So if you want to fit the y-intercepts or "b", on the y-axis you should go by 25's [0, 25, 50, 75, 100...]
If the x-axis does not have to follow the same pattern (25's), you should go by 5's [0, 5, 10, 15, 20...]
y = 7x + 50
y = 2x + 175
First I would plot the y-intercepts for each equation, then plot a few points with x = 5, 10, 15 Then draw a straight line.
The point where the two lines meet/cross paths is your solution. It should be (25, 225) The x-axis is the number of miles, and the y-axis is the total cost. So Truck driver A and B would arrive/be at the same place/meet in 25 miles at the same cost of $225
how many ways are there to list the digits {1,2,2,3,4,5,6} so that identical digits are not in consecutive position?
Answer: 2520 ways
Step-by-step explanation:
7!/2!
Answer:
no
Step-by-step explanation:
Write as a single integral in the form b f(x) dx. a 1 f(x) dx −3 + 4 f(x) dx 1 − −2 f(x) dx −3
Answer:
\int \limits_{-2}^{4} f(x) \, dx
Step-by-step explanation:
The objective is to write
[tex]\int \limits_{-3}^{1} f(x) \, dx + \int \limits_{1}^{4} f(x) \, dx - \int \limits_{-3}^{-2} f(x) \, dx[/tex]
as a single integral in the form
[tex]\int \limits_{a}^{b} f(x)\, dx[/tex].
We consider the segments [tex][-3, 1], [1, 4][/tex] and [tex][-3,-2][/tex]. If we combine the first and the second segment, we obtain
[tex][-3,1] \cup [1,4] = [-3,4][/tex]
Therefore, adding the first two integrals gives
[tex]\int \limits_{-3}^{1} f(x) \, dx + \int \limits_{1}^{4} f(x) \, dx = \int \limits_{-3}^{4} f(x) \, dx[/tex]
Now,we have
[tex]\int \limits_{-3}^{4} f(x) \, dx - \int \limits_{-3}^{-2} f(x) \, dx[/tex]
To subtract them, we need to find the difference of the segments [tex][-3,4][/tex] and [tex][-3,-2][/tex].
[tex][-3,4] \; \backslash \; [-3,-2] = [-2,4][/tex]
Therefore,
[tex]\int \limits_{-3}^{4} f(x) \, dx - \int \limits_{-3}^{-2} f(x) \, dx = \int \limits_{-2}^{4} f(x) \, dx[/tex]
Thus, the single integration of all the definite integral is mentioned below:
[tex]\int_{-2}^{4} f(x) dx[/tex]
Given the integral is,
[tex]\int_{-3}^{1} f(x) dx + \int_{1}^{4} f(x) dx - \int_{-3}^{-2} f(x)dx[/tex]
We need to change the whole definite integral into a single integral.
Now, we have the limits of integrations are [ - 3, 1 ], [ 1, 4 ], and [ - 3, -2 ].
Now, the first limit and second limit of integration are in addition, therefore combining forms of the integration are the union of these units.
Thus,
[ - 3, 1 ] U [ 1, 4 ] = [ -3 , 4 ]
Now, the second limit and third limit of integration are in subtraction, therefore combining forms of the integration are the intersection of these units.
Thus,
[ 1, 4 ] intersection [ - 3, -2 ] = [ null set]
Now, the combination of [ -3 , 4 ] and [ -null set] is [ - 2, 4].
Thus, the single integration of all the definite integral is mentioned below:
[tex]\int_{-2}^{4} f(x) dx[/tex]
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Locate the relative extremum and point of Inflection. Use a graphing utility to confirm your results. Y = 3x^2 ln x/2 relative extremum (x, y) = point of inflection (x, y) =
Answer:
Relative Extrema: \left(\frac{2}{\sqrt{e}},-\frac{6}{e}\right)
Step-by-step explanation:
There is only one extreme minimum point that is \left(\frac{2}{\sqrt{e}},-\frac{6}{e}\right) and there is no any point of reflection for this function.
You can find it in attached pictures of graphs.
2-41 The time to complete a construction project is normally distributed with a mean of 60 weeks and a standard deviation of 4 weeks. What is the probability the project will be finished in 62 weeks or less? 62-60/4=2/4=0.5=69146 What is the probability the project will be finished in 66 weeks or less? 66-60/4=6/4=1.5 What is the probability the project will take longer than 65 weeks?
The probability that the project will be finished in 62 weeks or less is 0.6915.
The probability that the project will be finished in 66 weeks or less is 0.9332.
The probability that the project will take longer than 65 weeks is 0.1056.
Given that:
The time to complete a construction project is normally distributed.
The mean is :
μ = 60 weeks
The standard deviation is:
σ = 4 weeks
The z-score is:
[tex]z=\frac{x-\mu }{\sigma }[/tex]
When x = 62,
[tex]z=\frac{62-60}{4}[/tex]
[tex]=0.5[/tex]
So, P(x ≤ 62) = P(z ≤ 0.5).
From the standard table, P(z ≤ 0.5) = 0.6915
When x = 66,
[tex]z=\frac{66-60}{4}[/tex]
[tex]=1.5[/tex]
So, P(x ≤ 66) = P(z ≤ 1.5).
From the standard table, P(z ≤ 1.5) = 0.9332
When x = 65,
[tex]z=\frac{65-60}{4}[/tex]
[tex]=1.25[/tex]
So, P(x > 65) = P(z > 1.25).
= 1 - P(z ≤ 1.25).
From the standard table, P(z ≤ 1.25) = 0.8944
So, P(x > 65) = P(z > 1.25)
= 1 - 0.8944
= 0.1056
Hence, the probabilities are 0.6915, 0.9332, and 0.1056 respectively.
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The probability the project will be finished in 62 weeks or less is approximately 69.15%. The probability the project will be finished in 66 weeks or less is approximately 93.32%. The probability the project will take longer than 65 weeks is approximately 10.56%.
Explanation:To find the probability that the project will be finished in 62 weeks or less, we need to calculate the z-score. The z-score formula is (x - μ) / σ where x is the value we are interested in, μ is the mean, and σ is the standard deviation. Plugging in the values, we get (62 - 60) / 4 = 0.5. Using a z-score table, we can look up the probability corresponding to a z-score of 0.5, which is approximately 0.6915 or 69.15%.
Similarly, to find the probability that the project will be finished in 66 weeks or less, we calculate the z-score: (66 - 60) / 4 = 1.5. Looking up a z-score of 1.5 in the table, we find the probability is approximately 0.9332 or 93.32%.
To find the probability that the project will take longer than 65 weeks, we can subtract the probability of it being finished in 65 weeks or less from 1. Using the z-score formula, we get (65 - 60) / 4 = 1.25. The probability of finishing in 65 weeks or less is the area to the left of this z-score, which is approximately 0.8944 or 89.44%. Subtracting from 1, we get the probability of taking longer than 65 weeks is approximately 0.1056 or 10.56%.
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There are 5 very different seats in a car. In how many ways can 5 different people be seated in the car for a road trip if only 2 of them know how to drive?
Answer:
48
Step-by-step explanation:
Let A and B be the two people who are able to drive. If A is driving, there are 4! ways to arrange the remaining peoplein the car seats. If B is driving, there are also 4! ways to arrange the remaining people. The number of arrangements 'n' is:
[tex]n=2*4!\\n=2*4*3*2*1\\n=48\ ways[/tex]
They can be arranged in 48 ways.
Question 5 (Fill-In-The-Blank Worth 1 points)
(05.05 MC)
A system of equations is shown below:
6x - 5y = 5 ,
3x + 5y = 4
The x-coordinate of the solution to this system of equations is
Numerical Answers Expected!
Answer:
The x-coordinate of the solution to this system of equations is 1.
Step-by-step explanation:
Given,
[tex]6x - 5y = 5\\\\3x + 5y = 4[/tex]
We have to find out the x-coordinate of the equation.
Solution,
Let [tex]6x-5y=5\ \ \ \ equation\ 1[/tex]
Again let [tex]3x+5y=4\ \ \ \ \ equation \ 2[/tex]
Now using elimination method we will solve the equations.
For this we will add equation 1 and equation 2 and get;
[tex](6x-5y)+(3x+5y)=5+4\\\\6x-5y+3x+5y=9\\\\9x=9[/tex]
Now on dividing both side by '9' we get;
[tex]\frac{9x}{9}=\frac{9}{9}\\\\x=1[/tex]
Hence The x-coordinate of the solution to this system of equations is 1.
1
ur welcome homie
poggers
"People who often attend cultural activities, such as movies, sports events and concerts, are more likely than their less cultured cousins to survive the next eight to nine years, even when education and income are taken into account, according to a survey by the University of Umea in Sweden" (American Health, April 1997, p. 20).
(a) Can this claim be tested by conducting a randomized experiment? Explain.
(b) On the basis of the study that was conducted, can we conclude that attending cultural events causes people to be likely to live longer? Explain.
(c) The article continued, "No one’s sure how Mel Gibson and Mozart help health, but the activities may enhance immunity or coping skills." Comment on the validity of this statement.
(d) The article notes that education and income were taken into account. Give two examples of other factors about the people surveyed that you think should also have been taken into account.
Answer:
Step-by-step explanation:
Given that a claim was made as
"People who often attend cultural activities, such as movies, sports events and concerts, are more likely than their less cultured cousins to survive the next eight to nine years, even when education and income are taken into account, according to a survey by the University of Umea in Sweden" (American Health, April 1997, p. 20).
a) Yes, this can be tested by conducting a randomized experiment. Selecing two groups of persons i who attend and other who do not attend and compare their life
This can be done from the data source already existing about persons who died recently under homogeneous conditions of environemnt
b) Yes, we can conclude provided random sample are taken and homogeneous conditions were followed
c) Yes, 100% true, though cannot say precisely how these helps as there is no scientific measurement, but it is a fact it improves health.
d) Other factors, are family history, bad habits, disease already present in the persons, etc.
Randomized experiments for this claim may not be feasible. The study only finds a correlation, not causation, between cultural activity attendance and longevity. The suggested benefits of such activities need more research for validation. Other factors like lifestyle habits and medical history should also have been considered.
Explanation:(a) Testing this claim through a randomized experiment might not be practically or ethically feasible. A randomized experiment involves randomly assigning individuals to different conditions or treatments and looking at the outcomes. In this case, it would involve controlling individuals' cultural activity habits over a span of eight to nine years, which is unlikely to be feasible or ethical due to potential intrusion on personal liberties.
(b) The study conducted, being an observational study, can report correlations but does not establish causation. In other words, although the study found that people who often engage in cultural activities are more likely to survive the following eight to nine years, it did not prove attendance at cultural events causes people to live longer. There might be other unnoticed factors or variables at play.
(c) The statement in the article is speculative. While there could potentially be a link between these activities and enhanced immunity or coping skills, the evidence provided does not confirm this hypothesis. Further experimental studies would be needed to validate this claim.
(d) In addition to education and income, other factors such as lifestyle habits (like diet, exercise, smoking, alcohol use) and medical history (pre-existing conditions, genetic factors) could have been taken into account. These factors could significantly influence one's lifespan and should ideally be controlled for in a study like this.
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Which represents a quadratic function?
f(x) = −8x3 − 16x2 − 4x
f (x) = three-quarters x 2 + 2x − 5
f(x) = StartFraction 4 Over x squared EndFraction minus StartFraction 2 Over x EndFraction + 1
f(x) = 0x2 − 9x + 7
Answer:
The answer to your question is the second option
Step-by-step explanation:
A Quadratic function is a polynomial of degree two. That means that the higher exponent is 2.
a) This option is incorrect because the highest power is 3 not two.
b) This option is the right answer, the highest power is 2, so, it is a quadratic function.
c) This option is incorrect, the highest power is -2.
d) This option is incorrect, the highest option is 1.
Answer:
Option 2 is the correct answer
Step-by-step explanation:
A quadratic function is a function in which the highest power to which the variable is raised is 2
1) f(x) = −8x3 − 16x2 − 4x
The given function is a cubic function because the highest power
to which the variable,x is raised is 3
2) f(x) = 3x²/4 + 2x - 5
The given function is a quadratic function because the highest power
to which the variable,x is raised is 2
3) f(x) = 4/x² - 2/x + 1
It can be rewritten as
f(x) = 4x^-2 - 2x^-1 + 1
The given function is not a quadratic function because the highest power to which the variable,x is raised is - 2
4) f(x) = 0x2 − 9x + 7
It can be rewritten as
f(x) = - 9x + 7
The given function is not a quadratic function because the highest power to which the variable,x is raised is 1
10- [6-2•2 + (8-3)]•2
Answer:
10-[6-4+(5)]×2
10-[2+5]×2
10-(7)×2
10-14= -4
You are certain to get 3 jacks when selecting 51 cards from a shuffled deck. Express the indicated degree of likelihood as a probability value between 0 and 1 inclusive.
The question pertains to calculating the probability of drawing exactly 3 jacks from 51 randomly drawn cards from a 52-card deck. While the probability is very high, it's not an absolute certainty. The exact calculations involve complex combinatorial mathematics.
Explanation:The subject of this question pertains to probability in mathematics, specifically to calculate the likelihood of drawing 3 jacks when selecting 51 cards from a shuffled deck of 52 cards.
First off, we need to understand that in a well-shuffled 52-card deck, there are 4 Jacks. Even if you select 51 out of 52 cards, there isn't a guarantee that you will select 3 jacks because the selection is random. The scenario you provided indicates a nearly certain event (since you're pulling nearly all the cards), but it still isn't an absolute certainty.
The exact probability computation for this kind of problem are more complex as they would involve combinatorial calculations. For simplicity, let's consider a similar but simpler scenario. Let's assume you are drawing just 4 cards instead. The probability of getting exactly 3 Jacks would be a combination of the probability of picking a Jack, and the probability of picking a non-Jack card. This would be calculated as (C(4,3) * C(48,1)) / C(52,4), with C representing the combination formula. This gives us how many ways we can draw 3 Jacks and a non-Jack divided by how many ways we can draw any 4 cards.
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Final answer:
The probability of drawing 3 jacks from a standard deck of 52 cards when selecting 51 is 1 (certainty), as it is a guaranteed event given the conditions.
Explanation:
The question asks about the probability of a certain event occurring when dealing with a standard deck of 52 cards. In this case, the event is being certain to get 3 jacks when selecting 51 cards out of 52. Since there are 4 jacks in the deck, and upon drawing 51 cards you're left with only 1 card that is not drawn, it is guaranteed that you'll have the 3 jacks among the drawn cards.
Hence, the probability is 1 (certainty), as there is only one card you're not drawing and 4 chances to have drawn a jack, which means you will always end up with all 3 jacks among the chosen 51 cards.
The exponential probability distribution is a discrete distribution that is often used to describe time between customer arrivals.
Answer:
True
Step-by-step explanation:
The time between customer arrivals is called inter-arrival time. According to Queueing Notation, the inter-arrival time can be model based on difference probability distribution. The probability distribution by which the inter-arrival time can be modeled include:
Exponential Distribution or Markov distributionConstant or DeterministicHyper - exponentialArbitrary or General distributionNot defined?
x-2/5x-10
Answer:
(x-2)/5(x-2)
cancel x-2 from the numerator and the denominator and the answer is 1/5
Equations - Item 2829
The circumference (C) of a circle is 16 cm. Which formula can you use to find the
diameter (d) if you know that C = 3.14d?
The formula is used to find the diameter of circle is: [tex]d = \frac{C}{3.14}[/tex]
The diameter of circle is 5.1 cm
Solution:
Given that,
Circumference (C) of a circle is 16 cm
The formula for circumference of circle when diameter is given is:
[tex]C = \pi d\\\\\pi \text{ is a constant equal to 3.14}\\\\C = 3.14d[/tex]
Rearrange the formula to get "d"
Divide both sides by 3.14
[tex]d = \frac{C}{3.14}[/tex]
The above formula is used to find the diameter of circle
Given that, circumeference = C = 16 cm
Substituting we get,
[tex]d = \frac{16}{3.14}\\\\d = 5.095 \approx 5.1[/tex]
Thus diameter of circle is 5.1 cm
The following scores represent the results of a midterm exam in Statistics class. 25 35 43 44 47 48 54 55 56 57 59 62 63 65 66 68 69 69 71 72 72 73 74 76 77 77 78 79 80 81 81 82 83 85 89 92 93 94 97 98 a) Find the lower and upper quartiles for the data. b) Find the interquartile range. c) Construct a boxplot for this data set.
Answer:
a.
lower Quartile= 57.5
Upper Quartile=81
b.
23.5
c.
box-plot is attached in excel file
Step-by-step explanation:
The data is arranged in ascending order so, the lower quartile denoted as Q1 can be calculated as under
Q1=((n+1)/4)th score=(41/4)th score=(10.25)th score
Q1=10th score+0.25(11th-10th)score
Q1=57+0.25(59-57)=57+0.5=57.5
Q1=57.5
The data is arranged in ascending order so, the third quartile denoted as Q3 can be calculated as under
Q3=(3(n+1)/4)th score=(3*41/4)th score=(30.75)th score
Q3=30th score+0.75(31th-30th)score
Q3=81+0.75(81-81)=81+0=81
Q3=81
b)
Interquartile range=IQR=Q3-Q1=81-57.5=23.5
IQR=23.5
c)
The box-plot is made in excel and it shows no outlier. The box-plot shows the 5-number summary(minimum-Q1-median-Q3-maximum) as 25-57.5-72-81-98.
Choose an American household at random, and let the random variable X be the number of cars (including SUVs and light trucks) they own. Here is the probability model if we ignore the few households that own more than 6 cars:
Number of cars X Probability 0 1 2 3 4 5 6
0.07 0.31 0.43 0.12 0.04 0.02 0.01
A housing company builds houses with two-car garages. What percent of households have more cars than the garage can hold?
Final answer:
To find the percent of households having more cars than a two-car garage can hold, sum the probabilities for 3, 4, 5, and 6 cars, which results in 19%.
Explanation:
The student is interested in finding out what percent of American households have more cars than a two-car garage can hold, with the given probability distribution for the number of cars owned. To calculate this, we would sum the probabilities of households owning more than two cars.
The probabilities of owning 3, 4, 5, and 6 cars are 0.12, 0.04, 0.02, and 0.01 respectively. Adding these probabilities together gives us the percent of households with more cars than the garage can hold:
0.12 + 0.04 + 0.02 + 0.01 = 0.19
Therefore, 19% of American households own more cars than a two-car garage can hold.
Determine whether the system of linear equations has one and only one solution, infinitely many solutions, or no solution. x − 3y = −1 4x + 3y = 26 one and only one solution infinitely many solutions no solution Find the solution, if one exists. (If there are infinitely many solutions, express x and y in terms of the parameter t. If there is no solution, enter NO SOLUTION.)
Answer:
There is only one solution, x = 5 and y = 2.
Step-by-step explanation:
To answer this question, we have to solve this system of equations.
We have that:
x - 3y = -1
4x + 3y = 26
Writing x as a function of y in the first equation, and replacing in the second, we have that:
x = 3y - 1
Replacing in the second
4x + 3y = 26
4(3y - 1) + 3y = 26
12y - 4 + 3y = 26
15y = 30
y = 2
Since we have 15y = 30, y = 2, there is only one solution.
If we had 0y = 0, there would be infinitely many solutions.
If we had 0y = a, a different of zero, there would be no solution.
Solving for x
x = 3y - 1 = 3*2 - 1 = 5
There is only one solution, x = 5 and y = 2.
Answer:it has one solution.
Step-by-step explanation:
The given system of equations is expressed as
x − 3y = −1 - - - - - - - - -1
4x + 3y = 26 - - - - - - - - 2
The first step would be to eliminate y by adding equation 1 to equation 2. It becomes
5x = 25
Dividing the left hand side and the right hand side of the equation by 5, it becomes
5x/5 = 25/5
x = 5
Substituting x = 5 into equation 1, it becomes
5 − 3y = −1
3y = 5 +1 = 6
Dividing the left hand side and the right hand side of the equation by 3, it becomes
3y/3 = 6/3
y = 2
Which relationship is a direct variation?
Answer:
A relationship is said to have direct variation when one variable changes and the second variable changes proportionally; the ratio of the second variable to the first variable remains constant. For example, when y varies directly as x, there is a constant, k, that is the ratio of y:x.
PLEASE HELP!!!
Carol paid $0.78 per liter for gas while driving across Canada. Find the cost per gallon to the nearest cent.
Please give a step by step
Answer:
2.95 cent
Step-by-step explanation:
1 gallon = 231 cubic inches
1 litre = 1000ml = 61.0237 cubic inches
1 galloon = 231 / 61.0237 = 3.7854118 liters
if Carol paid $0.78 per litre
1 galloon = 0.78 x 3.7854118 = 2.952621204 ≅ 2.95 cent
Sketch an approximate solution curve that passes through each of the indicated points.
dy/dx = e^(−0.01) xy²
Answer:
y=-2· e^(0.01)/ x²
Step-by-step explanation:
We calculate the given differential equation, we get
dy/dx = e^(−0.01) xy²
dy/y² = e^(−0.01) x dx
∫ y^(-2) dy= e^(−0.01) ∫ x dx
- y^(-1) = e^(−0.01) x²/2
-1/y= e^(−0.01) x²/2
y=-2/ e^(−0.01) x²
y=-2· e^(0.01)/ x²
We use the site desmos.com, to plot graph for the solution of the the given differential equation. We get a graph.
Suppose a wheel with radius 16 cm rolls in a straight line over a flat surface rotating a total of 5 radians. How far did the wheel travel?
Answer: 80cm
Step-by-step explanation:
Given:
Radius r = 16cm
Radial distance x = 5 radians
Radian is a measure of angles.
1 radian = 180°/π
To convert radial distance to linear distance
Linear distance = radial distance × radius
d = xr
d = 5 × 16cm
d = 80cm
For each part, give a relation that satisfies the condition. a. Reflexive and symmetric but not transitive b. Reflexive and transitive but not symmetric c. Symmetric and transitive but not reflexive
Answer:
For the set X = {a, b, c}, the following three relations satisfy the required conditions in (a), (b) and (c) respectively.
(a) R = {(a,a), (b,b), (c, c), (a, b), (b, a), (b, c), (c, b)} is reflexive and symmetric but not transitive .
(b) R = {(a, a), (b, b), (c, c), (a, b)} is reflexive and transitive but not symmetric .
(c) R = {(a,a), (a, b), (b, a)} is symmetric and transitive but not reflexive .
Step-by-step explanation:
Before, we go on to check these relations for the desired properties, let us define what it means for a relation to be reflexive, symmetric or transitive.
Given a relation R on a set X,
R is said to be reflexive if for every [tex]a \in X, (a,a) \in R[/tex].
R is said to be symmetric if for every [tex](a, b) \in R, (b, a) \in R[/tex].
R is said to be transitive if [tex](a, b) \in R[/tex] and [tex](b, c) \in R[/tex], then [tex](a, c) \in R[/tex].
(a) Let R = {(a,a), (b,b), (c, c), (a, b), (b, a), (b, c), (c, b)}.
Reflexive: [tex](a, a), (b, b), (c, c) \in R[/tex]
Therefore, R is reflexive.
Symmetric: [tex](a, b) \in R \implies (b, a) \in R[/tex]
Therefore R is symmetric.
Transitive: [tex](a, b) \in R \ and \ (b, c) \in R[/tex] but but (a,c) is not in R.
Therefore, R is not transitive.
Therefore, R is reflexive and symmetric but not transitive .
(b) R = {(a, a), (b, b), (c, c), (a, b)}
Reflexive: [tex](a, a), (b, b) \ and \ (c, c) \in R[/tex]
Therefore, R is reflexive.
Symmetric: [tex](a, b) \in R \ but \ (b, a) \not \in R[/tex]
Therefore R is not symmetric.
Transitive: [tex](a, a), (a, b) \in R[/tex] and [tex](a, b) \in R[/tex].
Therefore, R is transitive.
Therefore, R is reflexive and transitive but not symmetric .
(c) R = {(a,a), (a, b), (b, a)}
Reflexive: [tex](a, a) \in R[/tex] but (b, b) and (c, c) are not in R
R must contain all ordered pairs of the form (x, x) for all x in R to be considered reflexive.
Therefore, R is not reflexive.
Symmetric: [tex](a, b) \in R[/tex] and [tex](b, a) \in R[/tex]
Therefore R is symmetric.
Transitive: [tex](a, a), (a, b) \in R[/tex] and [tex](a, b) \in R[/tex].
Therefore, R is transitive.
Therefore, R is symmetric and transitive but not reflexive .
Relation from the set of two variables is subset of certain product. The relation for the condition are,
[tex]R_1\;\;\;\;\ (1,1), (1,2),,(2,1), (2,2),(2,3)((3,2),3,3)[/tex]
[tex]R_2\;\;\;\;\ (1,1), (2,2),,(3,3)(1,3)3,1)[/tex]
[tex]R_3\;\;\;\;\ (1,2),,(2,1), ,(2,3)((3,2)[/tex]
Relation-Relation from the set of two variables is subset of certain product. Relation are of three types-
ReflexiveSymmetricTransitive1) Reflexive and symmetric but not transitive -
Let a data set as,
[tex]X=1,2,3[/tex]
For the data set the relation can be given as,
[tex]R_1\;\;\;\;\ (1,1), (1,2),,(2,1), (2,2),(2,3)((3,2),3,3)[/tex]
[tex]R_1[/tex] is reflexive as it can be represent as [tex]R_1(a,a)[/tex] for,
[tex]a=1,2,3, \;\;\;\;\; [/tex]
[tex]a[/tex] ∈ [tex]X[/tex]
[tex]R_1[/tex] is symmetric as it can be represent as [tex]R_1(a,b)[/tex] for,
[tex]a,b \;\;\;\;(1,2) (2,1)[/tex]
[tex]a,b[/tex] ∈ [tex]X[/tex]
[tex]R_1[/tex] is not transitive as it can be represent as [tex]R_1\neq (a,c)[/tex] .
[tex]a,c\neq \;\;\;\;(1,3) (3,1)[/tex]
2) Reflexive and transitive but not symmetric
Let a data set as,
[tex]X=1,2,3[/tex]
For the data set the relation can be given as,
[tex]R_2\;\;\;\;\ (1,1), (2,2),,(3,3)(1,3)3,1)[/tex]
[tex]R_2[/tex] is reflexive as it can be represent as [tex]R_2(a,a)[/tex] for,
[tex]a=1,2,3, \;\;\;\;\; [/tex]
[tex]a[/tex] ∈ [tex]X[/tex]
[tex]R_1[/tex] is transitive as it can be represent as [tex]R_1(a,c)[/tex] for,
[tex]a,c \;\;\;\;(1,3) (3,1)[/tex]
[tex]a,c[/tex] ∈ [tex]X[/tex]
[tex]R_1[/tex] is not symmetric as it can be represent as [tex]R_1\neq (a,b)[/tex] .
[tex]a,b\neq \;\;\;\;(1,2) (2,1)[/tex]
3) Symmetric and transitive but not reflexive
Let a data set as,
[tex]X=1,2,3[/tex]
For the data set the relation can be given as,
[tex]R_3\;\;\;\;\ (1,2),,(2,1), ,(2,3)((3,2)[/tex]
[tex]R_1[/tex] is symmetric as it can be represent as [tex]R_3(a,b)[/tex] for,
[tex]a,b=(1,2),(2,1) \;\;\;\;\; [/tex]
[tex]a,b[/tex] ∈ [tex]X[/tex]
[tex]R_3[/tex] is transitive as it can be represent as [tex]R_3(a,c)[/tex] for,
[tex]a,c \;\;\;\;(1,3) (3,1)[/tex]
[tex]a,c[/tex] ∈ [tex]X[/tex]
[tex]R_1[/tex] is not reflexive as it can be represent as [tex]R_3\neq (a,a)[/tex] .
[tex]a,a\neq \;\;\;\;(1,1) [/tex]
Thus the relation for the condition are,
[tex]R_1\;\;\;\;\ (1,1), (1,2),,(2,1), (2,2),(2,3)((3,2),3,3)[/tex]
[tex]R_2\;\;\;\;\ (1,1), (2,2),,(3,3)(1,3)3,1)[/tex]
[tex]R_3\;\;\;\;\ (1,2),,(2,1), ,(2,3)((3,2)[/tex]
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Find the force exerted by the surface on the point of contact with the hammer head. Assume that the force the hammer exerts on the nail is parallel to the nail.
Answer:
716.3N
Step-by-step explanation:
Moment produced by force F = 150 N:
Mf = 150 * 30 = 4500 Ncm
The same moment is imparted at the nail.
Fn * 5 / sin (60) = 4500 Ncm
Fn = 779.423 N
Force exerted by surface on hammer pivot is:
Fx = 779.423 sin (30) - 150 = 239.7115 N
Fy = 779.423 cos (30) = 675 N
Fres = sqrt ( (Fx)^2 + (Fy)^2)
Fres = sqrt ( 675 ^2 + 239.7115^2)
Fres = 716.3 N
The force the hammer exerts on the nail is parallel to the nail. is 716.3 N
Force and momentFrom the given information, we ned to first calculate the moment as shown:
Moment produced by force F = 150 N:Since Moment = Force * distance:
Mf = 150 * 30 = 4500 NcmThe same moment is imparted at the nail.
5Fn/sin (60) = 4500 Ncm
Fn = 779.42N
Next is to calculate the force exerted by the surface on the hammer pivot.
Fx = 779.423 sin (30) - 150 = 239.7115 N
Fy = 779.423 cos (30) = 675 N
[tex]F = \sqrt{ ( (Fx)^2 + (Fy)^2)}\\F = \sqrt{ ( 675 ^2 + 239.7115^2)}\\F = 716.3N[/tex]
The force the hammer exerts on the nail is parallel to the nail. is 716.3 N
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Using data from 20 compact cars, a consumer group develops a model that predicts the stopping time for a vehicle by using its weight. You consider using this model to predict the stopping time for your large SUV. Explain why this is not advisable.
No, it is not advisable to predict the stopping time for your large SUV using model trained for compact cars.
Prediction means generating the values of the dependent variable using some specific models in machine learning.
Given that, the model is trained on 20 compact cars and the model is developed such that it predicts the stopping time for a vehicle by using its weight.
Here the dependent variable is stopping time which is required to be predicted. As the model is trained on compact cars that is medium size cars and if we expect the same model to predict stopping time for large SUV, then model is going to predict false stopping time as the weights for large SUV is quiet higher than the compact cars. So, model may consider it as an outlies and will lead to incorrect prediction.
Therefore, it is not advisable to use the same model for predicting the stopping time for your large SUV.
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Final answer:
Using a stopping time model developed from compact car data to predict the stopping time for a large SUV is not advisable due to differences in vehicle dynamics, which may lead to inaccurate results.
Explanation:
When a consumer group develops a model to predict the stopping time of a vehicle based on its weight, the model must be used within the context of the data from which it was derived. Using the model, which was built on data from 20 compact cars, to predict the stopping time of a larger SUV is not advisable due to differences in vehicle dynamics, size, weight distribution, and potentially different braking systems. Models are designed to be predictive within the range of data they are based on, and extrapolating them beyond that range can lead to inaccurate predictions. Specifically, the heavier mass of an SUV compared to compact cars means that it would likely have a longer stopping distance due to greater momentum, and this may not be represented in a model calibrated to lighter vehicles.
You are designing a rectangular enclosure with 2 rectangular sections separated by parallel walls. The interior wall has a length of 60 feet and the area of the enclosure is 1700 ft2. What amount of fencing is required for this project?
Answer:
236.68 feet needed for fencing
Explanation:
You are designing a rectangular enclosure with 2 rectangular sections separated by parallel walls
Let L be the length and W be the width
Perimeter of rectangle = 3(length )+ 4 (width)
[tex]P=3L+4W[/tex]
Area of the rectangle = length * width
[tex]A=2LW[/tex]
[tex]W= \frac{A}{2L}[/tex]
[tex]W= \frac{1700}{2(60)}=14.17[/tex]
Replace the values in perimeter
[tex]P=3L+4W[/tex]
[tex]P=3(60)+4(14.17)=236.68[/tex]
So 236.68 feet needed for fencing
Determine the distance between point (x1, y1) and point (x2, y2), and assign the result to pointsDistance. The calculation is: Ex: For points (1.0, 2.0) and (1.0, 5.0), pointsDistance is 3.0.
Answer:
void distance(int x1, int x2, int y1, int y2){
pointsDistance = sqrt((x2-x1)^(2) + (y2-y1)^(2));
}
Step-by-step explanation:
Suppose we have two points:
[tex]A = (x_{1}, y_{1})[/tex]
[tex]B = (x_{2}, y_{2}[/tex]
The distance between these points is:
[tex]D = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}[/tex]
So, for points (1.0, 2.0) and (1.0, 5.0)
[tex]D = \sqrt{(1 - 1)^{2} + (5 - 2)^{2}} = \sqrt{9} = 3[/tex]
I suppose this questions asks me to write a code, since i have to attribute the result to pointsDistance. I am going to write a C code for this, and you have to include the math.h library.
void distance(int x1, int x2, int y1, int y2){
pointsDistance = sqrt((x2-x1)^(2) + (y2-y1)^(2));
}
Answer:
pointsDistance = sqrt (pow(x2 - x1,2.0) + pow(y2 - y1,2.0));
Step-by-step explanation:
A general 2x2 diagonal matrix has the form(a00b). Thus the two unknown real numbers a b are needed to specify each 2x2 diagonal matrix. In Exercises 11 16, how many unknown real numbers are needed to specify each of the given matrices
1. An upper triangular 2x2 matrix?
2.) An m × n matrix?
Answer:
1. 3, and 2. m x n
Step-by-step explanation:
1. for an upper triangular 2x2 matrix i.e. (a,0,c,d), three (03) unknown elements a, c, and d are needed to be specified.
2. for m x n matrix, m*n elements are needed to be specified.
Final answer:
To specify an upper triangular 2x2 matrix, 3 unknown real numbers are needed. For an m × n matrix, m × n unknown real numbers are required.
Explanation:
The question asks how many unknown real numbers are needed to specify each of the given matrices: an upper triangular 2x2 matrix, and an m × n matrix.
1. An Upper Triangular 2x2 Matrix
An upper triangular matrix has the form:
(a, b)
(0, c)
Thus, to specify an upper triangular 2x2 matrix, 3 unknown real numbers are needed: a, b, and c.
2. An m × n Matrix
An m × n matrix has m rows and n columns. To specify such a matrix, one needs m × n unknown real numbers, representing each element in the matrix.
27. In constructing a confidence interval estimate of the population mean you decide to select 49 random observations to get your point estimate of the mean (sample mean). Your friend is also constructing a similar confidence interval estimate but decides to use a sample size of 36 random observations.
Which of the following is true?
a.) Your confidence interval estimate is narrower
b.) Your friend’s confidence interval estimate has a greater degree of confidence
c.) Your confidence interval estimate is wider
d.) Your confidence interval estimate has a greater degree of confidence
2.) The width of a confidence interval estimate for a proportion will be:
a.) Narrower for 99% confidence level than for a 95% confidence level
b.) Wider for a sample size of 100 than for a sample size of 75
c.) Narrower for 90% confidence level than for a 95% confidence level
d.) Narrower when the sample proportion is .50 than when the sample proportion is 20.
Answer:
1) a.) Your confidence interval estimate is narrower
2) c.) The width of a confidence interval estimate for a proportion will be narrower for 90% confidence level than for a 95% confidence level
Step-by-step explanation:
Confidence Interval can be stated as M±ME where
M is the sample meanME is the margin of errorMargin of Error determines the range of the confidence interval around the mean.
Margin of error (ME) of the mean can be calculated using the formula
ME=[tex]\frac{z*s}{\sqrt{N} }[/tex] where
z is the corresponding statistic in the given confidence levels is the standard deviation of the sample(or the population if it is known) N is the sample sizeFrom the formula we can reach the following conclusions:
As N increases, ME decreases.as confidence level increases, corresponding statistic increases, and thus margin of error increases.Since your sample size (49) is bigger than your friend's (36), your confidence interval is narrower, because margin of error is narrower.
Since the confidence level 90% has smaller statistic than the confidence level 95%, its confidence interval is narrower.
That is, we can estimate narrower confidence intervals with less confidence.
Data from a sample of citizens of a certain country yielded the following estimates of average TV viewing time per month for all the citizens. The times are in hours and minutes. (NA, not available)
Viewing Method May 2008 May 2007 Change (%)
Watching TV in the home
Watching timeshifted TV
Using the internet
Watching video on internet
127:49
5:31
26:41
2:34
121:59
3:52
24:17
NA
5
43
10
NA
Is the study descriptive or inferential?
A. Inferential, because the statistics are used to describe the sample
B. Descriptive, because the statistics are used to describe the sample
C. Descriptive, because the statistics are used to make an inference about the population
D. Inferential, because the statistics are used to make an inference about the population
Answer:
D. Inferential, because the statistics are used to make an inference about the population
Correct, the objective of this study is obtain information from the population with a sample and then use any method to estimate the population mean, the parameter of interest.
Step-by-step explanation:
An inferential study consists in take information about a population by a sample and use this information to see what would be the possible values for the population of interest
By the other hand a descriptive study is obtained from observing and measuring some variables of interest but without manipulate the data.
For this case we have sample averages for the viewing time per month.
Let's analyze one by one the possible options:
A. Inferential, because the statistics are used to describe the sample
False the study is inferential but the idea is not just obtain information about the sample, we want to see the population parameters not the statistics
B. Descriptive, because the statistics are used to describe the sample
False for this case we have averages calculated from the sample mean and is not possible to consider this study as descriptive.
C. Descriptive, because the statistics are used to make an inference about the population
False, the statistics are used to make an inference about the population, this statement is correct, but the problem is that this study is not descriptive.
D. Inferential, because the statistics are used to make an inference about the population
Correct, the objective of this study is obtain information from the population with a sample and then use any method to estimate the population mean, the parameter of interest.
The study is descriptive. It uses statistics to provide summaries about the average TV viewing times per month among citizens, without making any inferences about a larger population. Therefore, the correct option is B.
Explanation:The study in question is descriptive. Descriptive statistics are used to describe the main features of a collection of data in quantitative terms. They provide simple summaries about the sample and the measures. The data mentioned here are providing a summary of the average TV viewing times per month for the citizens. They describe various elements of interest within a particular set and there isn't any interpretation or inference being made about the larger population from which the sample was drawn. Therefore, the correct option is B. Descriptive, because the statistics are used to describe the sample.
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Two events are listed below:
A: {The home's roof is less than 10 years}
B: {The home has a security system}
Define the event :A^c ∪ BGroup of answer choices:O The home's roof is at least 10 years or it has a no security systemO The home's roof is at least 10 years and it has a security systemO The home's roof is less than 10 years or it has a security systemO The home's roof is at least 10 years or it has a security system
Answer:
Option 4) The home's roof is at least 10 years or it has a security system.
Step-by-step explanation:
We are given the following events in the question:
A: The home's roof is less than 10 years}
B: The home has a security system
We have to find the interpretation of event
[tex]A^c \cup B[/tex]
Now, [tex]A^c[/tex]
This represents the complement of A and consist of events other than A,
Thus,
[tex]A^c[/tex]: The home's roof is not less than 10 years or the home's roof is greater than equal to 10 years or the home's roof is at least 10 years.
The union of two sets is a new set that contains all of the elements that are in at least one of the two sets.
[tex]A^c \cup B[/tex]
Thus, it can be interpreted as the home's roof is at least 10 years or the home has a security system.
Option 4) The home's roof is at least 10 years or it has a security system
The event A^c ∪ B represents all homes with roofs that are at least 10 years or have a security system.
Explanation:When evaluating the union of the complement of event A and event B, which is denoted as Ac ∪ B, we are looking for all outcomes that are either in the complement of A or in B, or in both. The complement of event A, denoted as Ac, includes all outcomes not in A. In the context of the given events, this would mean the complement of event A (the home's roof is less than 10 years) includes homes with roofs that are at least 10 years. Event B is that the home has a security system. Therefore, Ac ∪ B represents all homes with roofs that are at least 10 years or have a security system (or both).
The data in below relates to characteristics of high-definition televisions A through E. Identify the individuals, variables, and data corresponding to the variables. Determine whether each variable is qualitative, continuous, or discrete.
Screen
Setup Size (in) Type Number of Channels Available
A 47 Projection 300
B 45 Plasma 118
C 60 Plasma 423
D 40 Plasma 269
E 43 Projection 290
Answer:
Step-by-step explanation:
Hello!
You have two variables of interest.
X: Setup size (inches)
Y: Type the number of channels available.
Qualitative variables are those who describe characteristics of the subject of study, for example, the eye color of a person.
Quantitative variables are those that count quantities, for example, the shoe size of a person.
Continuous and discrete variables are quantitative. The difference is that the continuous variables are those who count in a determined range of valours, but between two observed values, there are infinite possible outcomes, for example, the body temperature of a cat. The normal temperature of a cat is around 38ºC, using a normal thermometer you measure the body temperature of two cats and obtain the following values 37.8 and 37.9 if you change the thermometer to one designed to take more precise measurements, it is possible that you obtain more values, for example, 37.81 and 39.94 and with a more precise tool you may become temperatures with more digits, this means that within this two temperatures there are infinite values of temperature, only limited by the equipment available.
A discrete variable is a quantitative variable but between the values, these variables take there are no other possible observations, regardless of the method of equipment used. An example of a discrete variable is the amount of money in a pocket. If you have two bills in one pocket, one is a 10 dollar bill and the other is a 20 dollar bill, there are no possible values in between, you either have ten or twenty, there is not possible, in this example, to count 15 dollars.
Then the variable "Y: Type number of channels available." is quantitative discrete, it counts the number of channels and between each channel there is nothing.
The variable "X: Setup size (inches)", the "inch" is a unit of length, and these variables are usually continuos, but in this example, your variable describes the screen width of the televisions and the type of image definition. Both are characteristics of the TVs so the variable is a qualitative one.
I hope it helps!