Answer:
(a) Probability that at least one component needs service during the warranty period = 0.4044.
(b) Probability that exactly one of the components needs service during the warranty period = 0.3419.
Step-by-step explanation:
Given A1 be the event that the receiver functions properly throughout the warranty period.
A2 be the event that the speakers function properly throughout the warranty period.
A3 be the event that the CD player functions properly throughout the warranty period.
Also P(A1) = 0.91, P(A2) = 0.85, and P(A3) = 0.77.
Now P(A1)' means that the receiver need service during the warranty period which is 1 - P(A1) = 1 - 0.91 = 0.09.
Similarly,P(A2)' =1 - P(A2) =1 - 0.85 =0.15 and P(A3)' =1 - P(A3)=1 - 0.77 = 0.23
Note: The ' sign on the P(A1) represent compliment of A1 or not A1.
(a) The probability that at least one component needs service during the warranty period = 1 - none of the component needs service during the warranty period
And none of the component needs service during the warranty period means that all the three components functions properly during the warranty period .
So, Probability that at least one component needs service during the warranty period = 1 - P(A1) x P(A2) x P(A3) = 1 - (0.91 x 0.85 x 0.77) = 0.4044.
(b) Now to find the Probability that exactly one of the components needs service during the warranty period, there would be three cases for this:
Receiver needs service and other two does not need during the warranty period.Speaker needs service and other two does not need during the warranty period.CD player needs service and other two does not need during the warranty period.And we have to add these three cases to calculate above probability.
Probability that exactly one of the components needs service during the warranty period = P(A1)' x P(A2) x P(A3) + P(A1) x P(A2)' x P(A3) + P(A1) x P(A2) x P(A3)'
= 0.09 x 0.85 x 0.77 + 0.91 x 0.15 x 0.77 + 0.91 x 0.85 x 0.23
= 0.3419.
Answer:
(a) P (At least one component needs service) = 0.404
(b) P (Either component A₁ or A₂ or A₃) = 0.997
Step-by-step explanation:
Given:
[tex]A_{1}=[/tex] Event that the receiver functions properly throughout the warranty period.
[tex]A_{2}=[/tex] Event that the speakers function properly throughout the warranty period.
[tex]A_{3}=[/tex] Event that the CD player functions properly throughout the warranty period.
[tex]P(A_{1})=0.91,\ P(A_{2})=0.85\ and\ P(A_{3})=0.77[/tex]
(a)
Compute the probability that at least one component needs service during the warranty period:
P (At least one component needs service) = 1 - P (None of the one component needs service)
= 1 - {P ([tex]A_{1}[/tex]) × P ([tex]A_{2}[/tex]) × P ([tex]A_{3}[/tex])}
[tex]=1 - (0.91\times0.85\times0.77)\\=1-0.595595\\=0.404405\\\approx0.404[/tex]
Thus, the probability that at least one component needs service during the warranty period is 0.404.
(b)
Compute the probability that exactly one of the components needs service during the warranty period, i.e. P (Either A₁ or A₂ or A₃):
[tex]P(A_{1}\cup A_{2}\cup A_{3})=P(A_{1})+P(A_{2})+P(A_{3})-P(A_{1}\cap A_{2})-P(A_{2}\cap A_{3})-P(A_{3}\cap A_{1})+P(A_{1}\cap A_{2}\cap A_{3})\\=P(A_{1})+P(A_{2})+P(A_{3})-[P(A_{1})\tmes P(A_{2})]-[P(A_{2})\tmes P(A_{3})]-[P(A_{3})\tmes P(A_{1})] +[P(A_{1})\tmes P(A_{2})\times P(A_{3})]\\=0.91+0.85+0.77-(0.91\times0.85)-(0.85\times0.77)-(0.77\times0.91)+(0.91\times0.85\times0.77)\\=0.996895\\\approx0.997[/tex]
Thus, the probability that exactly one of the components needs service during the warranty period is 0.997
Suppose an arrow is shot upward on the moon with a velocity of 44 m/s, then its height in meters after tt seconds is given by h(t)=44t−0.83t2h(t)=44t-0.83t2. Find the average velocity over the given time intervals.
a. [3, 4]:
b. [3, 3.5]:
c. [3, 3.1]:
d. [3, 3.01]:
e. [3, 3.001]
Answer:
a. 38.19m/s
b. 38.605m/s
c. 38.937m/s
d. 39.0117m/s
e. 39.01917m/s
Step-by-step explanation:
The average velocity is defined as the relationship between the displacement that a body made and the total time it took to perform it. Mathematically is given by the next formula:
[tex]v_a_v_g = \frac{\Delta x}{\Delta t} =\frac{x_f-x_i}{t_f-t_i}[/tex]
Where:
[tex]x_f=Final\hspace{3}distance\hspace{3}traveled\\x_i=Initial\hspace{3}distance\hspace{3}traveled\\t_f=Final\hspace{3}time\hspace{3}interval\\t_i=Initial\hspace{3}time\hspace{3}interval[/tex]
a. Let's find h(3) and h(4) using the data provided by the problem:
[tex]h(3)=44(3)-0.83(3^2)=124.53=x_i\\h(4)=44(4)-0.83(4^2)=162.72=x_f[/tex]
The average velocity over the interval [3, 4] is :
[tex]v_a_v_g=\frac{162.72-124.53}{4-3} =38.19m/s[/tex]
b. Let's find h(3.5) using the data provided by the problem:
[tex]h(3.5)=44(3.5)-0.83(3.5^2)=143.8325=x_f[/tex]
The average velocity over the interval [3, 3.5] is :
[tex]v_a_v_g=\frac{143.8325-124.53}{3.5-3} =38.605m/s[/tex]
c. Let's find h(3.1) using the data provided by the problem:
[tex]h(3.1)=44(3.1)-0.83(3.1^2)=128.4237=x_f[/tex]
The average velocity over the interval [3, 3.1] is :
[tex]v_a_v_g=\frac{128.4237-124.53}{3.1-3} =38.937m/s[/tex]
d. Let's find h(3.01) using the data provided by the problem:
[tex]h(3.1)=44(3.01)-0.83(3.01^2)=124.920117=x_f[/tex]
The average velocity over the interval [3, 3.01] is :
[tex]v_a_v_g=\frac{124.920117-124.53}{3.01-3} =39.0117m/s[/tex]
e. Let's find h(3.001) using the data provided by the problem:
[tex]h(3.001)=44(3.001)-0.83(3.001^2)=124.5690192=x_f[/tex]
[tex]v_a_v_g=\frac{124.5690192-124.53}{3.001-3} =39.01917m/s[/tex]
The average velocity over a given time [3,4] is 38.19 m/sec, the average velocity over a given time [3,3.5] is 38.605 m/sec, the average velocity over a given time [3,3.1] is 38.937 m/sec and this can be determined by using the given data.
Given :
Suppose an arrow is shot upward on the moon with a velocity of 44 m/s, then its height in meters after tt seconds is given by [tex]\rm h(t) = 44t-0.83t^2[/tex].
a) [3,4]
At time t = 3 and 4, the value of h(t) is given below:
[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]
[tex]\rm h(4) = 44(4)-0.83(4)^2 = 176-13.28=162.72[/tex]
The average velocity is given by:
[tex]\rm v = \dfrac{162.72-124.53}{4-3}=38.19[/tex]
b) [3,3.5]
At time t = 3 and 3.5, the value of h(t) is given below:
[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]
[tex]\rm h(3.5) = 44(3.5)-0.83(3.5)^2 = 154-10.1675=143.8325[/tex]
The average velocity is given by:
[tex]\rm v = \dfrac{143.8325-124.53}{3.5-3}=38.605[/tex]
c) [3,3.1]
At time t = 3 and 3.1, the value of h(t) is given below:
[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]
[tex]\rm h(4) = 44(3.1)-0.83(3.1)^2 = 136.4-7.9763=128.4237[/tex]
The average velocity is given by:
[tex]\rm v = \dfrac{128.4237-124.53}{3.1-3}=38.937[/tex]
d) [3,3.01]
At time t = 3 and 3.01, the value of h(t) is given below:
[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]
[tex]\rm h(3.01) = 44(3.01)-0.83(3.01)^2 = 132.44-7.519883=124.920117[/tex]
The average velocity is given by:
[tex]\rm v = \dfrac{124.920117-124.53}{3.01-3}=39.0117[/tex]
e) [3,3.001]
At time t = 3 and 3.001, the value of h(t) is given below:
[tex]\rm h(3) = 44(3)-0.83(3)^2 = 132-7.47=124.53[/tex]
[tex]\rm h(3.001) = 44(3.001)-0.83(3.001)^2 = 132.044-7.47498083=124.5690192[/tex]
The average velocity is given by:
[tex]\rm v = \dfrac{124.5690192-124.53}{3.001-3}=39.01917[/tex]
For more information, refer to the link given below:
https://brainly.com/question/18153640
Assume that 1700 births are randomly selected and 857 of the births are girls. Use subjective judgment to describe the number of girls as _______
a. significantly high
b. significantly low
c. neither significantly low nor significantly high.
Answer:
c. neither significantly low nor significantly high.
Step-by-step explanation:
We have been given that 1700 births are randomly selected and 857 of the births are girls. We are asked to choose the correct option about number of girls using subjective judgement.
We could expect that there would be 850 boys and 850 girls on average among 1700 births.
In our given scenario there are 857 girls, so there would be 843 boys. 857 girls compared to 850 girls are close.
Now probability of choosing 857 girls or more from 1700 births is approximately 0.5041 or 50%, therefore, the outcome is neither significantly low nor significantly high.
The number of girls in the study, which is about 50.4% of the total births, does not significantly deviate from the expected natural ratio of 100 girls to 105 boys, and therefore is described as neither significantly low nor significantly high.
Explanation:When we talk about the natural ratio of births, we refer to the expected proportion of girls to boys in a population, under natural circumstances without human intervention. According to Newsweek, the natural ratio is 100 girls to 105 boys. If we look at the provided study data where 1700 births were randomly selected and 857 of the births are girls, we need to determine if this proportion of girls is unusually high or low compared to the natural ratio.
The expected number of girls, according to the natural ratio, would be about 50% of the population, given there is no gender preference or other factors affecting the gender birth rate. In this case, the proportion of girls is 857/1700, or approximately 50.4% which is very close to the expected 50%. Therefore, using subjective judgment, we would describe the number of girls in this study as c. neither significantly low nor significantly high.
It is important to remember that small deviations from the expected ratio can occur due to natural variability, and the number observed in this study does not seem to deviate significantly from the natural ratio.
In an experimental study on friendliness and tipping, every alternate customer to whom the waiter is extra friendly toward are referred to as the...A. Control group
B. Experimental group
C. Nonexperimental group
D. Dependent group
Answer:
A. Control group
Step-by-step explanation:
A control group is a group in an experiment or study that does not receive experimental procedure during such that it is then used as a benchmark to measure how the other tested subjects do.
An experimental group is a group in an experiment or study that receives an experimental procedure. The values of gotten from the test are recorded and the effect of independent variables on the dependent variables are determined.
Control experiment can be used to determine whether or not the customers are friendly. Experimental group will be the customers whom the waiter is extra friendly toward. The control group will be the alternate customer whom the waiter is not extra friendly toward.
According to a recent report, 60% of U.S. college graduates cannot find a full time job in their chosen profession. Assume 57% of the college graduates who cannot find a job are female and that 18% of the college graduates who can find a job are female. Given a male college graduate, find the probability he can find a full time job in his chosen profession? (See exercise 58 on page 220 of your textbook for a similar problem.)
Answer:
There is a 55.97% that a male can find a full time job in his chosen profession.
Step-by-step explanation:
We have these following probabilities:
A 60% probability that a college graduates cannot find a full time job in their chosen profession.
A 40% probability that a college graduates can find a full time job in their chosen profession.
57% of the college graduates who cannot find a job are female
43% of the college graduates who cannot find a job are male
18% of the college graduates who can find a job are female
82% of the college who can find a job are male.
Given a male college graduate, find the probability he can find a full time job in his chosen profession?
The total males are 43% of 60%(Those who cannot find a job) and 82% of 40%(Those who can find a job). So the percentage of males is [tex]P(M) = 0.43*0.60 + 0.82*0.40 = 0.586[/tex]
Those who are males and find a job in their chosen profession are 82% of 40%. So [tex]P(M \cap J) = 0.82*0.40 = 0.328[/tex]
[tex]P = \frac{P(M \cap J)}{P(M)} = \frac{0.328}{0.586} = 0.5597[/tex]
There is a 55.97% that a male can find a full time job in his chosen profession.
Which technique for gathering data (observational study or experiment) do you think was used in the following studies? Explain your answer. (a) The U.S. Census Bureau tracks population age. In 1900, the percentage of the population that was 19 years old or younger was 44.4%. In 1930, the percentage was 38.8%; in 1970, the percentage was 37.9%; and in 2000, the percentage in the age group was down to 28.5% (The First Measured Century, T. Caplow, L. Hicks, B. J. Wattenberg). This is an experiment because a treatment was deliberately imposed on the individuals in order to observe a possible change in the response or variable being measured. This is an observational study because a treatment was deliberately imposed on the individuals in order to observe a possible change in the response or variable being measured. This is an experiment because observations and measurements of individuals are conducted in a way that doesn't change the response or the variable being measured. This is an observational study because observations and measurements of individuals are conducted in a way that doesn't change the response or the variable being measured. (b) After receiving the same lessons, a class of 100 students was randomly divided into two groups of 50 each. One group was given a multiple-choice exam covering the material in the lessons. The other group was given an essay exam. The average test scores for the two groups were then compared. This is an experiment because observations and measurements of individuals are conducted in a way that doesn't change the response or the variable being measured. This is an observational study because observations and measurements of individuals are conducted in a way that doesn't change the response or the variable being measured. This is an observational study because a treatment was deliberately imposed on the individuals in order to observe a possible change in the response or variable being measured. This is an experiment because a treatment was deliberately imposed on the individuals in order to observe a possible change in the response or variable being measured.
Answer:
a. 4)
b. 4)
Step-by-step explanation:
Hello!
An observational study is one where the investigator has no control or intervenes on it. He just defines the variable of interest and merely collects and documents the information.
An experimental study or experiment is one where the investigator intervenes by defining the variable of interest and artificially manipulates the study factor. It also one of its characteristics the randomization of cases or subjects in groups (two or more, depending on what is the hypothesis of study).
(a) The U.S. Census Bureau tracks population age. In 1900, the percentage of the population that was 19 years old or younger was 44.4%. In 1930, the percentage was 38.8%; in 1970, the percentage was 37.9%; and in 2000, the percentage in the age group was down to 28.5% (The First Measured Century, T. Caplow, L. Hicks, B. J. Wattenberg).
This is an experiment because treatment was deliberately imposed on the individuals in order to observe a possible change in the response or variable being measured. This is an observational study because treatment was deliberately imposed on the individuals in order to observe a possible change in the response or variable being measured. This is an experiment because observations and measurements of individuals are conducted in a way that doesn't change the response or the variable being measured. This is an observational study because observations and measurements of individuals are conducted in a way that doesn't change the response or the variable being measured.In this item, the researcher merely obtained the records of the population that were 19 or younger over several years and compared the obtained percentages. This is a clear example of an observational study, the researcher did not manipulate any factor or variable, he just looked up the information and documented it.
(b) After receiving the same lessons, a class of 100 students was randomly divided into two groups of 50 each. One group was given a multiple-choice exam covering the material in the lessons. The other group was given an essay exam. The average test scores for the two groups were then compared.
This is an experiment because observations and measurements of individuals are conducted in a way that doesn't change the response or the variable being measured. This is an observational study because observations and measurements of individuals are conducted in a way that doesn't change the response or the variable being measured. This is an observational study because treatment was deliberately imposed on the individuals in order to observe a possible change in the response or variable being measured. This is an experiment because treatment was deliberately imposed on the individuals in order to observe a possible change in the response or variable being measured.In this item, the students were randomly assigned to one of two groups and each group was given a test, group one: multiple choice and group two: essay. This is an example of an experimental study, after the class, the researcher controlled the variable "type of test" giving one type to each group and the subjects were randomly selected fro the groups assuring that they will not receive biased results. And at the end of the experiment, the response variables "test scores" where compared.
I hope it helps!
Final answer:
The studies presented about U.S. Census Bureau population tracking and exam type comparison in a classroom represent an observational study and an experiment, respectively. The former is so because no treatment was applied, while the latter involved purposefully assigning different tests to observe outcomes.
Explanation:
When classifying the given studies as either an observational study or an experiment, it's important to understand the key characteristics of each. An observational study involves monitoring subjects without any intervention, while an experiment involves the deliberate imposition of a treatment to observe potential changes.
(a) The U.S. Census Bureau tracking population age is an example of an observational study because the data on the percentage of the population that was 19 years old or younger in various years was collected without any manipulation or treatment being applied to the subjects.
(b) Dividing a class into two groups and administering different types of exams constitutes an experiment. This is because a treatment (the type of exam) was deliberately imposed on the students to observe the difference in test scores, which we interpret as the outcome of interest.
Suppose a statistics teacher wants to know whether the numberof hours students spend studying in a group affects the finalcourse grade. In each part, explain whether the research methoddescribed is a randomized experiment or an observationalstudy.
a) Each student keeeps a log of the hours he or she spendsstudying in a group and reports the total after the course iscompleted.
b) Students are randomly assigned to study groups. The teachertells each group how often to meet. This varies from one hour theday before each exam to two hours per week.
c) Students voluntarily join groups based on how often thegroups will meet. The groups are designated as meeting weekly,meeting only before exam,s or meeting whenever enough members feelthat it is necessary.
Answer:a) observational study
b) Randomization
c)Observational study
Step-by-step explanation:
a) The participants in (a) are monitored closely and data collected are reviewed.
b) Participants are randomly assigned to work and study.
a) The research method described is an observational study. b) The research method described is a randomized experiment. c) The research method described is an observational study.
Explanation:a) The research method described is an observational study. The students are simply keeping a log of the hours they spend studying in a group and reporting it after the course is completed. The researcher is not manipulating any variables or assigning students to different groups.
b) The research method described is a randomized experiment. The students are randomly assigned to study groups and the teacher tells each group how often to meet. This allows for the manipulation of the independent variable (number of hours studying in a group) and the measurement of its effect on the dependent variable (final course grade).
c) The research method described is an observational study. The students voluntarily join groups based on how often the groups will meet. The researcher is not manipulating any variables or assigning students to different groups. The students' choice of groups is a naturally occurring phenomenon that is being observed.
Compare A and B in three ways, where Aequals52 comma 698 is the number of deaths due to a deadly disease in the United States in 2005 and Bequals17 comma 481 is the number of deaths due to the same disease in the United States in 2009. a. Find the ratio of A to B. b. Find the ratio of B to A. c. Complete the sentence: A is ____ percent of B.
Answer:
(a) Ratio of A to B = 17566 : 5827
(b) Ratio of B to A = 5827 : 17566
(c) A is 301.46% percent of B.
Step-by-step explanation:
We are given A = 52,698 number of deaths due to a deadly disease in the United States in 2005 and B = 17,481 number of deaths due to the same disease in the United States in 2009.
(a) Ratio of A to B = [tex]\frac{A}{B}[/tex] = [tex]\frac{52,698}{17,481}[/tex] = 17566 : 5827
(b) Ratio of B to A = [tex]\frac{B}{A}[/tex] = [tex]\frac{17,481}{52,698}[/tex] = 5827 : 17566
(c) Let A is x% of B so the equation formed will be;
A = x% of B
52,698 = x% of 17481
Therefore x = [tex]\frac{52,698}{17,481}*100[/tex] = 301.46%
Hence, A is 301.46% of B.
The ratio of A to B is 3:1, the ratio of B to A is 1:3, and A is approximately 301.45 percent of B.
Explanation:a. Find the ratio of A to B:
We can find the ratio of A to B by dividing A by B, which gives us 52,698/17,481. Evaluating this division gives a ratio of approximately 3:1, meaning that for every 3 deaths in 2005, there was 1 death in 2009.
b. Find the ratio of B to A:
To find the ratio of B to A, we divide B by A, which gives us 17,481/52,698. Simplifying this division gives a ratio of approximately 1:3, which is the inverse of the previous ratio.
c. Complete the sentence: A is ____ percent of B:
To find the percentage of A relative to B, we divide A by B, then multiply by 100. Evaluating this division gives us (52,698/17,481) * 100, which is approximately 301.45%. Therefore, we can complete the sentence by saying 'A is approximately 301.45 percent of B'.
Learn more about Comparison of death rates here:https://brainly.com/question/32659927
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In a study of environmental lead exposure and IQ, the data was collected from 148 children in Boston, Massachusetts. Their IQ scores at age of 10 approximately follow a normal distribution with mean of 115.9 and standard deviation of 14.2. Suppose one child had an IQ of 74. The researchers would like to know whether an IQ of 74 is an outlier or not.
Calculate the lower fence for the IQ data, which is the lower limit value that the IQ score can be without being considered an outlier. Keep a precision level of two decimal places for the lower fence.
Answer:
a) Lower inner fence = 77.6168 = 77.62 to 2 d.p
Lower outer fence = 48.9044 = 48.90 to 2 d.p
b) The probability of obtaining an IQ score value of 74 or less is P(x ≤ 74) is 0.00159
Step-by-step explanation:
Lower inner and outer fences are used to illustrate or write off extreme values of a data set (the outliers).
Lower inner fence = Q₁ – (1.5 × IQR)
Lower outer fence = Q₁ – (3 × IQR)
Q₁ = 25th percentile = lower quartile
IQR = Inter quartile Range = Q₃ - Q₁
Q₃ = 75th percentile = upper quartile
To calculate Q₁ for a normal distribution with only mean and standard deviation known,
We need the standardized score whose probability is 0.25 P(z) = 0.25
From the normal distribution table
z = (± 0.674)
z = (x - xbar)/σ
x = the value in the data we're interested in,
xbar = mean = 115.9
σ = standard deviation = 14.2
Lower quartile corresponds to (z = - 0.674)
- 0.674 = (x - 115.9)/14.2
Q₁ = X = 106.3292
The upper quartile, Q₃ corresponds to z = (+0.674)
Q₃ = 125.4708
IQR = 125.4708 - 106.3292 = 19.1416
Lower inner fence = Q₁ – (1.5 × IQR)
Lower outer fence = Q₁ – (3 × IQR)
Lower inner fence = 106.3292 - (1.5 × 19.1416) = 106.3292 - 28.7124 = 77.6168
Lower outer fence = 106.3292 – (3 × 19.1416) = 48.9044
b) The probability of obtaining an IQ score value of 74 or less is P(x ≤ 74)
We standardize 74 by obtaining its z-score
z = (x - xbar)/σ
z = (74 - 115.9)/14.2 = - 2.95
P(x ≤ 74) = P(z ≤ -2.95) = 0.00159 (Obtained from normal distribution tables)
Final answer:
The lower fence for the IQ data, which determines whether an IQ score is an outlier, is calculated as the mean minus two times the standard deviation. In this case, the lower fence is 87.5, which makes an IQ score of 74 an outlier as it falls significantly below this threshold.
Explanation:
To determine if an IQ score is an outlier, we often use the interquartile range (IQR) and calculate the fences. However, since the data is approximately normally distributed and we have the mean and standard deviation, we can also consider an IQ score to be an outlier if it falls more than two standard deviations from the mean. In this question, we do not have the IQR, so we'll use standard deviations to calculate the outlier threshold.
The mean IQ score is 115.9 and the standard deviation is 14.2. An outlier is typically defined as a value that is more than two or three standard deviations away from the mean. These thresholds are sometimes called the outer fences in statistical outlier detection. Using two standard deviations, we can calculate the lower limit as follows:
Lower Limit = Mean - 2 × Standard Deviation
Lower Limit = 115.9 - 2(14.2)
Lower Limit = 115.9 - 28.4
Lower Limit = 87.5
Therefore, an IQ score of 74 is considerably lower than the lower limit of 87.5, suggesting that it could indeed be considered an outlier.
Two students, X and Y, forgot to put their names on their exam papers. The professor knows that these two students do well on the exam with probabilities 0.8 and 0.4, respectively. After grading, the professor notices that X and Y forgot to put their names on their exams. One of their exams was done well and the other was done poorly. Given this information, and assuming that students worked independently of each other, what is the probability that the good exam belongs to student X
Answer:
The probability that the good exam belongs to student X is 0.8571.
Step-by-step explanation:
It is provided that the probability that X did well in the exam is, P (X) = 0.90 and the probability that X did well in the exam is, P (Y) = 0.40,
Compute the probability that exactly one student does well in the exam as follows:
[tex]P(Either\ X\ or\ Y\ did\ well)=P(X\cap Y^{c})+P(X^{c}\cap Y)\\=P(X)P(Y^{c})+P(X^{c})P(Y)\\=P(X)[1-P(Y)]+[1-P(X)]P(Y)\\=(0.80\times0.60)+(0.20\times0.40)\\=0.56[/tex]
Then the probability that X is the one who did well in the exam is:
[tex]P(X\ did\ well\ in\ the\ exam)=\frac{P(X\cap Y^{c})}{P(X\cap Y^{c})+P(X^{c}\cap Y)}\\ =\frac{P(X)[1-P(Y)]}{P(X\cap Y^{c})+P(X^{c}\cap Y)} \\=\frac{0.80\times0.60}{0.56}\\=0.857143\\\approx0.8571[/tex]
Thus, the probability that the good exam belongs to student X is 0.8571.
Suppose that the probability of a baseball player getting a hit in an at-bat is 0.3089. If the player has 25 at-bats during a week, what's the probability that he gets greater than 9 hits?
Answer:
[tex] P(X>9) = 0.3593[/tex]
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=25, p=0.3089)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
For this case we want this probability:
[tex] P(X >9)[/tex]
And we can use the complement rule like this:
[tex] P(X>9) = 1-P(X \leq 8)= 1-[P(X=0) + P(X=1) +....+P(X=8)] [/tex]And we can find the individual probabilities like this:
[tex] P(X=0) =(25C0)(0.3089)^0 (1-0.3089)^{25-0} =0.0000974[/tex]
[tex] P(X=1) =(25C1)(0.3089)^1 (1-0.3089)^{25-1} =0.0011[/tex]
[tex] P(X=2) =(25C2)(0.3089)^2 (1-0.3089)^{25-2}=0.00584[/tex]
[tex] P(X=3) =(25C3)(0.3089)^3 (1-0.3089)^{25-3}= 0.02[/tex]
[tex] P(X=4) =(25C4)(0.3089)^4 (1-0.3089)^{25-4}=0.049[/tex]
[tex] P(X=5) =(25C5)(0.3089)^5 (1-0.3089)^{25-5}=0.092[/tex]
[tex]P(X=6)=(25C6) (0.3089)^6 (1-0.3089)^{25-6} = 0.138[/tex]
[tex] P(X=7) =(25C7)(0.3089)^7 (1-0.3089)^{25-7}=0.167[/tex]
[tex] P(X=8) =(25C8)(0.3089)^8 (1-0.3089)^{25-8}=0.168[/tex]
And in order to do the operations we can use the following excel code:
"=1-BINOM.DIST(8,25,0.3089,TRUE)"
And we got:
[tex] P(X>9) = 0.3593[/tex]
In one week, Mohammed can knit 5 sweaters or bake 240 cookies. In one week Aisha can knit 15 sweaters or bake 480 cookies. Mohammed's opportunity cost knitting one sweater is: A. 480 cookies. B. 240 cookies. C. 48 sweaters. D. 1/48 of a cookie E. 48 cookies.
Answer:
We conclude, Mohammed's opportunity cost knitting one sweater is 48 cookies.
Step-by-step explanation:
We have that Mohammed can knit 5 sweaters or bake 240 cookies.
In one week Aisha can knit 15 sweaters or bake 480 cookies.
We calculate how much is Mohammed's opportunity cost knitting one sweater. We get
\frac{240}{5}= 48.
We conclude, Mohammed's opportunity cost knitting one sweater is 48 cookies.
A sports statistician is interested in determining if there is a relationship between the number of home team and visiting team losses and different sports. A random sample of 526 games is selected and the results are given below. Find the critical value chi Subscript alpha Superscript 2 to test the claim that the number of home team and visiting team losses is independent of the sport. Use alphaequals0.01. Round to three decimal places.
Answer:
The critical value would be: [tex]\chi^2_{crit}=11.345[/tex] and we use the following excel code to find it: "=CHISQ.INV(1-0.01,3)"
[tex]p_v = P(\chi^2_{3} >3.29)=0.349[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(3.29,3,TRUE)"
Since the p value is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance that the two variables are independent.
Step-by-step explanation:
Previous concepts
A chi-square goodness of fit test "determines if a sample data matches a population".
A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".
Assume the following dataset:
F B S Bs Total
home wins 39 156 25 83 303
Visitor wins 31 98 19 75 223
Total 70 254 44 158 526
We need to conduct a chi square test in order to check the following hypothesis:
H0: The number of home team and visiting team losses is independent of the sport.
H1: The number of home team and visiting team losses is dependent of the sport.
The level of significance assumed for this case is [tex]\alpha=0.01[/tex]
The statistic to check the hypothesis is given by:
[tex]\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]
The table given represent the observed values, we just need to calculate the expected values with the following formula [tex]E_i = \frac{total col * total row}{grand total}[/tex]
And the calculations are given by:
[tex]E_{1} =\frac{70*303}{526}=40.32[/tex]
[tex]E_{2} =\frac{254*303}{526}=146.32[/tex]
[tex]E_{3} =\frac{44*303}{526}=25.35[/tex]
[tex]E_{4} =\frac{158*303}{526}=91.02[/tex]
[tex]E_{5} =\frac{70*223}{526}=29.68[/tex]
[tex]E_{6} =\frac{254*223}{526}=107.68[/tex]
[tex]E_{7} =\frac{44*223}{526}=18.65[/tex]
[tex]E_{8} =\frac{158*223}{526}=66.98[/tex]
And the expected values are given by:
F B S Bs Total
home wins 40.32 146.32 25.35 91.02 303
Visitor wins 29.68 107.68 18.65 66.98 223
Total 70 254 44 158 526
And now we can calculate the statistic:
[tex]\chi^2 = \frac{(39-40.32)^2}{40.32}+\frac{(156-146.32)^2}{146.32}+\frac{(25-25.35)^2}{25.35}+\frac{(83-91)^2}{91}+\frac{(31-29.68)^2}{29.68}+\frac{(98-107.68)^2}{107.68}+\frac{(19-18.65)^2}{18.65}+\frac{(75-66.98)^2}{66.98} =3.29[/tex]Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=(rows-1)(cols-1)=(4-1)(2-1)=3[/tex]
The critical value would be: [tex]chi^2_{crit}=11.345[/tex] and we use the following excel code to find it: "=CHISQ.INV(1-0.01,3)"
And we can calculate the p value given by:
[tex]p_v = P(\chi^2_{3} >3.29)=0.349[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(3.29,3,TRUE)"
Since the p value is higher than the significance level we FAIL to reject the null hypothesis at 5% of significance that the two variables are independent.
g 4. You start a dog walking business. Define to be a random variable denoting how many dogs you walkthis week. The probability mass function (pmf),f(x), o'is defined as follows:
0 1 2 3 4 5 6 7 P(X = x) 0.14 0.12 0.15 0.23 0.18 0.09 0.08 0.01
(a) Verify that f(x) is a valid probability mass function.
(b) Find the probability that you will walk at least two dogs this week.
(c) Compute the expected number of dogs you will walk this week.
(d) Compute the expected value of X2.
(e) Compute Var[x] = E(X2) – (E[X])2.
Answer:
(a) Yes, f(x) is a valid probability mass function.
(b) The probability that you will walk at least two dogs this week = 0.74.
(c) The expected number of dogs you will walk this week = 3 dogs.
(d) The expected value of X2 = 11.29
(e) Var[x] = E(X2) – (E[X])2 = 3.2811
Step-by-step explanation:
We are given with the probability mass function (pmf),f(x), o'is defined as follows:
Firstly let X = Number of dogs you walk this week
X P(X = x)
0 0.14
1 0.12
2 0.15
3 0.23
4 0.18
5 0.09
6 0.08
7 0.01
(a) Now f(x) to be a valid probability mass function, two conditions should be met :
All values should be >= 0.Sum of all probabilities must be equal to 1.So, First condition is already met as all values are positive and for second condition = 0.14 + 0.12+ 0.15+ 0.23+ 0.18+ 0.09+ 0.08+ 0.01 = 1
Hence both the conditions are satisfied so f(x) is a valid probability mass function.
(b) Probability that we will walk at least two dogs this week = P(X>=2)
= 1 - P(X = 0) - P(X = 1) = 1 - 0.14 - 0.12 = 0.74
(c) To Compute the expected number of dogs you will walk this week we will use expectation formula which says:
E(X) = [tex]\frac{\sum X\times P(X=x)}{\sum P(X=x)}[/tex] = [tex]\frac{0*0.14 + 1*0.12 + 2*0.15 + 3*0.23 + 4*0.18 + 5*0.09 + 6*0.08 + 7*0.01}{1}[/tex]
= 2.83 or 3 after rounding off.
Therefore the expected number of dogs you will walk this week are 3 dogs.
(d) The expected value of X2 [E(X2)] = [tex]\frac{\sum X^{2} \times P(X=x)}{\sum P(X=x)}[/tex]
= [tex]\frac{0^{2} *0.14 + 1^{2} *0.12 + 2^{2} *0.15 + 3^{2} *0.23 + 4^{2} *0.18 + 5^{2} *0.09 + 6^{2} *0.08 + 7^{2} *0.01}{1}[/tex] = 11.29
(e) Var[x] = E(X2) – (E[X])2
We have E(X2) = 11.29 and E(X) = 2.83
Var[x] = [tex]11.29 - (2.83)^{2}[/tex] = 3.2811.
The random variable X represents the number of dogs walked in a week for a dog walking business. By verifying that the sum of the probabilities equals 1, we've confirmed the pmf is valid. The probability of walking at least two dogs is 0.74, and calculations for the expected number and variance involve using the given probabilities and applying the respective formulas.
Explanation:Understanding Probability Mass Functions
In the context of a dog walking business, let us define the random variable X as the number of dogs walked in a week. The values that X can take on are 0, 1, 2, 3, 4, 5, 6, and 7. The probabilities of these events are given by P(X = x), where x are the respective values X can take.
To verify if the given probability mass function (pmf) is valid, the sum of all probabilities P(X = x) should be equal to 1. We calculate the sum as follows:
0.14 (for X=0)0.12 (for X=1)0.15 (for X=2)0.23 (for X=3)0.18 (for X=4)0.09 (for X=5)0.08 (for X=6)0.01 (for X=7)Adding them gives 1 which confirms it is a valid pmf.
To find the probability of walking at least two dogs this week, which is P(X ≥ 2), we need to add up the probabilities for all events where X is 2 or more:
P(X ≥ 2) = P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7)
= 0.15 + 0.23 + 0.18 + 0.09 + 0.08 + 0.01
= 0.74
To compute the expected number of dogs walked in a week, or E[X], we use the formula:
E[X] = Σ x * P(X = x)
For E[X2], we calculate the expectation of X-squared by multiplying each value of x by itself and then by its probability, before summing these products.
Finally, the variance Var[X] of the random variable X is given by E(X2) - (E[X])2. This involves first calculating E[X2] as explained, then squaring the previously found E[X] value, and subtracting that square from E[X2].
An experiment was performed upon rats to investigate the effect of ingesting Alar (a chemical sprayed on apple trees to keep fruit from dropping before ripe) upon subsequent cancer rates.
The following variables were measured:
gender (0=female, 1=male); weight (g); dose of Alar (nil, low, high); and number of tumors.
Which of the following is FALSE?
A) Gender is categorical; dose is ordinal
B) Gender is discrete; weight is continuous
C) Number of tumors is categorical
D) Dose is discrete
E) Weight is continuous
Answer:
Option C and D are false
Step-by-step explanation:
All the mentioned option are correct in the given scenario except option C and D.
The reason is that dose is categorized as nil, low and high so, dose is categorical variable. Also, number of tumors is quantitative variable because it can be meaningfully interpreted in numerical form. The number if tumors is discrete quantitative variable.
Now consider all options
A) Gender is categorical ; dose is ordinal
This option is true because gender can be categorized into male and female and also dose is ordinal because it has order i.e. nil,low and high.
B) Gender is discrete; weight is continuous
This option is false because gender can be a discrete variable and weight is continuous variable because it is measurable. So, the statement is true.
Option C and D are already discussed an option E is discussed in option B.
The false statement is C) Number of tumors is categorical. The number of tumors is a discrete variable that involves countable values, not a categorical variable.
Explanation:In this experiment, we have different types of variables. The statement C) Number of tumors is categorical is false. The number of tumors is actually a discrete variable as it involves countable values. The other statements are correct: A) Gender is categorical; dose is ordinal, as gender is divided into male and female, which is a categorical classification, and the dose is ranked as nil, low, high which makes it an ordinal variable. Statement B) Gender is discrete; weight is continuous is also correct because gender is a discrete variable (only two possible values, male or female), and weight is a continuous variable as it can take any value within a certain range. Statement D) Dose is discrete is correct, as the dose can only take certain values (nil, low, high) it is considered a discrete variable. Lastly, E) Weight is continuous is accurate as Weight can take any value within a range and it involves measurement.
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Ngoc needs to mix a 10% fungicide solution with a 50% fungicide solution to create 200 millileters of a 26% solution. How many millileters of each solution must Ngoc use?
Answer:
80milliliters of the 10% fungicide solution 170milliliters of the 50% fungicide solutionStep-by-step explanation:
let A represent the amount of the 10% fungicide solution
let B represent the amount of the 50% fungicide solution
B milliliters = 50% of B = (50/100)A = 0.5B
Total milliliters = 26% of 200 milliliters = 0.26 * 200 = 52
A milliliters + B milliliters = 200 milliliters
0.1A + 0.5B = 52
A + B = 2000.1A + 0.5B = 52
using substitution method to solve A and B
from equation 1 A = 200-Binsert A = 200 - B in equation 20.1(200-B) + 0.5B = 52
20 - 0.1B + 0.5B = 52
20 + 0.4B = 52
0.4B = 52 -20 = 32
0.4B = 32
B = 32/0.4 = 80
since B = 80
A = 200 -B = 200 - 80 = 120
80milliliters of the 10% fungicide solution 170milliliters of the 50% fungicide solutionAnswer: he must use 120 milliliters of the 10% solution and 80 milliliters of the 50% solution.
Step-by-step explanation:
Let x represent the amount of 10% fungicide solution that Ngoc must use.
Let y represent the amount of 50% fungicide solution that Ngoc must use.
The total volume of the fungicide solution that he wants to create is 200 milliliters. It means that
x + y = 200
Ngoc needs to mix a 10% fungicide solution with a 50% fungicide solution to create 200 millileters of a 26% solution. This means that
(10/100 × x) + (50/100 × y) = (26/100 × 200)
0.1x + 0.5y = 52 - - - - - - - - - -1
Substituting x = 200 - y into equation 1, it becomes
0.1(200 - y) + 0.5y = 52
20 - 0.1y + 0.5y = 52
- 0.1y + 0.5y = 52 - 20
0.4y = 32
y = 32/0.4 = 80 milliliters
x = 200 - y = 200 - 80
x = 120 milliliters
The standard deviation, the range, and the interquartile range (IQR) summarize the variability of the data. a. Why is the standard deviation usually preferred over the ranges? b. Why is the IQR sometimes preferred to the standard deviation? c. What is an advantage of the standard deviation over the IQR?
Answer:
Step-by-step explanation:
a) The standard deviation is usually preferred over the range because it is calculated from all of the data and will not be impacted as much as the range when they are outliers,and the standard deviation uses all of the data.
b)The IQR sometimes referred to the standard deviation when there is an outlier because the IQR is less sensitive to this features than standard deviation.
that is the IQR is not affected by an outlier,while the standard deviation is affected by an outlier.
c)The advantage of the standard deviation over the IQR is the standard deviation takes into account the values of all observation,while the IQR uses only some of the data.
An instructor in a college class recently gave an exam that was worth a total of 100 points. The instructor inadvertently made the exam harder than he had intended. The scores were very symmetric, but the average score for his students was 43 and the standard deviation of the scores was 5 points. The instructor is considering 2 different strategies for rescaling the exam results: Method 1:Add 17 points to everyone's score. Method 2:Multiply everyone's score by 1.7. Which of the following are true?
A) Method 1 will increase the standard deviation of the students' scores
B) Method 2 will increase the standard deviation of the students' scores.
C) Method 1 will decrease the standard deviation of the students' scores.
D) Method 2 will decrease the standard deviation of the students' scores.
Answer:
B) Method 2 will increase the standard deviation of the students' scores.
Step-by-step explanation:
Given that an instructor in a college class recently gave an exam that was worth a total of 100 points.
The average score for his students was 43 and the standard deviation of the scores was 5 points.
And now he is considering two different strategies for rescaling the exam results of which:
Method 1 = Add 17 points to everyone's score.
Method 2 = Multiply everyone's score by 1.7 .
And we have to check what will be the impact of these methods on the standard deviation of the students' scores.
For this let us consider a simple example to understand this:
Firstly, Formula for calculating Standard Deviation = [tex]\sqrt{\frac{\sum (X-Xbar)^{2}}{n-1}}[/tex]
Suppose,
X X - Xbar [tex](X - Xbar)^{2}[/tex]
3 3 - 6 = -3 -3 * -3 = 9
5 5 - 6 = -1 -1 * -1 = 1
10 10 - 6 = 4 4 * 4 = 16
Mean of above data, Xbar = [tex]\frac{3+ 5+10}{3}[/tex] = 6
Standard Deviation of data = [tex]\sqrt{\frac{26}{3-1} }[/tex] = 3.6055
Now let us suppose that we multiply each value of above data with 2 so the new data will be:
X X - Xbar [tex](X - Xbar)^{2}[/tex]
3*2 = 6 6 - 12 = -6 -6 * -6 = 36
5*2 = 10 10 - 12 = -2 -2 * -2 = 4
10*2 =20 20 - 12 = 8 8 * 8 = 64
Mean of new data, Xbar = [tex]\frac{6+ 10+20}{3}[/tex] = 12
Standard Deviation of new data = [tex]\sqrt{\frac{104}{3-1} }[/tex] = 7.2111
Hence, we see that when we multiply any value to the data the standard deviation will increase and in other words it will multiplied by that value which value we multiplied with each data value i.e. when we multiply each data value with 2 the standard deviation also get multiplied by as
3.6055 * 2 = 7.2111
Therefore option B is correct that Method 2 will increase the standard deviation of the students' scores.
And on the other hand Similarly by adding any constant to the data the Standard Deviation will remain same. Therefore Method 1 will have no impact on standard deviation of the students' scores.
Sally’s income has increased by 10% and she decides to change her consumption of macaroni and cheese from 10 boxes to 8 boxes. Her income elasticity of demand is ____ and the good is a _____ type of good.
Answer:
-2; Inferior good
Step-by-step explanation:
Given that,
Initial Quantity = 10 boxes
New Quantity = 8 boxes
Percentage increase in Sally's income = 10%
Change in consumption:
= 8 boxes - 10 boxes
= - 2 boxes
Percentage change in quantity demanded:
= (Change in quantity demanded ÷ Initial quantity) × 100
= (-2 ÷ 10) × 100
= - 20%
Therefore,
Income elasticity of demand:
= percentage change in quantity demanded ÷ Percentage change in income
= - 20% ÷ 10
= -2
Inferior goods are generally have a negative income elasticity of demand which means that an increase in the income of the consumer will lead to reduce the quantity demanded for inferior good and vice versa.
Hence, the good is a inferior type of good.
Use the concept thaty = c, −[infinity] < x < [infinity],is a constant function if and only ify' = 0to determine whether the given differential equation possesses constant solutions.9xy' + 5y = 10
Answer:
Yes, one of the solutions of this differential equation is a constant solution equation.
Step-by-step explanation:
9xy' + 5y = 10.
If y' = 0, 9xy' = 0
5y = 10
y = 2 = c.
So, for all real values of x such that -∞ < x < ∞, 9xy' will be 0 and one of the solutions of the differential equation will be y = 2.
Hope this helps!
A woman sued a computer keyboard manufacturer, charging that her repetitive stress injuries were caused by the keyboard. The injury awarded about $3.5 million for pain and suffering, but the court then set aside that award as being unreasonable compensation. In making this determination, the court identified a "normative" group of 27 similar cases and specified a reasonable award as one within two standard deviations of the mean of the awards in the 27 cases. The 27 awards were (in $1000s) 36, 62, 73, 114, 139, 140, 148, 154, 238, 290, 340, 410, 600, 750, 750, 750,1050, 1100, 1135, 1150, 1200, 1200, 1250, 1578, 1700, 1825, and 2000, from which?xi = 20,182, ?xi2 = 24,656,384.What is the maximum possible amount that could be awarded under the two-standard-deviation rule? (Round your answer to the nearest whole number.)
Answer:
variance = (27*24656384-20182^2)/(27*26) =368104.3
standard devaition SD= sqrt(368104.3) =606.716
maximum possible amount that could be awarded under the two-standard-deviation rule = mean +2*SD
= (20182/27)+(2*606.716)
= 1960.913
=$1960913
The perimeter of the window of the camper shell is 130 in. Find the length of one of the shorter sides of the window.
in
You can't deduce the length of a side from the perimeters of a rectangle.
Say that [tex]s[/tex] and [tex]S[/tex] are, respectively, the short and long side of the rectangle.
So, we know that
[tex]2s+2S=130 \iff 2(s+S)=130 \iff s+S=65[/tex]
But we can't solve exactly for [tex]s[/tex] nor for [tex]S[/tex], unless more information is given.
The length of one of the shorter sides of the window cannot be explicitly determined without additional information. However, considering the window has a rectangular shape, the length of the shorter side should be less than half of the total perimeter, in this case, less than 65 inches.
Explanation:To find the length of one of the shorter sides of the window, we must first understand that the perimeter of a rectangle is calculated by the formula: 2*length + 2*width = Perimeter.
However, the problem doesn't specify the dimensions of the box, but it does tell us that one pair of sides (the length and the width) are not equal. This suggests that the window is a rectangle. If we assume that the length of the window is longer than the width (length > width), then the shorter sides of the rectangle (the widths) will be of equal length.
Without more details, we can't calculate the exact measurement of the length of one of the shorter sides, but we can say that it should be less than half of the total perimeter, which is less than 65 inches.
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A simple model for the shape of a tsunami is given by dW/dx = W √(4 − 2W), where W(x) > 0 is the height of the wave expressed as a function of its position relative to a point offshore.
By inspection, find all constant solutions of the DE. (Enter your answers as a comma-separated list.)
Answer:
a) [tex] W=0,2[/tex]
b) [tex] W = 2 [1- tanh^2 (x+c)] = 2 sech^2 (x+c)[/tex]
Step-by-step explanation:
Part a
For this case we have the following differential equation:
[tex] W \sqrt{4-2W}=0[/tex]
If we square both sides we got:
[tex] W^2 (4-2W) =0[/tex]
And we have two possible solutions for this system [tex] W=0, W=2[/tex]
So then that represent the constant solutions for the differential equation.
So then the solution for this case is :
[tex] W=0,2[/tex]
Part b: Solve the differential equation in part (a)
For this case we can rewrite the differential equation like this:
[tex] \frac{dW}{dx} =W \sqrt{4-2W}[/tex]
And reordering we have this:
[tex] \frac{dW}{W \sqrt{4-2W}} = dx[/tex]
Integrating both sides we got:
[tex] \int \frac{dW}{W \sqrt{4-2W}} = \int dx[/tex]
Using CAS for the left part we got:
[tex] -tanh^{-1} (\frac{1}{2} \sqrt{4-2W})= x+c[/tex]
We can multiply both sides by -1 we got:
[tex] tanh^{-1} (\frac{1}{2} \sqrt{4-2W})=-x-c[/tex]
And we can apply tanh in both sides and we got:
[tex] \frac{1}{2} \sqrt{4-2W} = tanh(-x-c)[/tex]
By properties of tanh we can rewrite the last expression like this:
[tex]\frac{1}{2} \sqrt{4-2W} = -tanh(x+c)[/tex]
We can square both sides and we got:
[tex] \frac{1}{4} (4-2W) = tanh^2 (x+c) [/tex]
[tex] 1-\frac{1}{2}W = tanh^2 (x+c)[/tex]
And solving for W we got:
[tex] W = 2 [1- tanh^2 (x+c)] = 2 sech^2 (x+c)[/tex]
And that would be our solution for the differential equation
a new toy is regularly priced at $26.99 is on sale for 3/4 off. write an expression to represent the price oc the toy,p minus 34 of the price. then, combine like terms to simplify the expression
Answer:
[tex]T(p) = \frac{1}{4}p[/tex]
The new price is $6.75
Step-by-step explanation:
The new price of the toy after the discount (T) is given by the original price (p) subtracted by the discounted amount (3/4 of p):
[tex]T(p) = p-\frac{3}{4}p \\T(p) = (1-\frac{3}{4})p\\T(p) = \frac{1}{4}p[/tex]
If the original price was $26.99, the new price is:
[tex]T= \frac{1}{4}*\$26.99\\T=\$6.75[/tex]
A dead body was found within a closed room of a house where the temperature was a constant 65° F. At the time of discovery the core temperature of the body was determined to be 85° F. One hour later a second measurement showed that the core temperature of the body was 80° F. Assume that the time of death corresponds to t = 0 and that the core temperature at that time was 98.6° F. Determine how many hours elapsed before the body was found.
Answer:
1 hr 52 minutes
Step-by-step explanation:
As per Newton law of cooling we have
[tex]T(t) = T_s +(T_0-T_s)e^{-kt}[/tex]
where T0 is the initial temperature of the body
Ts = temperature of surrounding
t = time lapsed
k = constant
Using this we find that T0 = 98.6 : Ts= 65
Let x hours be lapsed before the body was found.
Then we have
[tex]T(x) = 65 +(98.6-65)e^{-kx} = 85\\e^{-kx}=\frac{20}{33.8} =0.5917[/tex]
Next after 1 hour temperature was 80
[tex]T(x+1) = 65+33.6(e^{-k(x+1)}=80\\e^{-k(x+1) =0.4464[/tex]
Dividing we get
[tex]e^k = 1.325408\\k = 0.2817[/tex]
Substitute this in
[tex]e^{-kx} =0.5917\\x=\frac{ln 0.5917}{-k} \\=1.863[/tex]
approximately 1 hour 52 minutes have lapsed.
Calculate the infant mortality rate (per 1,000 live births) from the following data:
a. Number of infant deaths under 1 year in the United States during 1991 = 36,766
b. Number of live births during 1991 = 4,111,000
Answer:
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Find the direction cosines and direction angles of the given vector. (Round the direction angles to two decimal places.) a = 5, 9, 3 cos(α) = cos(β) = cos(γ) = α = ° β = ° γ = °
Answer:
Step-by-step explanation:
given is a vector as (5,9,3)
a = (5,9,3)
To find out direction cosines
First let us calculate modulus of vector a
[tex]||a|| =\sqrt{5^2+9^2+3^2} \\=\sqrt{25+81+9} \\=\sqrt{115}[/tex]
Direction ratios are (5,9,3)
Magnitude of vector a = [tex]\sqrt{115}[/tex]
So direction cosines would be
[tex](\frac{5}{\sqrt{115} } ,\frac{9}{\sqrt{115} },\frac{3}{\sqrt{115} })[/tex]
Angles would be
[tex](\alpha, \beta, \gamma) = arccos ((\frac{5}{\sqrt{115} } ,\frac{9}{\sqrt{115} },\frac{3}{\sqrt{115} })[/tex]
=cos inverse (0.4662, 0.8393, 0.2798)
= (62.21, 32.93,32,94)
Find the distance between the points (-5, -10) and (2, 4).
Math item stem image
CLEAR CHECK
4.58
12.12
15.65
21
Answer:
15.65
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 (-5, -10) and (2, 4)
[tex]D = \sqrt{(2 - (-5))^{2} + (4 - (-10))^{2}}[/tex]
[tex]D = \sqrt{7^{2} + 14^{2}[/tex]
[tex]D = 15.65[/tex]
So the correct answer is:
15.65
In one type of state lottery 5 distinct numbers are picked from 1,2,3,...,40 uniformly at random.
(a) Describe a sample space and a probability measure P to model this experiment.
(b) What is the probability that out of the 5 picked numbers exactly three will be even?
The sample space consists of all sets of 5 numbers that can be picked from 40, and each outcome has an equal chance of occurring. The probability that exactly three out of five numbers are even is calculated as the combination of choosing three even numbers and two odd numbers, divided by the total number of outcomes.
Explanation:(a) In this case, the sample space consists of all possible outcomes of the experiment. There are 40 distinct numbers, and we're picking 5. The total number of distinct sets of 5 that can be chosen from 40 is given by the combination formula C(n, k) = n! / [k!(n - k)!]. So, in this case, C(40, 5), which would give the total number of outcomes. The probability measure P assigns each outcome a probability. Since we're choosing the numbers randomly and uniformly, each outcome has an equal probability of occurring, which is 1 / C(40, 5).
(b) To find the probability that exactly three out of the five numbers are even, we first note that there are 20 even numbers and 20 odd numbers in the set 1, 2, ..., 40. The number of ways to choose three even numbers is C(20, 3) and the number of ways to choose two odd numbers is C(20, 2). The probability of this happening is thus [C(20, 3) * C(20, 2)] / C(40, 5).
Learn more about Probability here:https://brainly.com/question/22962752
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Identify the type I error and the type II error that correspond to the given hypothesis. The percentage of adults who have a job is equal to 88 %. Identify the type I error. Choose the correct answer below. A. Fail to reject the null hypothesis that the percentage of adults who have a job is equal to 88 % when that percentage is actually different from 88 %. B. Fail to reject the null hypothesis that the percentage of adults who have a job is equal to 88 % when the percentage is actually equal to 88 %. C. Reject the null hypothesis that the percentage of adults who have a job is equal to 88 % when that percentage is actually equal to 88 %. D. Reject the null hypothesis that the percentage of adults who have a job is equal to 88 % when that percentage is actually different from 88 %. Identify the type II error. Choose the correct answer below. A. Reject the null hypothesis that the percentage of adults who have a job is equal to 88 % when that percentage is actually different from 88 %. B. Reject the null hypothesis that the percentage of adults who have a job is equal to 88 % when the percentage is actually equal to 88 %. C. Fail to reject the null hypothesis that the percentage of adults who have a job is equal to 88 % when the percentage is actually equal to 88 %. D. Fail to reject the null hypothesis that the percentage of adults who have a job is equal to 88 % when that percentage is actually different from 88 %.
Answer:
Type I error: The correct option is (C).
Type II error: The correct option is (D).
Step-by-step explanation:
The type-I-error is the probability of rejecting the null hypothesis when the null hypothesis is true.
The type-II-error is the probability of filing to reject the null hypothesis when in fact it is false.
The hypothesis in this problem can be defined as follows:
Null hypothesis (H₀): The percentage of adults who have a job is equal to 88%.
Alternate Hypothesis (Hₐ): The percentage of adults who have a job is different from 88%.
Type I error:The type-I-error in this case will be committed when we conclude that the percentage of adults who have a job is different from 88% when in fact it is equal to 88%.
Type II error:The type-II-error in this case will be committed when we conclude that the percentage of adults who have a job is equal to 88% when in fact it is different than 88%.
Indicate in standard form the equation of the line passing through the given points.
L(5.0), M(0,5)
Answer:
y = - x + 5
Step-by-step explanation:
L(5.0), M(0,5)
y = mx + b
m = (5 - 0) / (0 - 5) = 5 / -5 = - 1
b = y - mx = 5 - ((-1) x 0) = 5
y = - x + 5