an income and expense statement. Skip and Jacke put together the following worksheet The following is soene additional infermation. - For income tems, they knew yearly amounts. For expenses, in soene cases they knew the annal amount (for example, jackie's commiter train pass). They listed those amousts and noted them as such. - Unfortunately, for other expenses, they provided tast month's payment, stating that the amount diant change much from month to monthi. - Based an their enpioyonprovided health insurance, they incur a ceoay of 335 per visit. They ectimated that they savin doctor (combined) 7 times during the year. - lackle nstumated that she spends about $175 a month on cothing and $200 a moren en groceries. - They went en a worwegian Grulse trip that cost 32,200 . They took an addalenal 8500 in spend no money that was their enly vacaton. signment Chapter 02 Using Financial Statements and Budgets Prepare an annual income and expense statement for the Garcias as of December 31 . Round all dollar amounts to the nearest whole dofar, and do not include an item in another category if a category aiready exists for it. If an item does not reguine a baiance, be sure to insert a zero to avold a loss of crei. manta Fhantae os theinm Cinanrial Cratemante and Pisdeate Bonita Industries budgets on an annual basis for its fiscal year. The following beginning and ending inventory levels are planned for the fiscal year of July 1, 2021 to June 30, 2022: Three kilos of raw materials are needed to produce each unit of finished product. If Bonita Industries plans to produce 570000 units during the 2021-2022 fiscal year, how many kilos of materials will the company need to purchase for its production during the year? 1709000 1711000 1710000 1714000 what device is used inside the dome of the pantheon to lighten its weight? If a price change causes total revenue to change in the opposite direction, demand is elastic inelastic perfect inelastic unitary elastic Question 17 (1 point) Compared to coffee, we would expect the cross elasticity of demand for tea to be negative, but positive for cream. tea to be positive, but negative for cream. both tea and cream to be positive. both tea and cream to be negative. The survey has bias. (a) Determine the type of bias. (b) Suggest a remedy. A poliing organization conducts a study to estimate the percentage of households that have pets. It mails a questionnaire to 1555 randomly selected households across the country and asks the head of each household if he or she has pets. Of the 1555 households selected, 50 responded. (a) Which of these best describos the blas in the survoy? Sampling bias Response bias Nonresponse biass Undercoverage blas (b) How can the bias be remedied? The survey has bias. (a) Determine the type of bias. (b) Suggest a remedy. A polling organization conducts a study to estimate the percentage of households that have pets. It mails a questionnaire to 1555 randomly selected households across the country and asks the head of each household if he or she has pets. Of the 1555 households selected, 50 responded. Underopverage bias (b) How can the blas be remedied? A. The polling organization should mail the questionnaire to each person in the households.