A student who has created a linear model is disappointed to find that herR2 value is a very low 13%. a) Does this mean that a linear model is not appropriate? Explain. b) Does this model allow the student to make accurate predictions? Explain.

a) The surface area, S, as a function of radius r is given by S = 2πrh + 2πr².

b) The value of S (surface area) approaches infinity as r approaches infinity.

a) The surface area of a closed cylindrical can consists of two parts: the lateral surface area and the two circular base areas.

Let's denote the height of the cylinder as 'h'.

The volume V of the cylinder is given by:

V = πr²h.

We want to express the surface area S in terms of the radius r.

The lateral surface area is given by:

Lateral Surface Area = 2πrh.

Each circular base has an area of πr².

So, the total surface area S is:

S = 2πrh + 2πr².

Hence,  the total surface area S is S = 2πrh + 2πr².

b) As r approaches infinity, let's analyze the behavior of the surface area S:

S = 2πrh + 2πr².

As r gets larger and larger, the term 2πr² dominates the expression.

This is because the term 2πrh (the lateral surface area) is proportional to both 'r' and 'h', whereas the term 2πr² (the circular base areas) is solely proportional to 'r²'.

So, as 'r' approaches infinity, the value of 2πr² will also approach infinity, and this will greatly outweigh the influence of the term 2πrh.

Therefore, the value of S will tend toward infinity as 'r' approaches infinity.

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a) S(r) = 2πr^2 + (2V/r)

b) As r→∞, S(r)→∞, the surface area increases without bound, so it approaches infinity.

a) To find the surface area, S, of a closed cylindrical can as a function of its radius r, you can use the formula for the surface area of a closed cylinder:

S = 2πr^2 + 2πrh

Where:

S is the surface area,

π (pi) is a mathematical constant approximately equal to 3.14159,

r is the radius of the cylinder,

and h is the height (or length) of the cylinder.

However, you mentioned that the volume V is fixed, so we can express the height (h) of the cylinder in terms of its radius (r) and fixed volume (V). The volume of a cylinder is given by:

V = πr^2h

Solving for h:

h = V / (πr^2)

Now, substitute this expression for h into the formula for the surface area:

S = 2πr^2 + 2πr(V / (πr^2))

Simplify:

S = 2πr^2 + 2V/r

So, the surface area, S, as a function of the radius r for a closed cylindrical can with a fixed volume V is:

S(r) = 2πr^2 + 2V/r

b) As r approaches infinity, let's analyze what happens to the value of S(r):

S(r) = 2πr^2 + 2V/r

The first term, 2πr^2, is a quadratic term in r. As r becomes very large (approaching infinity), this term dominates the expression, and S(r) grows without bound. In other words, it approaches infinity.

The second term, 2V/r, is inversely proportional to r. As r becomes very large, this term approaches zero. However, the first term dominates the behavior of S(r), so the overall behavior is dominated by the growth of the quadratic term.

So, as r approaches infinity, the value of S(r) approaches infinity (i.e., S(r) tends toward infinity).

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Answer:

Step-by-step explanation:

Triangle BCD is a right angle triangle.

From the given right angle triangle

BC represents the hypotenuse of the right angle triangle.

Taking 45 degrees as the reference angle,

BC represents the adjacent side of the right angle triangle.

BD represents the opposite side of the right angle triangle.

To determine BC, we would apply trigonometric ratio

Cos θ = adjacent side/hypotenuse side. Therefore,

Cos 45 = BC/2√2

√2/2 = BC/2√2

BC = 2√2 × √2/2

BC = 2

Answer:

e = 0.6342

S = 0.808

γsat = 126.207 lb/ft³

Step-by-step explanation:

Given

VT = 0.320 ft³

WT = 38.9 lb

w = 19.2% = 0.192

Gs = 2.67

then we apply

γ = WT / VT

γ = 38.9 lb / 0.320 ft³

γ = 121.5625 lb/ft³

then we get γdry as follows

γdry = γ / (1 + w)

γdry = 121.5625 lb/ft³ / (1 + 0.192)

γdry = 101.982 lb/ft³

the void ratio (e) can be obtained applying this equation

γdry = Gs*γw / (1 + e)

101.982 = 2.67*62.42 / (1 + e)

⇒ e = 0.6342

We can get the degree of saturation (S) as follows

S*e = Gs*w

S = Gs*w / e

S = 2.67*0.192 / 0.6342

S = 0.808

The saturated unit weight (γsat) will be obtained applying this formula

γsat = (Gs + e)*γw / (1 + e)

γsat = (2.67 + 0.6342)*62.42 lb/ft³/ (1 + 0.6342)

γsat = 126.207 lb/ft³

Answer:

the pair of equations y = 3, z = 7 represent the intersection of two plans, The set of points is (0,3,7) and the line is parallel to x axis.

Step-by-step explanation:

Consider the provided equation.

y=3 represents a vertical plane which is in xy plane.

Z=7 represents a horizontal plane which is parallel to xy plane

The both planes are perpendicular to each other and intersect.

y=3 and z=7 is the intersection of two plans, where the value of x is zero y=3 and z=7.

The set of points is (0,3,7) and the line is parallel to x axis.

Final answer:

The pair of equations y = 3 and z = 7 represents a set of points in three-dimensional space where the y-coordinate is always 3 and the z-coordinate is always 7.

Explanation:

The pair of equations y = 3 and z = 7 represents a set of points in three-dimensional space where the y-coordinate is always 3 and the z-coordinate is always 7. In other words, any point that satisfies both equations will have a y-value of 3 and a z-value of 7, regardless of the x-coordinate.

Answer:

d. Cluster

Step-by-step explanation:

Random: Random is asking a group of people from a population. For example, to estimate the proportion of Buffalo residents who are Bills fans, you ask 100 Buffalo residents and estimate to the entire population.

Systematic: Similar to random. For example, you want to estimate something about a population, and your sample is every 5th people you see on the street.

Cluster:Divides the population into groups, with geographic characteristics.. Each element is the groups is used. Suppose you want to study the voting choices of Buffalo Bills players. You can divide into offense, defense and special teams, and ask each player of these 3 groups.

Stratified: Done on a group of clusters, that is, from each cluster(group), a number of people are selected.

In this problem, we have that:

All employees at three stores of a large retail chain were asked to fill out a survey.

Divided by clusters(stores).

So the orrect answer is:

d. Cluster

The type of sampling described in the question, where all employees at three selected stores are surveyed, is an example of cluster sampling. This method divides the population into groups, or 'clusters', and includes all members from selected clusters in the study.

The scenario described seems to be an example of cluster sampling. In cluster sampling, all subjects (in this case, employees at the stores) within selected groups (the three stores in this case) are studied. The selection is not random or systematic, but based on grouping. It is also not a stratified sample as we are not choosing samples proportionally from subgroups within the population.

In this case, the whole population is divided into groups (stores). Each group is called a cluster. All the members (employees) of the selected groups (stores) are included in the study. It is important to note that unlike stratification, where samples are taken from each group, in cluster sampling, we study the entire group.

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Answer:

c). Two tailed test

Step-by-step explanation:

The given hypothesis are

Null hypothesis: H0:μ= 1.7

Alternative hypothesis: H1:μ≠ 1.7

The alternative hypothesis demonstrates that  mean number of children are not 1.7 in 2000. This means that mean number of children can be greater than 1.7 or mean number of children can be less than 1.7. Thus, the given alternative hypothesis indicates the two tailed test.

Answer:

51

Step-by-step explanation:

Substitute the value of a

3(2)³+10(2)²-3(2)-7

3(8)+10(4)-3(2)-7

24+40-6-7

64-13

51

Answer:

a) [tex] \frac{ds}{dt}= v(t) = 3t^2 -18t +15[/tex]

b) [tex] v(t=3) = 3(3)^2 -18(3) +15=-12[/tex]

c) [tex] t =1s, t=5s[/tex]

d)  [tex] [0,1) \cup (5,\infty)[/tex]

e) [tex] D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46[/tex]

And we take the absolute value on the middle integral because the distance can't be negative.

f) [tex] a(t) = \frac{dv}{dt}= 6t -18[/tex]

g) The particle is speeding up [tex](1,3) \cup (5,\infty)[/tex]

And would be slowing down from [tex][0,1) \cup (3,5)[/tex]

Step-by-step explanation:

For this case we have the following function given:

[tex] f(t) = s = t^3 -9t^2 +15 t[/tex]

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

[tex] \frac{ds}{dt}= v(t) = 3t^2 -18t +15[/tex]

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

[tex] v(t=3) = 3(3)^2 -18(3) +15=-12[/tex]

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

[tex] 3t^2 -18 t +15 =0[/tex]

We can divide both sides of the equation by 3 and we got:

[tex] t^2 -6t +5=0[/tex]

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

[tex] (t-5)*(t-1) =0[/tex]

And for this case we got [tex] t =1s, t=5s[/tex]

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

[tex] t^2 -6t +5 >0[/tex]

[tex] (t-5) *(t-1)>0[/tex]

So then the intervals positive are [tex] [0,1) \cup (5,\infty)[/tex]

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

[tex] D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6[/tex]

And if we replace we got:

[tex] D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46[/tex]

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

[tex] a(t) = \frac{dv}{dt}= 6t -18[/tex]

Part g and h

The particle is speeding up [tex](1,3) \cup (5,\infty)[/tex]

And would be slowing down from [tex][0,1) \cup (3,5)[/tex]

b) The velocity after 3 seconds is -12 feet per second. c) The particle is at rest when t = 1 and t = 5. d) The particle is moving in the positive direction between the critical points of the velocity function. e) The total distance traveled during the first 6 seconds can be found by integrating the absolute value of the velocity function. f) The acceleration at time t is given by the derivative of the velocity function. h) The particle is speeding up when its acceleration is positive and slowing down when its acceleration is negative.

b) To find the velocity after 3 seconds, we need to find the derivative of the function f(t). The derivative of f(t) is v(t), the velocity function. So, v(t) = f'(t), which is equal to 3t^2 - 18t + 15. Now, to find v(3), we substitute t = 3 into the velocity function:

v(3) = 3(3)^2 - 18(3) + 15

= 27 - 54 + 15

= -12 feet per second

c) The particle is at rest when its velocity is zero. So, to find when the particle is at rest, we need to find the time when v(t) = 0:

0 = 3t^2 - 18t + 15

Solving this quadratic equation, we find that the particle is at rest when t = 1 and t = 5

d) The particle is moving in the positive direction when its velocity is positive. So, we need to find the time intervals when v(t) > 0. We can do this by finding the critical points of the velocity function and determining the sign of v(t) in between those critical points. By analyzing the sign of v(t), we can determine the intervals when the particle is moving in the positive direction.

(e) To find the total distance traveled during the first 6 seconds, we need to find the definite integral of the absolute value of the velocity function from 0 to 6:

Distance = ∫06 |v(t)| dt

(f) The derivative of the velocity function v(t) gives us the acceleration function a(t), which is the rate of change of velocity. So, a(t) = v'(t), which is equal to 6t - 18.

(h) The particle is speeding up when its acceleration is positive, and slowing down when its acceleration is negative. So, we need to find the intervals when a(t) > 0 and a(t) < 0 to determine when the particle is speeding up and when it is slowing down.

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To calculate the resultant force and its direction given two forces at angles, you decompose each force into x and y components, sum these components separately to find the resultant vector, and then use Pythagorean theorem and inverse tangent to find magnitude and direction.

To find the magnitude of the resultant force and the angle it makes with the positive x-axis, given forces at 45 degrees and 30 degrees below the x-axis with magnitudes 28 lb and 12 lb respectively, we break each force into its x and y components. For force a at 45 degrees, the components are 28cos(45) in the x-direction and 28sin(45) in the y-direction. For force b at -30 degrees, the components are 12cos(-30) in the x-direction and 12sin(-30) in the y-direction, since it is below the x-axis.

To find the resultant force (Fres), we add the x-components and y-components separately: Fres,x = 28cos(45) + 12cos(-30) and Fres,y = 28sin(45) + 12sin(-30). The total magnitude is calculated using the Pythagorean theorem: |Fres| = sqrt(Fres,x² + Fres,y²). The angle θ with the positive x-axis is found using the inverse tangent of the y-component over the x-component (θ = atan(Fres,y/Fres,x)).

Answer:

a ) 0.1403604645 and 0.1368

b) 0.3464961 and 0.3485

c) 0.802671982 and 0.8018

Step-by-step explanation:

Y~ B (15,0.45)

Y~ N (15*0.45, 15*0.45*0.55) = Y~ N (6.75, 3.7125)

a) P(Y=5) = 15C5 (0.45)^5 * (0.55)^10 = 0.1403604645

   For normal approximation

P(Y = 5 ) = P ( 4.5 < Y < 5.5 )  ......... continuity correction

Hence,

[tex]P ( 4.5 < Y < 5.5 ) = P ( \frac{4.5 - 6.75}{\sqrt{3.7125} } < Z < \frac{5.5 - 6.75}{\sqrt{3.7125} } ) = P ( -1.16775 < Z < -0.64875 )[/tex]

The probability P ( 4.5 < Y < 5.5 ) = 0.1368

b) P(Y>7) = 15C8 (0.45)^ 8 (0.55)^7 + 15C9 (0.45)^9 * (0.55)^6 + 15C10 (0.45)^10 * (0.55)^5 + 15C11 (0.45)^11 * (0.55)^4 + 15C12 (0.45)^12 * (0.55)^3 + 15C13 (0.45)^13 * (0.55)^2 + 15C14 (0.45)^14 * (0.55) + (0.45)^15

= 0.3464961

  For normal approximation

P(Y > 7 ) = P (Y > 7.5 )  ......... continuity correction

Hence,

[tex]P (Y > 7.5) = P (Z > \frac{7.5-6.75}{\sqrt{3.7125} } ) = P (Z > 0.389249)\\[/tex]

The probability P ( Y>7.5 ) = 0.3485

c) P (4 < Y < 10) = 15C5 (0.45)^5 (0.55)^10 + 15C6 (0.45)^ 6 (0.55)^9 + 15C7 (0.45)^7 (0.55)^8 + 15C8 (0.45)^ 8 (0.55)^7 + 15C9 (0.45)^9 * (0.55)^6

= 0.802671982

For normal approximation

P( 4 < Y < 10 ) = P (4.5< Y < 9.5 )  ......... continuity correction

Hence,

[tex]P ( 4.5 < Y < 9.5 ) = P ( \frac{4.5 - 6.75}{\sqrt{3.7125} } < Z < \frac{9.5 - 6.75}{\sqrt{3.7125} } ) = P ( -1.167748416 < Z < 1.427248064 )[/tex]

The probability P (4.5< Y < 9.5 ) = 0.8018

Answer:

Sample mean from population A has probably more accurate estimate of its population mean than the sample mean from population B.

Step-by-step explanation:

To yield a more accurate estimate of the population mean, margin of error should be minimized.

margin of error (ME) of the mean can be calculated using the formula

ME=[tex]\frac{z*s}{\sqrt{N} }[/tex] where

for a given confidence level, and the same standard deviation, as the sample size increases, margin of error decreases.

Thus, random sample of 50 people from population A, has smaller margin of error than the sample of 20 people from population B.

Therefore, sample mean from population A has probably more accurate estimate of its population mean than the sample mean from population B.

Final answer:

In statistics, a larger sample size generally leads to a more accurate estimate of the population mean.

Explanation:

Population A: Sample size = 50, SD = 6.0

Population B: Sample size = 20, SD = 6.0

In this case, the sample mean from Population A (larger sample size) will likely yield a more accurate estimate of its population mean compared to Population B (smaller sample size) due to the larger sample size resulting in a more reliable estimation.

Answer:

Step-by-step explanation:

The given simultaneous equations are expressed as

y = 2x + 7 - - - - - - - - - -1

y = -2x - 5 - - - - - - - - - - 2

We would apply the method of substitution.

The first step is to equate equation 1 to equation 2. It becomes

2x + 7 = - 2x - 5

Next step is to add 2x to the left hand side and the right hand side of the equation. It becomes

2x + 2x + 7 = - 2x - 2x - 5

4x + 7 = - 5

Next step is to subtract 7 from the left hand side and the right hand side of the equation. It becomes

4x + 7 - 7 = - 5 - 7

4x = - 12

Next step is to divide the left hand side and the right hand side of the equation by 4. It becomes

4x/4 = -12/4

x = - 3

Substituting x = - 3 into equation 1, it becomes

y = 2 × - 3 + 7 = - 6 + 7

y = 1

Answer:

0

Step-by-step explanation:

1/3m - 1 - 1/2n

1/3(21) - 1 - 1/2(12)

7 - 1 - 6 = 0

Answer:

Option d)  Is an alternative method to reducing fractions to the lowest terms.

Step-by-step explanation:

Cancellation Method:

Thus, cancellation method

Option d)  Is an alternative method to reducing fractions to the lowest terms.

The cancellation method is a technique used to reduce fractions to their lowest terms by dividing both the numerator and denominator by their greatest common factor, making option (d) the correct answer.

The cancellation method in mathematics is a technique used to simplify fractions by dividing both the numerator and denominator by their greatest common factor. The goal of this method is to reduce fractions to their lowest terms. In practice, this method involves identifying factors common to both the numerator and the denominator and 'canceling' them out by dividing each by that factor. The cancellation method does not raise fractions to the highest terms, nor does it involve multiplying a number evenly by the numerator and denominator; rather, it simplifies fractions.

The answer to the student's question is option (d), as the cancellation method is indeed an alternative method to reducing fractions to the lowest terms. When working with unit conversions, for example, it's essential to cancel units correctly to ensure the correct units in the final answer. This process requires attention to detail to avoid inverting ratios and ending up with incorrect units.

Answer:

Therefore, the final form

p+/-E = 0.666+/-0.333

Step-by-step explanation:

Given:

Confidence interval = 0.333 < p < 0.999

To express the confidence interval in the forn p+/-E, where;

p is the midpoint of the confidence interval

E is the error.

The midpoint of the confidence interval is

p = (0.333+0.999)/2 = 1.332/2

p = 0.666

The error can be calculated using the formula:

Error = interval width/2

E = (0.999-0.333)/2 = 0.666/2

E = 0.333

Therefore, the final form

p+/-E = 0.666+/-0.333

The confidence interval 0.333< p <0.999 is expressed in the form p ± E as 0.666 ± 0.333. The midpoint of the interval is calculated by adding the two bounds and dividing by 2, and the distance from this point to either end of interval is calculated by subtracting the lower limit from the midpoint.

The confidence interval 0.333< p <0.999 can be expressed in the form p ± E by calculating the middle point of the interval (p), and the distance from the middle point to either end of the interval (E). To calculate the midpoint, add the two bounds and divide the result by 2. Thus, p = (0.999 + 0.333) / 2 = 0.666. Then, calculate E by subtracting the lower limit from p. So, E = 0.666 - 0.333 = 0.333. So, the confidence interval can be written as p ± E, or 0.666 ± 0.333.

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Answer:

A. $45.83

Step-by-step explanation:

This problem can be solved by a simple rule of three.

Each year has 12 months. The company will pay $550 per year for your health insurance. This means that in 12 months, the company will pay $550.

What is the monthly amount they will pay?

This is how much they are going to pay in one month. So:

1 month - $x

12 months - $550

[tex]12x = 550[/tex]

[tex]x = \frac{550}{12}[/tex]

[tex]x = 45.83[/tex]

So the correct answer is:

A. $45.83

Answer:

Step-by-step explanation:

Let X be the time for any customer to call at any time to make reservation in Nite Time Inn.

Given that X is exponential with mean = 4 minutes

We are to find the probability

(a) 3 minutes or less.

=[tex]P(X\leq 3)=1-e^{-3/4} =0.5276[/tex]

(b) 4 minutes or less.

[tex]=P(X\leq 4)\\=1-e^{-4/4} =0.6321[/tex]

(c) 5 minutes or less.

[tex]=P(X\leq 5)\\=1-e^{-5/4} =0.7135[/tex]

(d) longer than 5 minutes.

=1-P(X≤5) = 0.2865

Answer:

Step-by-step explanation:

-2x + 7x + 5 = 0

-2x +7x = 5

5x = 5

divide both side by 5

5x/5 = 5/5

x = 1

Answer:

There is a 62% probability that the student will be awarded at least one of the two scholarships.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that the student gets a scolarship from Agency A.

B is the probability that the student gets a scolarship from Agency B.

We have that:

[tex]A = a + (A \cap B)[/tex]

In which a is the probability that the student will get an scolarship from agency A but not from agency B and [tex]A \cap B[/tex] is the probability that the student will get an scolarship from both agencies.

By the same logic, we have that:

[tex]B = b + (A \cap B)[/tex]

What is the probability that the student will be awarded at least one of the two scholarships?

This is

[tex]P = a + b + (A \cap B)[/tex]

We have that:

[tex]A = 0.55, B = 0.40[/tex]

If the student is awarded a scholarship from Agency A, the probability that the student will be awarded a scholarship from Agency B is 0.60.

This means that:

[tex]\frac{A \cap B}{A} = 0.6[/tex]

[tex]A \cap B = 0.6A = 0.6*0.55 = 0.33[/tex]

----------

[tex]A = a + (A \cap B)[/tex]

[tex]0.55 = a + 0.33[/tex]

[tex]a = 0.22[/tex]

--------

[tex]B = b + (A \cap B)[/tex]

[tex]0.40 = b + 0.33[/tex]

[tex]b = 0.07[/tex]

Answer:

[tex]P = a + b + (A \cap B) = 0.22 + 0.07 + 0.33 = 0.62[/tex]

There is a 62% probability that the student will be awarded at least one of the two scholarships.

The probability that the student will be awarded at least one of the two scholarships is 0.73.

To calculate the probability that the student will be awarded at least one of the two scholarships, we can use the following formula:

P(at least one scholarship) = 1 - P(no scholarships)

The probability that the student will not be awarded a scholarship from either agency is the product of the probability that the student will not be awarded a scholarship from Agency A and the probability that the student will not be awarded a scholarship from Agency B.

The probability that the student will not be awarded a scholarship from Agency A is

1 - 0.55 = 0.45.

The probability that the student will not be awarded a scholarship from Agency B is

1 - 0.40 = 0.60.

Therefore, the probability that the student will not be awarded a scholarship from either agency is

0.45 * 0.60 = 0.27.

Therefore, the probability that the student will be awarded at least one of the two scholarships is

1 - 0.27 = 0.73.

Another way to calculate this probability is to use the following formula:

P(at least one scholarship) = P(scholarship from Agency A) + P(scholarship from Agency B) - P(scholarship from both agencies)

We already know the probability that the student will be awarded a scholarship from each agency. The probability that the student will be awarded a scholarship from both agencies is the product of the probability that the student will be awarded a scholarship from Agency A and the probability that the student will be awarded a scholarship from Agency B given that the student was awarded a scholarship from Agency A.

The probability that the student will be awarded a scholarship from Agency B given that the student was awarded a scholarship from Agency A is 0.60.

Therefore, the probability that the student will be awarded a scholarship from both agencies is

0.55 * 0.60 = 0.33.

Therefore, the probability that the student will be awarded at least one of the two scholarships is

0.55 + 0.40 - 0.33 = 0.73.

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Answer:

a. Pie chart

This one is the correct option since we use a pie chart when we have categories in the data. And for this case we don't have any category defined at the begin so for this reason the pie chart would be not useful for this case.

Step-by-step explanation:

For this case our variable of interest is the average parking time of its students. And we have 280 values for these times. So then the variable of interest is quantitative.

Which of the following graphs should not be used to display information concerning the students parking times?

a. Pie chart

This one is the correct option since we use a pie chart when we have categories in the data. And for this case we don't have any category defined at the begin so for this reason the pie chart would be not useful for this case.

b.Stem-and-leaf display

That incorrect since the Stem and leaf plot is useful when we want to plot quantitative data.

c. Histogram

That incorrect since the Histogram is ideal when we want to plot quantitative data and analyze the distribution of the data.

d. Box plot

That incorrect since the Box plot is ideal and useful when we want to plot quantitative data and see central tendency measures.

Final answer:

A pie chart should not be used to display the average parking times of students because it cannot effectively represent the distribution or variability of parking times, unlike a histogram, box plot, or stem-and-leaf display.

Explanation:

The question revolves around determining which graph would be inappropriate for displaying the average parking times of students at a university. The options are: a Pie chart, a Stem-and-leaf display, a Histogram, and a Box plot. To assess the suitability of each graph, we need to understand what type of data we have and what information we wish to convey with the graph.

A pie chart is typically used to show parts of a whole. However, in this scenario, we're interested in the distribution of parking times, not how a single parking time compares as a fraction of the total. Therefore, a pie chart would not effectively represent the variability or distribution of parking times.

Both a histogram and a box plot are well-suited for displaying the distribution of quantitative data such as parking times, making them good choices for this context. A stem-and-leaf display is also a valid option, especially for providing a quick visual summary that includes actual data points, which can be useful for identifying specific patterns or outliers.

In conclusion, the pie chart is the graph that should not be used to display information concerning the students' parking times. It simply doesn't align with the objective of analyzing the distribution or average of the parking times, which is best visualized through the other mentioned graphical methods.

Answer:

D

Step-by-step explanation:

Answer:

d = (max distance / max time)t + 0.

Step-by-step explanation:

Earthquake travels 55km of rock in 25 seconds, the relationship between d, the distance traveled in km, and t, the time elapsed in seconds is:

d = (max distance / max time)t + 0

d = (55 km / 25 sec) * t

  = 2.2t

Answer:

2.2km/secs

Step-by-step explanation:

Speed = distance /time

= 55km/25secs

= 2.2km/secs

let P(-2,0) & Q(0,2)

slope = 2-0/0+2

= 1

the slope is 1 which will be 45°

Answer: slope = 1

Step-by-step explanation:

The formula for finding slope is given as :

slope = [tex]\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

[tex]x_{1}[/tex] = -5

[tex]x_{2}[/tex] = 3

[tex]y_{1}[/tex] = -3

[tex]y_{2}[/tex] = 5

substituting the values into the formula , we have :

slope = [tex]\frac{5-(-3)}{3-(-5)}[/tex]

slope = [tex]\frac{5+3}{3+5}[/tex]

slope = [tex]\frac{8}{8}[/tex] = 1

Therefore : the slope of the line is 1

Answer:

Therefore, Δx=5/n, when have n intervals.

Step-by-step explanation:

From exercise we have interval [0,5]. So the length of the given interval is     5-0=5. Since all intervals [x0,x1],[x1,x2],…,[xn−1,xn]  are equal in width.

We know that their width is Δx. We conclude that width of each subinterval Δx in terms of the number of subintervals n is equal 5/n.

Therefore, Δx=5/n, when have n intervals.

Answer:

[tex]z=\frac{(35621-33498)-0}{\sqrt{\frac{4310^2}{100}+\frac{4310^2}{120}}}}=3.637[/tex]  

[tex]p_v =P(z>3.637)=0.000138[/tex]  

Comparing the p value with the significance level [tex]\alpha=0.01[/tex] we see that [tex]p_v<<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for the US residents with High school diplome is significantly higher than those with GED.  

Step-by-step explanation:

Data given and notation

[tex]\bar X_{1}=35621[/tex] represent the mean for sample 1  

[tex]\bar X_{2}=33498[/tex] represent the mean for sample 2  

[tex]\sigma_{1}=4310[/tex] represent the population standard deviation for 1  

[tex]\sigma_{2}=4310[/tex] represent the population standard deviation for 2

[tex]n_{1}=100[/tex] sample size for the group 2  

[tex]n_{2}=120[/tex] sample size for the group 2  

[tex]\alpha=0.01[/tex] Significance level provided

z would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for US residents (sample 1) is higher than the mean for sample 2, the system of hypothesis would be :  

Null hypothesis:[tex]\mu_{1}-\mu_{2}\leq0[/tex]  

Alternative hypothesis:[tex]\mu_{1} - \mu_{2}> 0[/tex]  

We have the population standard deviation's, so for this case is better apply a z test to compare means, and the statistic is given by:  

[tex]z=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}[/tex] (1)  

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

With the info given we can replace in formula (1) like this:  

[tex]z=\frac{(35621-33498)-0}{\sqrt{\frac{4310^2}{100}+\frac{4310^2}{120}}}}=3.637[/tex]  

P value  

Since is a right tailed test the p value would be:  

[tex]p_v =P(z>3.637)=0.000138[/tex]  

Comparing the p value with the significance level [tex]\alpha=0.01[/tex] we see that [tex]p_v<<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for the US residents with High school diplome is significantly higher than those with GED.  

The equivalent statements to 'If the sky is not clear, then you don't see the stars' are 'You don't see the stars unless the sky is clear', 'Clear sky is necessary to see the stars', and 'You see the stars only if the sky is clear'. Other statements make assumptions that are not present in the original.

The original statement, 'If the sky is not clear, then you don't see the stars' is a conditional statement that refers to the necessary conditions for seeing the stars. In this logic, the clear sky is a necessity to see the stars.

There are several statements equivalent to the original one, according to the principles of logic namely:

However, the statements 'Clear sky is necessary and sufficient for seeing the stars' and 'Clear sky is sufficient to see the stars' are not necessarily equivalent to the original because they assume that a clear sky is all that's needed to see the stars, ignoring the other conditions like absence of light pollution or the time of the day. Whereas, the original statement does not make this assumption.

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Answer:

3 and 9 are parallelograms

other two are quadrilaterals

Step-by-step explanation:

Answer with Step-by-step explanation:

a.[tex]x=b^p[/tex]

[tex]y=b^q[/tex]

Taking both sides log

[tex]log x=plog b[/tex]

Using identity:[tex]logx^y=ylogx[/tex]

[tex]p=\frac{logx}{log b}=log_b x[/tex]

Using identity:[tex]log_x y=\frac{log y}{log x}[/tex]

[tex]log y=qlog b[/tex]

[tex]q=\frac{log y}{log b}=log_b y[/tex]

b.[tex]xy=b^pb^q[/tex]

We know that

[tex]x^a\cdot x^b=x^{a+b}[/tex]

Using identity

[tex]xy=b^{p+q}[/tex]

c.[tex]log_b(xy)=log_b(b^{p+q})[/tex]

[tex]log_b(xy)=(p+q)log_b b[/tex]

Substitute the values then we get

[tex]log_b(xy)=(log_b x+log_b y)[/tex]

By using [tex]log_b b=1[/tex]

Hence, [tex]log_b(xy)=log_b x+log_b y[/tex]

To prove the property log b(xy) = log bx + log by, we let x = [tex]b^p[/tex]and y = [tex]b^q[/tex], express xy in terms of the base b and the exponents p and q, and then use the properties of logarithms to show the equality.

The student is asking to prove the logarithmic property − log b(xy) = log bx + log by. Here's a step-by-step explanation:

Let x = bp and y = bq. To solve for p and q, take the logarithm base b of both sides. Thus, p = logbx and q = logby.

Using the property of exponents, xy = bp*bq = bp+q.

Now compute logb(xy). According to the logarithmic property, logb(bp+q) = p + q. Since p = logbx and q = logby, then logb(xy) = logbx + logby.

Therefore, we have proven the given logarithmic property using the steps provided in the question.

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Answer:

a = L

b = MT^(-1)

c = LT^(-1)

k = MT^(-2)

f = MLT^(-2)

S = T^(-1)

Step-by-step explanation:

x (0) = a

x is denoted by displacement in vibration analysis hence attains units of x.

Hence, a = L

b is the damping coefficient:

[tex]b = \frac{F}{\frac{dx}{dt} } \\= MLT^(-2) / LT^(-1)\\= MT^(-1)[/tex]

x'(0) = c

dx/dt = velocity hence c attains the units of velocity

c = LT^(-1)

Coefficient k is the stiffness:

[tex]k = \frac{F}{x} = \frac{MLT^(-2)}{L} = MT^(-2)[/tex]

Coefficient f is the magnitude of the exciting force

[tex]F = m*acceleration = MLT^(-2)[/tex]

Coefficient S is the angular frequency

angular frequency is displacement in radians per seconds; hence,

S = T^(-1)

Answer:

Step-by-step explanation:

a) {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}

[tex]\Omega[/tex]=6*6=36

b)A=(2*3)=6

P(A)=6/36=1/6

c) B=6+5+4+3+2+1=21

P(B)=21/36

d) P(A|B) - an event where either a 3 or 4 is rolled first and is followed by an even number and their sum goes over 7

P(A|B)=3

e) Not always, not sure how to explain, I'm not good with English Math

f) Same as above

The sample space for rolling two die has 36 outcomes. The probability of rolling a three or four, followed by an even number is 1/6, while the probability of the sum of the rolls not exceeding seven is 7/12. These two events are not mutually exclusive, but they are independent.

a. The sample space for rolling two dice consists of 36 possible outcomes, as there are six possible outcomes for the first die and six for the second die, and 6*6=36.

b. Event A happens when we roll a three or four first, followed by an even number. There are 4 such outcomes: (3,2), (3,4), (3,6), (4,2), (4,4), and (4,6). The probability of event A occurring is therefore 6/36 = 1/6.

c. Event B happens when the sum of the two rolls is at most seven. There are 21 outcomes: (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(5,1),(5,2),(6,1). So, the probability of event B is 21/36 = 7/12.

d. P(A|B) represents the probability of event A occurring given that event B has already occurred. There are 4 outcomes in B that are also in A: (3,2), (3,4), (4,2), (4,3). Hence, P(A|B) = 4/21.

e. Events A and B are not mutually exclusive, as they can both occur in the same trial (e.g., when the dice rolls are (3,2), (3,4), (4,2), or (4,3)).

f. A and B are independent because the probability of A doesn't change whether B occurs or not, and vice versa. The fact that P(A|B) = P(A) confirms this.

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