Answer:
a) 16% of students have an SAT math score greater than 615.
b) 2.5% of students have an SAT math score greater than 715.
c) 34% of students have an SAT math score between 415 and 515.
d) [tex]Z = 1.05[/tex]
e) [tex]Z = -1.10[/tex]
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the empirical rule.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Empirical rule
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
[tex]\mu = 515, \sigma = 100[/tex]
(a) What percentage of students have an SAT math score greater than 615?
615 is one standard deviation above the mean.
68% of the measures are within 1 standard deviation of the mean. The other 32% are more than 1 standard deviation from the mean. The normal probability distribution is symmetric. So of those 32%, 16% are more than 1 standard deviation above the mean and 16% more then 1 standard deviation below the mean.
So, 16% of students have an SAT math score greater than 615.
(b) What percentage of students have an SAT math score greater than 715?
715 is two standard deviations above the mean.
95% of the measures are within 2 standard deviations of the mean. The other 5% are more than 2 standard deviations from the mean. The normal probability distribution is symmetric. So of those 5%, 2.5% are more than 2 standard deviations above the mean and 2.5% more then 2 standard deviations below the mean.
So, 2.5% of students have an SAT math score greater than 715.
(c) What percentage of students have an SAT math score between 415 and 515?
415 is one standard deviation below the mean.
515 is the mean
68% of the measures are within 1 standard deviation of the mean. The normal probability distribution is symmetric, which means that of these 68%, 34% are within 1 standard deviation below the mean and the mean, and 34% are within the mean and 1 standard deviation above the mean.
So, 34% of students have an SAT math score between 415 and 515.
(d) What is the z-score for student with an SAT math score of 620?
We have that:
[tex]\mu = 515, \sigma = 100[/tex]
This is Z when X = 620. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{620 - 515}{100}[/tex]
[tex]Z = 1.05[/tex]
(e) What is the z-score for a student with an SAT math score of 405?
We have that:
[tex]\mu = 515, \sigma = 100[/tex]
This is Z when X = 405. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{405 - 515}{100}[/tex]
[tex]Z = -1.10[/tex]
a. Approximately 15.87% of students have an SAT math score greater than 615.
b. Approximately 2.28% of students have an SAT math score greater than 715.
c. 68% percentage of students have an SAT math score between 415 and 515
d. 1.05 is the z-score for student with an SAT math score of 620
e. -1.1 is the z-score for a student with an SAT math score of 405
To answer these questions, we can use the properties of a bell-shaped distribution and the empirical rule. The empirical rule states that for a bell-shaped distribution:Approximately 68% of the data falls within one standard deviation of the mean. Approximately 95% of the data falls within two standard deviations of the mean.Approximately 99.7% of the data falls within three standard deviations of the mean.
Given information: Mean (μ) = 515 and Standard Deviation (σ) = 100
(a) To find this, we need to calculate the z-score for a score of 615 and then find the percentage of data above that z-score using the standard normal distribution table.
z-score = (X - μ) / σ
z-score = (615 - 515) / 100
z-score = 1
Using the standard normal distribution table (or calculator), we find that approximately 84.13% of the data is below a z-score of 1. Since the distribution is symmetric, the percentage above the z-score of 1 is also approximately 100% - 84.13% = 15.87%.
Therefore, approximately 15.87% of students have an SAT math score greater than 615.
(b) We repeat the same process for a score of 715.
z-score = (715 - 515) / 100
z-score = 2
Using the standard normal distribution table (or calculator), we find that approximately 97.72% of the data is below a z-score of 2. The percentage above the z-score of 2 is approximately 100% - 97.72% = 2.28%.
Therefore, approximately 2.28% of students have an SAT math score greater than 715.
(c) We can use the empirical rule to find the percentage of students within one standard deviation of the mean and then subtract that from 100% to find the percentage between 415 and 515.
Percentage between 415 and 515 ≈ 68%
(d) We can calculate the z-score as follows:
z-score = (620 - 515) / 100
z-score = 1.05
(e) We can calculate the z-score as follows:
z-score = (405 - 515) / 100
z-score = -1.1
Remember that a negative z-score indicates a value below the mean.
Note: These calculations assume a normal distribution, and the actual percentages may vary slightly due to the discrete nature of test scores and rounding in calculations.
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Twenty-one telephones have just been received at an authorized service center. Seven of these telephones are cellular, seven are cordless, and the other seven are corded phones. Suppose that these components are randomly allocated the numbers 1, 2, . . . , 21 to establish the order in which they will be serviced.a. What is the probability that all the cordless phones are among the first fourteen to be serviced?b. What is the probability that after servicing fourteen of these phones, phones of only two of the three types remain to be serviced?c. What is the probability that two phones of each type are among the first six serviced?
Answer:
a) P=1/116280
b) P=143/38760
c) P=441/2584
Step-by-step explanation:
We have seven of these telephones are cellular, seven are cordless, and the other seven are corded phones.
a) We calculate the number of possible combinations
{21}_C_{14}=\frac{21!}{14! · (21-14)!}=116280
The number of favorable combinations is 1.
Therefore, the probability is
P=1/116280
b) We calculate the number of possible combinations
{21}_C_{14}=\frac{21!}{14! · (21-14)!}=116280
We calculate the number of favorable combinations
{14}_C_{7}=\frac{14!}{7! · (14-7)!}=429
Therefore, the probability is
P=429/116280
P=143/38760
c) We calculate the number of possible combinations
{21}_C_{6}=\frac{21!}{6! · (21-6)!}=54264
We calculate the number of favorable combinations
{7}_C_{2} · {7}_C_{2} · {7}_C_{2} =
=\frac{7!}{2!·(7-2)!} · \frac{7!}{2!·(7-2)!} · \frac{7!}{2!·(7-2)!}
=21 · 21 · 21=9261
Therefore, the probability is
P=9261/54264
P=441/2584
A radioactive isotope is unstable, and will decay by emitting a particle, transforming into another isotope. The assumption used to model such situations is that the amount of radioactive isotope decreases proportionally to the amount currently present.(a) Let N(t) designate the amount of the radioactive material present at time t and let N0 = N(0). Write and solve the IVP for radioactive decay of a radioactive material.(b) The half-life of a radioactive material is the time required for it to reach one-half of the original amount. What is the half-life of a material that in one day decays from 12 mg to 9 mg?
Answer:
Part a: The equation is [tex]N=N_{0}e^{-kt}[/tex]
Part b: The half life of the material is 2.4 days.
Step-by-step explanation:
Part a
The relation is given as
[tex]\frac{dN}{dt}=-kN[/tex]
Rearranging the equation gives
[tex]\frac{dN}{N}=-kdt[/tex]
Integrating and simplifying the equation as
[tex]\int \frac{dN}{N}=\int (-kdt)\\ln N=-kt+lnC\\N=e^{-kt+lnC}\\N=Ce^{-kt}[/tex]
This is the equation of radioactive decay of the radioactive material. For estimation of C consider following IVP where t=0,N=N_o so the equation becomes
[tex]N=Ce^{-kt}\\N_0=Ce^{-k(0)}\\N_0=Ce^{0}\\N_0=C[/tex]
Now substituting the value of C in the equation gives
[tex]N=N_{0}e^{-kt}[/tex]
This is the relation of concentration of unstable radioactive material at a given time .
Part b
From the given data the equation becomes as
[tex]N=N_{0}e^{-kt}\\9=12\times e^{-k*1}\\e^{-k}=\frac{9}{12}[/tex]
Now half like is defined as the time when the quantity is exact half, i.e. N/N_o =0.5 so
[tex]N=N_{0}e^{-kt}\\6=12\times e^{-k*t}\\e^{-kt}=\frac{6}{12}\\{e^{-k}}^t}=0.5\\(0.75)^t=0.5\\So\, t\, is\, given\, as \\t=\frac{ln 0.5}{ln 0.75}\\t=2.4 \, days[/tex]
So the half life of the material is 2.4 days.
Which equation shows a true relationship between the angle, arc length, and area of the sector shown?
Answer:c
Step-by-step explanation:because the arc length is x that is variable and multiplying it by the the formula gives the right answers.
Consider the points below. P(0, −4, 0), Q(5, 1, −3), R(5, 2, 1) (a) Find a nonzero vector orthogonal to the plane through the points P, Q, and R. Correct: Your answer is correct. (b) Find the area of the triangle PQR.
a) A nonzero vector orthogonal to the plane through the points P, Q, and R is [tex]PQ= < 5, 5, -3 >[/tex] and [tex]PR= < 5, 6, 1 >[/tex].
b) The area of the triangle PQR is 32.09 square units.
Given points are P(0, -4, 0), Q(5, 1, -3), T(5, 2, 1)
a) [tex]PQ= < 5, 1, -3 > - < 0, -4, 0 >[/tex]
[tex]PQ= < 5-0, 1-(-4), -3-0 >[/tex]
[tex]PQ= < 5, 5, -3 >[/tex]
[tex]PR= < 5, 2, 1 > - < 0, -4, 0 >[/tex]
[tex]PR= < 5-0, 2-(-4), 1-0 >[/tex]
[tex]PR= < 5, 6, 1 >[/tex]
b) The cross product of PQ and PR is
[tex]PQ\times PR=\left[\begin{array}{ccc}i&j&k\\5&5&-3\\5&6&1\end{array}\right][/tex]
[tex]PQ\times PR=(5+18)i-(5+15)j+(30-25)k[/tex]
[tex]PQ\times PR=23i-20j+10k[/tex]
The magnitude of [tex]PQ\times PR[/tex] is
[tex]|PQ\times PR| =|23i-20j+10k|[/tex]
[tex]|PQ\times PR| =\sqrt{23^2+(-20)^2+10^2}[/tex]
[tex]|PQ\times PR| =\sqrt{529+400+100}[/tex]
[tex]|PQ\times PR| =\sqrt{1029}[/tex]
[tex]|PQ\times PR| =32.09[/tex]
Therefore,
a) A nonzero vector orthogonal to the plane through the points P, Q, and R is [tex]PQ= < 5, 5, -3 >[/tex] and [tex]PR= < 5, 6, 1 >[/tex].
b) The area of the triangle PQR is 32.09 square units.
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Final answer:
To find a nonzero vector orthogonal to the plane through points P, Q, and R, calculate the cross product of two vectors in the plane. The area of triangle PQR can be found by dividing the length of the cross product vector by 2.
Explanation:
To find a nonzero vector orthogonal to the plane through points P, Q, and R, we can calculate the cross product of two vectors in the plane. Let's take vectors PQ and PR. Using the formula for cross product, we have:
PQ x PR = (5 - 0, 1 - (-4), -3 - 0) = (5, 5, -3)
Therefore, the vector (5, 5, -3) is orthogonal to the plane through points P, Q, and R.
To find the area of triangle PQR, we can use the length of the cross product vector and divide it by 2. The length of the cross product vector PQ x PR is:
|PQ x PR| =√(5² + 5² + (-3)²) = sqrt(59)
Dividing by 2, we get the area of triangle PQR as:
Area = √(59) / 2
For the month of May in a certain city, the probability that the weather on a given day is cloudy is 0.77. Also in the month of May in the same city, the probability that the weather on a given day is cloudy and snowy is 0.27. What is the probability that a randomly selected day in May will be snowy if it is cloudy?The probability is approximately nothing. (Round to three decimal places as needed.)
Answer:
0.3506 is the required probability.
Step-by-step explanation:
We are given the following in the question:
A: Weather on a given day is cloudy
B: Weather on a given day is snowy
[tex]P(A) = 0.77\\P(A\cap B) = 0.27[/tex]
We have to find the probability that the randomly selected day in May will be snowy if it is cloudy.
That is we have to evaluate P(B|A)
[tex]P(B|A) = \dfrac{P(B\cap A)}{P(A)}\\\\P(B|A) = \dfrac{0.27}{0.77}\\\\P(B|A) = 0.3506[/tex]
Thus, 0.3506 is the probability hat a randomly selected day in May will be snowy if it is cloudy.
This week, one of the topics is describing sampling distributions. In project 2 you will be using female heights to answer a variety of questions. As practice for this project, please address the bulleted items below for male heights. It will be necessary to know that the average height for men is assumed to be 70 inches with a standard deviation of 4 inches. To receive full credit, you also need to comment on two other posts from your classmates (you can just state whether or not you agree with their solution and if not, what you did differently).George Washington was 6 feet tall. Find the z-score for George Washington. Find the probability that a randomly selected individual will be as tall or taller than George Washington. Interpret both the z-score and the probability in a sentence.
Answer:
a) 0.5
b) 0.309
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 70 inches
Standard Deviation, σ = 4 inches
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Height of George Washington = 6 feet = [tex]6\times 12 = 72\text{ inches}[/tex]
x = 72
a) z-score
[tex]z_{score} = \displaystyle\frac{72-70}{4} = 0.5[/tex]
Interpretation:
This, mean that George Washington is 0.5 standard deviation more than the mean height of for men.
b) P(individual will be as tall or taller than George Washington)
[tex]P( x \geq 72) = P( z \geq 0.5)[/tex]
[tex]= 1 - P(z < 0.5)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x \geq 72) = 1 - 0.691 = 0.309 = 30.9\%[/tex]
Interpretation:
0.309 is the probability that a randomly selected individual will be as tall or taller than George Washington.
U.S. craft-beer breweries (breweries that make fewer than 6 million barrels annually and are less than 25% owned by big breweries) have been doing a booming business. The number of these small breweries from 2008 through 2012 can be modeled by using a quadratic function of the form f(t) = at² + bt + c where a, b, and c are constants and t is measured in years, with t = 0 corresponding to 2008.
(a) Find a, b, and c if f(0) = 1547, f(2) = 1802, and f(4) = 2403.
(b) Use the model obtained in part (a) to estimate the number of craft-beer breweries in 2016, assuming that the trend continued. (Round your answer to the nearest integer.)
a) The values of a, b and c are [tex]\frac{173}{4}, 41[/tex] and [tex]1547[/tex] respectively.
b) The number of craft beer breweries in 2016 will be 4643.
The quadratic equation holds an important significance in mathematics and is used in many situations, like calculating area or finding profit. When two linear expressions get multiplied, the resultant expression is a quadratic expression.
The number of craft beer breweries is modeled by a quadratic equation [tex]f(t)=at^2+bt+c[/tex].
In 2008, t=0.
a) Here, it is given that [tex]f(0)=1547[/tex], [tex]f(2)=1802[/tex] and [tex]f(4)=2403[/tex].
Now, use these values to find a, b and c as follows:
[tex]f(0)=a(0)^2+b(0)+c[/tex]
[tex]1547=c[/tex]
[tex]f(2)=a(2)^2+2b+c[/tex]
[tex]1802=4a+2b+1547[/tex]
[tex]255=4a+2b[/tex]
[tex]f(4)=a(4)^2+b(4)+c[/tex]
[tex]2403=16a+4b+1547[/tex]
[tex]856=16a+4b[/tex]
[tex]214=4a+b[/tex]
Now, setting the equations [tex]4a+2b=255[/tex] and [tex]4a+b=214[/tex] equal to each other:
[tex]4a=255-2b[/tex]
[tex]4a=214-b[/tex]
[tex]255-2b=214-b[/tex]
[tex]b=255-214[/tex]
[tex]b=41[/tex]
Now plugging the value of b into the equation [tex]4a+b=214[/tex], we get
[tex]4a+b=214[/tex]
[tex]4a=214-b[/tex]
[tex]a=\frac{214-b}{4}[/tex]
[tex]a=\frac{214-41}{4}[/tex]
[tex]a=\frac{173}{4}[/tex]
So, we have [tex]a=\frac{173}{4}[/tex], [tex]b=41[/tex] and [tex]c=1547[/tex]
b) Using the values calculated in part (a), the quadratic equation becomes
[tex]f(t)=\frac{173}{4}t^2+41t+1547[/tex]
Now, in 2016, the value of t=8
So, the number of craft beer breweries in 2016 will be calculated by [tex]f(8)[/tex]:
[tex]f(8)=\frac{173}{4}(8)^2+41(8)+1547[/tex]
[tex]= 2768+328+1547[/tex]
[tex]= 4643[/tex]
Therefore, the number of craft beer breweries in 2016 will be 4643.
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We first find the constants by substituting in the provided points into the quadratic equation to create a series of equations. Solving them gives us a = 61.5, b = 47, and c = 1547. To predict the number of breweries in 2016, we substitute t = 8 into the function to get approximately 3455 breweries.
Explanation:To find the constants a, b, and c, we will substitute the given points (0, 1547), (2, 1802), and (4, 2403) into the equation f(t) = at² + bt + c. This will give us three equations:
1547 = a(0)² + b(0) + c 1802 = a(2)² + b(2) + c 2403 = a(4)² + b(4) + cSolving these gives us a = 61.5, b = 47, and c = 1547. Therefore the quadratic function is: f(t) = 61.5t² + 47t + 1547.
For (b), we want to find the number of breweries in 2016, which is 8 years from 2008. Substituting t = 8 into our quadratic function gives us f(8) = 61.5(8²) + 47(8) + 1547 which is approximately 3455. So, the model estimates there will be around 3455 craft-beer breweries in 2016.
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A researcher wants to obtain a sample of 30 preschool children consisting of 10 two-year-old children, 10 three-year-old, and 10 four-year-old children. Assuming that the children are obtained from local daycare centers, this researcher should use ____ sampling.
Answer:
Quota sampling
Step-by-step explanation:
Quota sampling is a type of sampling method in which a representative data is collected from a group. It is a non probability based sampling method, a type of stratified sampling. Quota sampling is required when researchers are working with limited time and are without a sampling frame. Quota sampling method helps to determine the characteristics of a subgroup under research. It is an highly accurate method of sampling.
Find the vector represented by the directed line segment with initial point A(1, −4, 1) and terminal point B(−2, 5, 4). SOLUTION By the definition, the vector corresponding to AB is
The vector represented by the directed line segment with initial point A(1, −4, 1) and terminal point B(−2, 5, 4) is (-3, 9, 3).
To find the vector represented by the directed line segment with initial point A(1, −4, 1) and terminal point B(−2, 5, 4), we subtract the coordinates of A from the coordinates of B. Subtraction of vectors is equivalent to adding a negative vector, so we have:
AB = B - A = (-2 - 1, 5 - (-4), 4 - 1) = (-3, 9, 3)
The vector AB is represented as (-3, 9, 3).
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Three people are running for president of a class. The results of a poll indicate that the first candidate has an estimated 37% chance of winning and the second candidate has an estimated 44% chance of winning. What is the probability that the third candidate will win?
Answer:
0.19
Step-by-step explanation:
The are three candidate running for president and we know that probability of winning for first candidate and the probability of winning for second candidate and we have to find the probability of winning for third candidate
P(C1)=0.37
P(C2)=0.44
P(C3)=?
We know that sum of probabilities is always 1. So,
P(C1)+P(C2)+P(C3)=1
0.37+0.44+P(C3)=1
P(C3)=1-0.37-0.44
P(C3)=0.19
Thus, the probability of winning for third candidate is 0.19.
A high-tech company wants to estimate the mean number of years of college education its employees have completed. A sample of 15 employees had a mean of 4 years with a standard deviation of .7 years. Find a 95% confidence interval for the true mean.
Answer:
[tex]4 - 2.14 \frac{0.7}{\sqrt{15}}=3.61[/tex]
[tex]4 + 2.14 \frac{0.7}{\sqrt{15}}=4.39[/tex]
The 95% confidence interval is given by (3.61;4.39)
Step-by-step explanation:
Notation and definitions
n=15 represent the sample size
[tex]\bar X=4[/tex] represent the sample mean
[tex]s=0.7[/tex] represent the sample standard deviation
m represent the margin of error
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Calculate the critical value tc
In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. The degrees of freedom are given by:
[tex]df=n-1=15-1=14[/tex]
We can find the critical values in excel using the following formulas:
"=T.INV(0.025,14)" for [tex]t_{\alpha/2}=-2.14[/tex]
"=T.INV(1-0.025,14)" for [tex]t_{1-\alpha/2}=2.14[/tex]
The critical value [tex]tc=\pm 2.14[/tex]
Calculate the margin of error (m)
The margin of error for the sample mean is given by this formula:
[tex]m=t_c \frac{s}{\sqrt{n}}[/tex]
[tex]m=2.14 \frac{0.7}{\sqrt{15}}=0.387[/tex]
Calculate the confidence interval
The interval for the mean is given by this formula:
[tex]\bar X \pm t_{c} \frac{s}{\sqrt{n}}[/tex]
And calculating the limits we got:
[tex]4 - 2.14 \frac{0.7}{\sqrt{15}}=3.61[/tex]
[tex]4 + 2.14 \frac{0.7}{\sqrt{15}}=4.39[/tex]
The 95% confidence interval is given by (3.61;4.39)
Suppose there is a correlation of 0.87 between the length of time a person is in prison and the amount of aggression the person displays on a psychological inventory. This means that spending a longer amount of time in prison causes people to become more aggressive. a. Trueb. False
Answer:
b. False
See explanation below
Step-by-step explanation:
The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.
And in order to calculate the correlation coefficient we can use this formula:
[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]
For this case we know that r =0.87. But we can't conclude that the linear correlation coefficient represent a cause-effect relationship, since the linear correlation coefficient measures the linear dependency between two variables, but we can have other types of association between the variables and that's not measured by the Pearson correlation coeffcient.
So for this case the answer woudl be
b. False
Final answer:
The claim that a correlation of 0.87 between time in prison and aggression implies causation is false because correlation does not equal causation. There may be other factors influencing aggression, and the correlation observed might be spurious.
Explanation:
The statement that a correlation of 0.87 between the length of time a person is in prison and the amount of aggression displayed means that spending a longer time in prison causes people to become more aggressive is false.
Correlation does not imply causation. Just because two variables are correlated does not mean that one variable causes the change in the other. There could be other factors or variables that contribute to the aggression, which are not accounted for in simply observing the correlation.
Negative experiences do increase aggression, as people are more prone to aggress when experiencing negative emotions such as frustration, pain, or being in a bad mood. In a prison context, the aggression could also stem from the social structure of prison life and not merely the length of time served.
The Stanford prison experiment by Philip Zimbardo is a notable study highlighting how quickly people can turn to aggressive behaviors based on the roles they are assigned within a prison environment.
New York City is the most expensive city in the United States for lodging. The room rate is $204 per night (USA Today, April 30, 2012). Assume that room ratesa mally distributed with a standard deviation of $55. a. What is the probability that a hotel room costs $225 or more per night? b. What is the probability that a hotel room costs less than $140 per night? c. What is the probability that a hotel room costs between $200 and $300 per night? d. What is the cost of the most expensive 20% of hotel rooms in New York City?
Answer:
a. 0.35197 or 35.20%; b. 0.1230 or 12.30%; c. 0.48784 or 48.78%; d. $250.20 or more.
Step-by-step explanation:
In general, we can solve this question using the standard normal distribution, whose values are valid for any normally distributed data, provided that they are previously transformed to z-scores. After having these z-scores, we can consult the table to finally obtain the probability associated with that value. Likewise, for a given probability, we can find, using the same table, the z-score associated to solve the value x of the equation for the formula of z-scores.
We know that the room rates are normally distributed with a population mean and a population standard deviation of (according to the cited source in the question):
[tex] \\ \mu = \$204[/tex] (population mean)
[tex] \\ \sigma = \$55[/tex] (population standard deviation)
A z-score is the needed value to consult the standard normal table. It is a transformation of the data so that we can consult this standard normal table to obtain the probabilities associated. The standard normal table has a mean of 0 and a standard deviation of 1.
[tex] \\ z_{score}=\frac{x-\mu}{\sigma}[/tex]
After having all this information, we can proceed as follows:
What is the probability that a hotel room costs $225 or more per night?1. We need to calculate the z-score associated with x = $225.
[tex] \\ z_{score}=\frac{225-204}{55}[/tex]
[tex] \\ z_{score}=0.381818[/tex]
[tex] \\ z_{score}=0.38[/tex]
We rounded the value to two decimals since the cumulative standard normal table (values for cumulative probabilities from negative infinity to the value x) to consult only have until two decimals for z values.
Then
2. For a z = 0.38, the corresponding probability is P(z<0.38) = 0.64803. But the question is asking for values greater than this value, then:
[tex] \\ P(z>038) = 1 - P(z<0.38)[/tex] (that is, the complement of the area)
[tex] \\ P(z>038) = 1 - 0.64803[/tex]
[tex] \\ P(z>038) = 0.35197[/tex]
So, the probability that a hotel room costs $225 or more per night is P(x>$225) = 0.35197 or 35.20%, approximately.
What is the probability that a hotel room costs less than $140 per night?We follow a similar procedure as before, so:
[tex] \\ z_{score}=\frac{x-\mu}{\sigma}[/tex]
[tex] \\ z_{score}=\frac{140-204}{55}[/tex]
[tex] \\ z_{score}=\frac{140-204}{55}[/tex]
[tex] \\ z_{score}= -1.163636 \approx -1.16[/tex]
This value is below the mean (it has a negative sign). The standard normal tables does not have these values. However, we can find them subtracting the value of the probability obtained for z = 1.16 from 1, since the symmetry for normal distribution permits it. Then, the probability associated with z = -1.16 is:
[tex] \\ P(z<1.16) = 0.87698[/tex]
[tex] \\ P(z<-1.16) = 1 - 0.87698 [/tex]
[tex] \\ P(z<-1.16) = 0.12302 \approx 0.1230[/tex]
Then, the probability that a hotel room costs less than $140 per night is P(x<$140) = 0.1230 or 12.30%.
What is the probability that a hotel room costs between $200 and $300 per night?[tex] \\ z_{score}=\frac{x-\mu}{\sigma}[/tex]
The z-score and probability for x = $200:
[tex] \\ z_{score}=\frac{200-204}{55}[/tex]
[tex] \\ z_{score}= -0.072727 \approx -0.07[/tex]
[tex] \\ P(z<0.07) = 0.52790[/tex]
[tex] \\ P(z<-0.07) = 1 - 0.52790 [/tex]
[tex] \\ P(z<-0.07) = 0.47210 \approx 0.4721[/tex]
The z-score and probability for x = $300:
[tex] \\ z_{score}=\frac{300-204}{55}[/tex]
[tex] \\ z_{score}=1.745454[/tex]
[tex] \\ P(z<1.75) = 0.95994[/tex]
[tex] \\ P(z<1.75) - P(z<-0.07) = 0.95994-0.47210 [/tex]
[tex] \\ P(z<1.75) - P(z<-0.07) = 0.48784 [/tex]
Then, the probability that a hotel room costs between $200 and $300 per night is 0.48784 or 48.78%.
What is the cost of the most expensive 20% of hotel rooms in New York City?A way to solve this is as follows: we need to consult, using the cumulative standard normal table, the value for z such as the probability is 80%. This value is, approximately, z = 0.84. Then, solving the next equation for x:
[tex] \\ z_{score}=\frac{x-\mu}{\sigma}[/tex]
[tex] \\ 0.84=\frac{x-204}{55}[/tex]
[tex] \\ 0.84*55=x-204[/tex]
[tex] \\ 0.84*55 + 204 =x[/tex]
[tex] \\ x = 250.2[/tex]
That is, the cost of the most expensive 20% of hotel rooms in New York City are of $250.20 or more.
This answer uses the normal distribution to calculate probabilities and thresholds in hotel room rates. The main steps are to convert room rates to z-scores, look up probabilities or percentiles in the standard normal distribution, and convert back to room rates if necessary.
Explanation:To answer these questions, we can use the properties of the normal distribution. First, we convert the given hotel room rate to a z-score by subtracting the mean and dividing by the standard deviation. Then we can look up these z-scores in a standard normal distribution table (or use a calculator with a normal distribution function) to get probabilities.
a. To find the probability that a hotel room costs $225 or more per night, we convert $225 to a z-score: z = ($225-$204)/$55 = 0.38. The probability of getting a z-score of 0.38 or more (which means a cost of $225 or more) is 0.3520. So the probability that a room costs $225 or more per night is 0.3520 or 35.20%.
b. To find the probability that a hotel room costs less than $140 per night, we convert $140 to a z-score: z = ($140-$204)/$55 = -1.16. The probability of getting a z-score of -1.16 or less (which means a cost of $140 or less) is 0.1230. So the probability that a room costs $140 or less per night is 0.1230 or 12.30%.
c. To find the probability that a hotel room costs between $200 and $300 per night, we convert $200 to a z-score (z1 = -0.073) and $300 to a z-score (z2 = 1.745). We then subtract the probabilities: P($200 < Rate < $300) = P(z1 < Z < z2) = P(Z < z2) - P(Z < z1) = 0.9591 - 0.4713 = 0.4878. So the probability that a room costs between $200 and $300 per night is 0.4878 or 48.78%.
d. To find the cost of the most expensive 20% of hotel rooms in New York City, we look for the z-score corresponding to the 80th percentile (because everything above this point is the top 20%). This z-score is 0.84. We then convert this z-score back to a cost using the mean and standard deviation: Cost = Mean + z*SD = $204 + 0.84*$55 = $250.20. So the cost of the most expensive 20% of hotel rooms in New York City is $250.20 per night.
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When someone is on trial for suspicion of committing a crime, the hypotheses are: H0: innocent; Ha: guilty. Which of the following is correct? Group of answer choices Type II error is convicting an innocent person. Type I error is acquitting a guilty person. Type II error is acquitting an innocent person. Type I error is convicting an innocent person.
Answer:
Option 4) Type I error is convicting an innocent person.
Step-by-step explanation:
We are given the following in the question:
[tex]H_{0}: \text{ Innocent}\\H_A: \text{ Guilty}[/tex]
Type II error:
It is the error of accepting the null hypothesis given it is false.Thus, with the given scenario type II error will be freeing a guilty person.
That is acquitting a guilty person.
Type I error:
It is the error of rejecting null hypothesis given it is true.Thus, with respect to given scenario type I error is convicting an innocent person.
Thus, the correct answer is:
Option 4) Type I error is convicting an innocent person.
Suppose that we use Euler's method to approximate the solution to the differential equation:
dy/dx=x¹y; y(0.2)=7.
Let f(x,y)=x¹/y.
We let x0=0.2 and y0=7 and pick a step size h=0.2. Euler's method is the following algorithm. From xₙ and yₙ, our approximations to the solution of the differential equation at the nth stage, we find the next stage by computing
xₙ₊₁ = xₙ + h, yₙ₊₁ = yₙ + h⋅f(xₙ, yₙ )
Complete the following table:
n xₙ yₙ
0 0.4 1
1 0.6 1.08
2 0.8 1.22814
3 1
4 1.2
5 1.4
1) The exact solution can also be found using the separation of variables.
It is y(x)=?
2) Thus the actual value of the function at point x = 1.4.
y(1.4)=?
The exact solution to given differential equation is y(x) = Ce^(x^2/2). Using Euler's method with step size of 0.2, the solution approximates to 1.696 at x = 1.4; while the exact solution gives y(1.4) ~= 7.63.
Explanation:The differential equation given is of the form dy/dx = x*y. The exact solution to this equation can be found using the method of separation of variables. Integrating both sides, we get y(x) = Ce^(x^2/2) where C is the integration constant. Given that y(0.2) = 7, we can find C = 7*exp(-0.02).
Moving on to Euler's method, the next stage can be found via the algorithm x(n+1) = x(n) + h and y(n+1) = y(n) + h*f(x(n), y(n)). The function f(x,y) = x / y is provided. Completing the table for x = 1, y can be calculated as y(1) = y(0.8) + h*f(0.8, y(0.8)) = 1.22814 + 0.2(0.8/1.22814) = 1.3924. Following the same steps until x = 1.4 gives y(1.4) ~= 1.696.
In part 2), with x = 1.4 in the exact solution we found earlier, y(1.4) = 7*exp(-0.02) * exp(1.4^2/2) ~= 7.63.
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In a set of 12 devices 4 are defective. Assume that all of the defective and all of the functional devices are indistinguishable. How many linear orderings are there in which no two defective devices are consecutive
Answer: 126 orderings
Step-by-step explanation:
Here let's suppose that total devices is given by n= 12
and defective ones are given by m= 4
Now, to find the number of orderings in which no two defective devices are consecutive is given by following relation.
¹²⁻⁴⁺¹₄C
= ⁹₄C
= 126 orderings
A single card is drawn at random from each of six well-shuffled decks of playing cards. Find the probability that all six cards drawn are different.
Answer:
0.74141
Step-by-step explanation:
There are 52 cards in total
since each card is different,
the probability = number of favorable cards / total number of outcomes
P(C₁) = number of favorable cards / total number of outcomes
= [tex]\frac{52}{52}[/tex]
P(C₂, C₁) = number of favorable cards / total number of outcomes
= [tex]\frac{51}{52}[/tex]
P(C₃,C₁ ∩ C₂) = [tex]\frac{50}{52}[/tex]
P(C₄,C₁ ∩ C₂ ∩ C₃) = [tex]\frac{49}{52}[/tex]
P(C₅,C₁ ∩ C₂ ∩ C₃ ∩ C₄) = [tex]\frac{48}{52}[/tex]
P(C₆,C₁ ∩ C₂ ∩ C₃ ∩ C₄ ∩ C₅) = [tex]\frac{47}{52}[/tex]
General multiplication rule
P(A) =P(C₁ ∩ C₂ ∩ C₃ ∩ C₄ ∩ C₅ ∩ C₆)
= [tex]\frac{52}{52}. \frac{51}{52}. \frac{50}{52}. \frac{49}{52}. \frac{48}{52} .\frac{47}{52}[/tex]
= 8,808,975 / 11,881,376
= 0.74141
In order to determine whether or not a variable type of process is in control (whereoutput can be measured, not just classified in a binary fashion—i.e., good or bad), you must use which type of chart(s)? (Check all that apply.)A) p-chartB) c-chartC) x-bar chartD) fishbone diagramE) R-chart
Answer:
i dont know
Step-by-step explanation:
just working
A computer sends a packet of information along a channel and waits for a return signal acknowledging that the packet has been received. If no acknowledgment is received within a certain time, the packet is re-sent.
Let X represent the number of times the packet is sent. Assume that the probability mass function of X is given by
p(x)={ cx for x=1,2,3,4, or 5
{ 0 otherwise
where c is a constant.
Find P(X = 2).
Answer:
Step-by-step explanation:
given that a computer sends a packet of information along a channel and waits for a return signal acknowledging that the packet has been received. If no acknowledgment is received within a certain time, the packet is re-sent
Let X represent the number of times the packet is sent. Assume that the probability mass function of X is given by
p(x)={ cx for x=1,2,3,4, or 5
{ 0 otherwise
where c is a constant.
Let us write pdf as follows
x 1 2 3 4 5 Total
p c 2c 3c 4c 5c 15c
Since total probability = 1
we get
[tex]c=\frac{1}{15}[/tex]
P(X=2)=2c = [tex]\frac{2}{15}[/tex]
For the given position vectors r(t) compute the unit tangent vector T(t) for the given value of t .
A) Let r(t)=(cos(5t),sin(5t)).
Then T(?4)= (___,___)
B) Let r(t)=(t2,t3).
Then T(4)= (___,___)
C) Let r(t)=e5ti+e?4tj+tk.
Then T(?4)=__i+__j+__k
Answer:
a) [tex] T(t) = \frac{<-5 sin(5t), 5cos(5t)>}{5}= <-sin(5t), cos(5t)>[/tex]
[tex] T(4) = <-sin(20), cos(20)>[/tex]
b) [tex] T(t) = \frac{<t^2, 3t^2>}{8\sqrt{37}}[/tex]
[tex] T(4) = <\frac{2\sqrt{37}}{37},\frac{6\sqrt{37}}{37} >[/tex]
c) [tex] T(t) = \frac{<5e^{5t}, -4e^{-4t}, 1>}{2425825977}[/tex]
[tex] T(4) = \frac{1}{2425825977}<5e^{50}, -4e^{-16},1 >[/tex]
Step-by-step explanation:
The tangent vector is defined as:
[tex] T(t) = \frac{r'(t)}{|r'(t)|}[/tex]
Part a
For this case we have the following function given:
[tex] r(t) = <cos(5t), sin(5t)>[/tex]
The derivate is given by:
[tex] r'(t) = <-5 sin(5t), 5cos(5t)>[/tex]
The magnitude for the derivate is given by:
[tex] |r'(t)| = \sqrt{25 sin^2(5t) +25 cos^2 (5t)}= 5\sqrt{cos^2 (5t) + sin^2 (5t)} =5[/tex]
And then the tangent vector for this case would be:
[tex] T(t) = \frac{<-5 sin(5t), 5cos(5t)>}{5}= <-sin(5t), cos(5t)>[/tex]
And for the case when t=4 we got:
[tex] T(4) = <-sin(20), cos(20)>[/tex]
Part b
For this case we have the following function given:
[tex] r(t) = <t^2, t^3>[/tex]
The derivate is given by:
[tex] r'(t) = <2t, 3t^2>[/tex]
The magnitude for the derivate is given by:
[tex] |r'(t)| = \sqrt{4t^2 +9t^4}= t\sqrt{4 + 9t^2} [/tex]
[tex] |r'(4)| = \sqrt{4(4)^2 +9(4)^4}= 4\sqrt{4 + 9(4)^2} = 4\sqrt{148}= 8\sqrt{37}[/tex]
And then the tangent vector for this case would be:
[tex] T(t) = \frac{<t^2, 3t^2>}{8\sqrt{37}}[/tex]
And for the case when t=4 we got:
[tex] T(4) = <\frac{2\sqrt{37}}{37},\frac{6\sqrt{37}}{37} >[/tex]
Part c
For this case we have the following function given:
[tex] r(t) = <e^{5t}, e^{-4t} ,t>[/tex]
The derivate is given by:
[tex] r'(t) = <5e^{5t}, -4e^{-4t}, 1>[/tex]
The magnitude for the derivate is given by:
[tex] |r'(t)| = \sqrt{25e^{10t} +16e^{-8t} +1} [/tex]
[tex] |r'(t)| = \sqrt{25e^{10*4} +16e^{-8*4} +1} =2425825977 [/tex]
And then the tangent vector for this case would be:
[tex] T(t) = \frac{<5e^{5t}, -4e^{-4t}, 1>}{2425825977}[/tex]
And for the case when t=4 we got:
[tex] T(4) = \frac{1}{2425825977}<5e^{50}, -4e^{-16},1 >[/tex]
To compute the unit tangent vector T(t) for the given position vector r(t) at a given value t, take the derivative of r(t) with respect to t and divide the resulting vector by its magnitude.
Explanation:To compute the unit tangent vector T(t) for the given position vector r(t) at a given value t, we need to take the derivative of r(t) with respect to t and then divide the resulting vector by its magnitude.
For part A, r(t) = (cos(5t), sin(5t)), so r'(t) = (-5sin(5t), 5cos(5t)). Plugging in t = -4, we get r'(-4) = (-5sin(-20), 5cos(-20)). To find T(-4), we divide r'(-4) by its magnitude.
T(-4) = (-5sin(-20)/sqrt((-5sin(-20))^2 + (5cos(-20))^2), 5cos(-20)/sqrt((-5sin(-20))^2 + (5cos(-20))^2)).
A batch contains 31 bacteria cells. Assume that 12 of the cells are not capable of cellular replication. Six cells are selected at random, without replacement, to be checked for replication. Round your answers to four decimal places (e.g. 98.7654).What is the probability that all six cells of the selected cells are able to replicate
Answer:
0.0369 = 3.69% probability that all six cells of the selected cells are able to replicate.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes
The combinations formula is important in this problem:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
Desired outcomes
31 cells
19 are able to replicate.
We pick six.
The order is not important, that is, if it had been 2 cells, cell A and cell B would be the same outcome as cell B and cell A. So we use the combinations formula to find the number of desired outcomes.
It is a combination of 6 from 19(cells who are able to replicate).
[tex]D = C_{19,6} = \frac{19!}{6!(19 - 6)!} = 27132[/tex]
Total outcomes
31 cells
6 are picked.
[tex]T = C_{31,6} = \frac{31!}{6!(31 - 6)!} = 736281[/tex]
What is the probability that all six cells of the selected cells are able to replicate?
[tex]P = \frac{D}{T} = \frac{27132}{736281} = 0.0369[/tex]
0.0369 = 3.69% probability that all six cells of the selected cells are able to replicate.
Answer:
The answer would be 98.8
Step-by-step explanation:
Resin being you must round simply to the nearest hundreds
If the mean weight of all students in a class is 165 pounds with a variance of 234.09 square pounds, what is the z-value associated with a student whose weight is 140 pounds?
Answer:
[tex]Z = -1.63[/tex]
Step-by-step explanation:
The z-score measures how many standard deviation a score X is above or below the mean.
it is given by the following formula:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
In which
[tex]\mu[/tex] is the mean, [tex]\sigma[/tex] is the standard deviation
In this problem, we have that:
Mean weight of all students in a class is 165 pounds, so [tex]\mu = 165[/tex]
Variance of 234.09 square pounds. The standard deviation is the square root of the variance. So [tex]\sigma = \sqrt{234.09} = 15.3[/tex]
What is the z-value associated with a student whose weight is 140 pounds?
Z when X = 140. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{140 - 165}{15.3}[/tex]
[tex]Z = -1.63[/tex]
To find the z-value associated with a student whose weight is 140 pounds, use the formula for calculating z-score. Substitute the values of mean, standard deviation, and the student's weight into the formula.
Explanation:To find the z-value associated with a student whose weight is 140 pounds, we need to use the formula for calculating z-score:
z = (x - μ) / σ
Step 1: Calculate the standard deviation by taking the square root of the variance. In this case, the standard deviation is √234.09. Step 2: Calculate the z-score using the given formula by substituting the values of x (140 pounds), μ (mean weight of the class - 165 pounds), and σ (standard deviation). Step 3: Solve the equation to find the z-score.
The z-value associated with a student whose weight is 140 pounds can be calculated using the given formula and the information provided.
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Find the balance of $2,500 deposited at 3% compounded annually for 3 years
Answer:
The balance will be $2,731.82.
Step-by-step explanation:
This problem can be solved by the following formula:
[tex]A = P(1 + r)^{t}[/tex]
In which A is the final amount(balance), P is the principal(the deposit), r is the interest rate and t is the time, in years.
In this problem, we have that:
[tex]P = 2500, r = 0.03 t = 3[/tex]
We want to find A
So
[tex]A = P(1 + r)^{t}[/tex]
[tex]A = 2500(1 + 0.03)^{3}[/tex]
[tex]A = 2,731.82[/tex]
The balance will be $2,731.82.
Answer:
2731.8
Step-by-step explanation:
2500 deposited at 3 percent interest annually equation form
2500(1+(3/100))
for 3 years power the bract by 3
2500(1+(3/100))^3=2731.81
aka 2732
An accounting professor wishing to know how many MBA students would take a summer elective in international accounting did a survey of the class she was teaching. Which kind of sample is this?
Answer: Convenience sample.
Step-by-step explanation:
Convenience sample is also known as grab or accident or opportunity samples. It is a example of non probability sample that involves selecting of subjects because of the proximity, convenience and accessibility a researcher as to them. This type of samples are not reliable for data gathering when it involves a very large sample space, let's say a global audience.
2. Lab groups of three are to be randomly formed (without replacement) from a class that contains five engineers and four non-engineers. (8pts) (a) How many different lab groups are possible
Answer:
The number of different lab groups possible is 84.
Step-by-step explanation:
Given:
A class consists of 5 engineers and 4 non-engineers.
A lab groups of 3 are to be formed of these 9 students.
The problem can be solved using combinations.
Combinations is the number of ways to select k items from a group of n items without replacement. The order of the arrangement does not matter in combinations.
The combination of k items from n items is: [tex]{n\choose k}=\frac{n!}{k!(n-k)!}[/tex]
Compute the number of different lab groups possible as follows:
The number of ways of selecting 3 students from 9 is = [tex]{n\choose k}={9\choose 3}[/tex]
[tex]=\frac{9!}{3!(9 - 3)!}\\=\frac{9!}{3!\times 6!}\\=\frac{362880}{6\times720}\\ =84[/tex]
Thus, the number of different lab groups possible is 84.
To verify and compare data across variables, data can be broken down and studied by subgroups in a process called _____. A. convenience sampling B. stratification C. cross tabulation D. non probability sampling E. back-translation
Answer:
B. stratification
Step-by-step explanation:
Stratification of data is a process in which data is agrupated into some groups or strats depending on its value.
This action is helpful when the person has a lot of data that can be clasified on several variables that will make their analysis easier to manage.
Answer:convenience sampling
Step-by-step explanation:
This is the type of sampling where the first available primary data source will be used without additional requirements.
In a study, the researcher tries to use sugar intake to predict an individual's weight. However, when the researcher adds exercise as another IV into the study, he finds out that the effect of sugar intake on weight has decreased. Which of the following statement can be true?
Select one:
a. Exercise is a suppressor.
b. Exercise doesn't have effects on individuals' weight
c. Exercise is a mediator.
d. All the statements are correct.
Answer:
a. Exercise is a suppressor.
Step-by-step explanation:
A variable that diminishes the effect of another variable on the outcome is known as a suppressor. In this particular experiment, when exercise is introduced, it was observed that when exercise was introduced, there was a decrease on the effect of sugar intake on weight. Therefore, exercise is a suppressor.
A test has the capability to detect a defect with 98% accuracy. It also tends to indicate a defect when there is none about 1% of the time. About 1 in 2000 test subjects has a defect. What is the probably that a subject chosen at random and is indicated to have a defect, really does have a defect? (0.25 points)
Answer:
OUR ANSWER IS =0.0467
Step-by-step explanation:
P(detect/defect)=0.98
P(detect/not defect)=0.01
P(defect)=1/2000=0.0005
P(not defect)=1-0.0005=0.9995
P(defect/detect)=P(detect/defect)*P(defect)/[P(detect/defect)*P(defect)+P(detect/not defect)*P(not defect)]
=0.98*0.0005/[0.98*0.0005+0.01*0.9995]
=0.00049/[0.00049+0.009995]
=0.00049/0.010485
=0.0467
Suppose that vehicles taking a particular freeway exit can turn right (R), turn left (L), or go straight (S). Consider observing the direction for each of three successive vehicles. (Enter your answers in set notation. Enter EMPTY or ∅ for the empty set.)
(a) List all outcomes in the event A that all three vehicles go in the same direction.
(b) List all outcomes in the event B that all three vehicles take different directions.
(c) List all outcomes in the event C that exactly two of the three vehicles turn right.
(d) List all outcomes in the event D that exactly two vehicles go in the same direction.
(e) List outcomes in D'.
Answer:
a)
A={RRR,SSS,LLL}
b)
B={RSL,SLR,LRS,RLS,SRL,LSR}
c)
C={RRL,RLR,LRR,RRS,SRR,RSR}
d)
D={RRL, RRS, RLR, RLL, RSR, RSS, LRR, LRL, LLR, LLS, LSL, LSS, SRR, SRS, SLL, SLS, SSR, SSL}
e)
D'={RRR, RLS, RSL, LRS, LLL, LSR, SRL, SLR, SSS}
Step-by-step explanation:
Sample space=S= {RRR, RRL, RRS, RLR, RLL, RLS, RSR, RSL, RSS, LRR, LRL, LRS, LLR, LLL, LLS, LSR, LSL, LSS, SRR, SRL, SRS, SLR, SLL, SLS, SSR, SSL, SSS}
a)
Let A be the event that all three vehicles go in same direction. It means that the all three vehicles go to right or left or straight. Thus, event A can be represented as
A={RRR,SSS,LLL}
b)
Let B be the event that all three vehicles go in different direction. It means that the three vehicles can go to
1. Right, Straight, Left
2. Straight, Left, Right
3. Left, Right, Straight
4. Right, Left, Straight
5. Straight, Right, Left
6. Left, Straight, Right
Thus, event B can be represented as
B={RSL,SLR,LRS,RLS,SRL,LSR}
c)
Let C be the event that exactly two of three vehicles turn right. It means that the three vehicles can go to
1. Right, Right , Left
2. Right , Left, Right
3. Left, Right, Right
4. Right, Right , Straight
5. Straight, Right,Right
6. Right , Straight, Right
Thus, event C can be represented as
C={RRL,RLR,LRR,RRS,SRR,RSR}
d)
Let D be the event that exactly two of three vehicles go in same direction. It means that the three vehicles can go to
1. Right, Right, Left
2. Right, Right, Straight
3. Right, Left, Right
4. Right, Left, Left
5. Right, Straight, Right
6. Right, Straight, Straight
7. Left, Right, Right
8. Left, Right, Left
9. Left, Left, Right
10. Left, Left, Straight
11. Left, Straight, Left
12. Left, Straight, Straight
13. Straight, Right, Right
14. Straight, Right, Straight
15. Straight, Left, Left
16. Straight, Left, Straight
17. Straight, Straight, Right
18.Straight,Straight, Left
Thus, event D can be represented as
D={RRL, RRS, RLR, RLL, RSR, RSS, LRR, LRL, LLR, LLS, LSL, LSS, SRR, SRS, SLL, SLS, SSR, SSL}
e)
D'=S-D= {RRR, RRL, RRS, RLR, RLL, RLS, RSR, RSL, RSS, LRR, LRL, LRS, LLR, LLL, LLS, LSR, LSL, LSS, SRR, SRL, SRS, SLR, SLL, SLS, SSR, SSL, SSS} -{RRL, RRS, RLR, RLL, RSR, RSS, LRR, LRL, LLR, LLS, LSL, LSS, SRR, SRS, SLL, SLS, SSR, SSL}
D'={RRR, RLS, RSL, LRS, LLL, LSR, SRL, SLR, SSS}
Final answer:
For a freeway exit where vehicles can go right, left, or straight, we determine the combination of directions for various events: Event A with all vehicles going the same direction has 3 outcomes; Event B with all vehicles taking different directions has 6 outcomes; Event C with two vehicles turning right has 6 outcomes; Event D with two vehicles going in the same direction has several outcomes depending on combinations; Event D' coincides with Event B.
Explanation:
When analyzing the outcomes of three successive vehicles taking different actions at a freeway exit, we can approach the given scenarios methodically:
(a) Event A: where all three vehicles go in the same direction. The outcomes are {RRR}, {LLL}, and {SSS}.To further explain, for (a), since there are three options for direction and all must be the same, we have exactly three outcomes when they are all identical. For (b), all vehicles must take a unique direction, and since there are three vehicles and three directions, we have 3! (factorial) permutations of the outcome, resulting in 6 possibilities. For (c), two vehicles must turn right and the third vehicle must take either of the two remaining directions, which can be arranged in 3 choose 2 (3C2 = 3) ways for each of the two non-right directions, making 6 outcomes total. For (d), it's similar to (c), but we include all same-direction pairs, resulting in 3C2 combinations for each pair of directions. Lastly, (e) is simply the case where no two outcomes are the same, hence it is the same as (b).
Find the measure of the angle θ between u and v. Express the answer in radians rounded to two decimal places, if it is not possible to express it exactly.
Answer:
The question is incomplete as some details are missing; Here is the complete question; If vector u = 3i and v = 4i + 4j, Find the measure of the angle θ between u and v. Express the answer in radians rounded to two decimal places, if it is not possible to express it exactly.
The measure of the angle θ between u and v = 0.785radians
Step-by-step explanation:
The detailed steps is as shown in the attachment.