Answer:
note:
solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment
Circle H is inscribed with quadrilateral D E F G. Angle E is 123 degrees. The measure of arc D E is 73 degrees. What is the measure of arc EF in circle H? 41° 50° 114° 173°
Answer:
41 degrees
Step-by-step explanation:
step 1
Find the measure of arc DGF
we know that
The inscribed angle is half that of the arc comprising
so
[tex]m\angle E=\frac{1}{2}[arc\ DGF][/tex]
we have
[tex]m\angle E=123^o[/tex]
substitute
[tex]123^o=\frac{1}{2}[arc\ DGF][/tex]
[tex]arc\ DGF=246^o[/tex]
step 2
Find the measure of arc EF
we know that
[tex]arc\ DGF+arc\ DE+arc\ EF=360^o[/tex] ----> by complete circle
substitute the given values
[tex]246^o+73^o+arc\ EF=360^o[/tex]
[tex]319^o+arc\ EF=360^o[/tex]
[tex]arc\ EF=360^o-319^o=41^o[/tex]
Answer: A)41 Degrees
Step-by-step explanation:
It was right on edu 2023
In a large, randomly mating population with no forces acting to change gene frequencies, the frequency of homozygous recessive individuals for the character extra-long eyelashes is 90 per 1000, or 0.09. What percentage of the population carries this trait but displays the dominant phenotype, short eyelashes
Answer:
16.38%
Step-by-step explanation:
Given
Frequency of homozygous recessive individuals for the character extra-long eyelashes is 90 per 1000
Let p = Probability of homozygous recessive individuals having character extra-long eyelashes = 0.09
p + q = 1 where q = Probability of homozygous recessive individuals not having character extra-long eyelashes
0.09 + p = 1
p = 1 - 0.09
p = 0.91
The frequency of the carrier individual is calculated by npq
Where n = mating partners = 2
Frequency = 2 * 0.09 * 0.91
Frequency = 0.1638 or 16.38%
Answer:
The percentage of the population carries this trait but displays the dominant phenotype, short eyelashes is the frequency of hetrozygous individuals which is
0.42 or 42 %
Step-by-step explanation:
To solve the question, we note that
p²+2pq+q²=1 and p + q =1
where
p = frequency of the dominant allele
q= frequency of the recessive allele
p² = frequency of homozygous dominant allele
q² = frequency of homozygous recessive allele
2·p·q = frequency of hetrozygous individuals
The frequency of the homozygous recessive individuals q² is 0.09
Therefore the frequency of q = √(0.09) = 0.3, therefore p = 1 - 0.3 =0.7
The percentage of the population carries this trait but displays the dominant phenotype, short eyelashes is the frequency of hetrozygous individuals or 2×p×q = Ae = 2*0.3*0.7 = 0.42
→ 0.42× 100 = 42 %
To assess the opinion of students at The Ohio State University about campus safety, a reporter for the student newspaper interviews 15 students she meets walking on the campus late at night who are willing to give their opinion. If the reporter instead decided to interview every fifth student that walks by, the method of sampling would be: a. simple random sampling. b. stratified random sampling. c. systematic sampling. d. a census.
Answer:
(C) Systematic Sampling
Step-by-step explanation:
The reporter decides to interview every fifth student of Ohio State University that passes by. This is referred to as Systematic Sampling.
In Systematic Sampling, the members of the population are place in an ordered list and the members of the sample are taken at a stated interval. This interval is called the Sampling Interval.
Although the members of the sample are random, they are chosen at stated intervals.
Of all airline flight requests received by a certain discount ticket broker, 90% are for domestic travel (D) and 10% are for international flights (I). Let x be the number of requests among the next three requests received that are for domestic flights. Assuming independence of successive requests, determine the probability distribution of x. (Hint: One possible outcome is DID, with the probability (0.9)(0.1)(0.9)
Answer:
Step-by-step explanation:
Given that of all airline flight requests received by a certain discount ticket broker, 90% are for domestic travel (D) and 10% are for international flights (I).
Let x be the number of requests among the next three requests received that are for domestic flights.
X can take values as 0,1, 2 or 3.
Since independence of successive requests is assumed we find that X having two outcomes and independence
X is binomial with n =3 and p= 0.90 (constant for each trial)
PDF of X would be
[tex]P(X=x) = 3Cx (0.9)^r (0.1)^{3-r} , r=0,1,2,3[/tex]
Thus pdf of X is
X 0 1 2 3
p 0.001 0.027 0.243 0.729
Final answer:
The probability distribution of x, where x is the number of domestic flight requests among the next three requests, follows a binomial distribution with n=3 and p=0.9. It can be calculated using the formula for binomial probabilities, considering the value of x can be 0, 1, 2, or 3. We get P = 0.081.
Explanation:
The question involves calculating the probability distribution of the random variable x, which represents the number of domestic flight requests among the next three requests received by a discount ticket broker. Since each request is independent and only two outcomes are possible (domestic or international), a binomial distribution can be used to determine the probabilities for x. For instance, the probability for one particular sequence of requests such as DID (domestic, international, domestic) is calculated as (0.9)(0.1)(0.9), which is the product of the probabilities for each request in that sequence. To find the probability distribution of x, we consider all possible sequences of three requests and their associated probabilities, which are determined by multiplying the individual probabilities for each request in the sequence following the rules of binomial distribution.
The random variable x can take on the following values: 0, 1, 2, or 3, representing the number of domestic flight requests out of three. We calculate probabilities for each of these values using the binomial formula:
[tex]P(x=0) = (0.1)^3\\P(x=1) = 3 * (0.9)(0.1)^2\\P(x=2) = 3 * (0.9)^2(0.1)\\P(x=3) = (0.9)^3[/tex]
P = 0.081.
A player of a video game is confronted with a series of opponents and has an 80% probability of defeating each one. Success with any opponent is independent of previous encounters. Until defeated, the player continues to contest opponents. a. What is the probability mass function of the number of opponents contested in a game? b. What is the probability that a player defeats at least two opponents in a game? c. What is the expected number of opponents contested in a game? d. What is the probability that a player contests four or more opponents in a game? e. What is the expected number of game plays until a player contests four or more opponents?
Answer:
(a) The PMF of X is: [tex]P(X=k)=(1-0.20)^{k-1}0.20;\ k=0, 1, 2, 3....[/tex]
(b) The probability that a player defeats at least two opponents in a game is 0.64.
(c) The expected number of opponents contested in a game is 5.
(d) The probability that a player contests four or more opponents in a game is 0.512.
(e) The expected number of game plays until a player contests four or more opponents is 2.
Step-by-step explanation:
Let X = number of games played.
It is provided that the player continues to contest opponents until defeated.
(a)
The random variable X follows a Geometric distribution.
The probability mass function of X is:
[tex]P(X=k)=(1-p)^{k-1}p;\ p>0, k=0, 1, 2, 3....[/tex]
It is provided that the player has a probability of 0.80 to defeat each opponent. This implies that there is 0.20 probability that the player will be defeated by each opponent.
Then the PMF of X is:
[tex]P(X=k)=(1-0.20)^{k-1}0.20;\ k=0, 1, 2, 3....[/tex]
(b)
Compute the probability that a player defeats at least two opponents in a game as follows:
P (X ≥ 2) = 1 - P (X ≤ 2)
= 1 - P (X = 1) - P (X = 2)
[tex]=1-(1-0.20)^{1-1}0.20-(1-0.20)^{2-1}0.20\\=1-0.20-0.16\\=0.64[/tex]
Thus, the probability that a player defeats at least two opponents in a game is 0.64.
(c)
The expected value of a Geometric distribution is given by,
[tex]E(X)=\frac{1}{p}[/tex]
Compute the expected number of opponents contested in a game as follows:
[tex]E(X)=\frac{1}{p}=\frac{1}{0.20}=5[/tex]
Thus, the expected number of opponents contested in a game is 5.
(d)
Compute the probability that a player contests four or more opponents in a game as follows:
P (X ≥ 4) = 1 - P (X ≤ 3)
= 1 - P (X = 1) - P (X = 2) - P (X = 3)
[tex]=1-(1-0.20)^{1-1}0.20-(1-0.20)^{2-1}0.20-(1-0.20)^{3-1}0.20\\=1-0.20-0.16-0.128\\=0.512[/tex]
Thus, the probability that a player contests four or more opponents in a game is 0.512.
(e)
Compute the expected number of game plays until a player contests four or more opponents as follows:
[tex]E(X\geq 4)=\frac{1}{P(X\geq 4)}=\frac{1}{0.512}=1.953125\approx 2[/tex]
Thus, the expected number of game plays until a player contests four or more opponents is 2.
Refer to the accompanying data set and use the 30 screw lengths to construct a frequency distribution. Begin with a lower class limit of 0.470 in, and use a class width of 0.010 in. The screws were labeled as having a length of 1 divided by 2 in. Does the frequency distribution appear to be consistent with the label? Why or why not?
Screw lengths (inches)
0.478
0.503
0.507
0.478
0.488
0.493
0.508
0.479
0.506
0.502
0.509
0.493
0.495
0.485
0.509
0.505
0.498
0.485
0.497
0.485
0.498
0.515
0.501
0.502
0.497
0.489
0.509
0.491
0.505
0.499
Complete the frequency distribution below.
Length (in)
Frequency
0.470 -
−
−
−
Answer:
A) Total frequency = 30
B) Yes, the distribution is consistent with the label because the frequencies are greatest when the lengths are closest to the labeled size of 1/2 inches which is 0.5 inches.
Step-by-step explanation:
Since the class limit width is 0.010 from the question, we arrive at;
Class Limits Frequency
Length(Inches)
0.470 - 0.479 3
0.480 - 0. 489 5
0.490 - 0.499 9
0.500 - 0.509 12
0.510 - 0.519 1
Adding the frequency, total = 30
The frequency distribution for the screws shows that most lengths are around 0.5 inches, consistent with the label. The classes and frequencies support this observation. The analysis is based on screw measurements provided.
To construct a frequency distribution of the screw lengths, we start with a lower class limit of 0.470 inches and use a class width of 0.010 inches. Now, let's count the number of screws in each class:
0.470 - 0.479: 30.480 - 0.489: 50.490 - 0.499: 100.500 - 0.509: 100.510 - 0.519: 2The frequency distribution is consistent with the given label of 0.5 inches as most of the data are centered around the 0.5 inch length.
A study of the checkout lines at the Safeway Supermarket in the South Strand area revealed that between 4 and 7 P.M. on weekdays there is an average of four customers waiting in line. What is the probability that you visit Safeway today during this period and find:
a. No customers are waiting?
b. Four customers are waiting?
c. Four or fewer are waiting?
d. Four or more are waiting?
Answer:
Step-by-step explanation:
Given that a study of the checkout lines at the Safeway Supermarket in the South Strand area revealed that between 4 and 7 P.M. on weekdays there is an average of four customers waiting in line.
Let X be the number of customers waiting in line
X is Poisson with parameter = 4
the probability that you visit Safeway today during this period and find:
a. No customers are waiting
[tex]P(X=0) = 0.018316[/tex]
b. Four customers are waiting?
[tex]=P(x=4) =0.193567[/tex]
c. Four or fewer are waiting?
=[tex]P(X\leq 4) = 0.628837[/tex]
d. Four or more are waiting
=[tex]P(X\geq 4)=0.56653[/tex]
You have one type of chocolate that sells for $2.40/lb and another type of chocolate that sells for $9.90/lb. You would like to have 30 lbs of a chocolate mixture that sells for $4.20/lb. How much of each chocolate will you need to obtain the desired mixture
Final answer:
To obtain the desired mixture, you will need 24 lbs of the first type of chocolate and 6 lbs of the second type of chocolate.
Explanation:
To find the amount of each chocolate needed to obtain the desired mixture, we can set up a system of equations. Let's say x represents the pounds of the first type of chocolate and y represents the pounds of the second type of chocolate. We know that the total weight of the mixture is 30 lbs, so we have the equation:
x + y = 30
We also know that the cost per pound of the mixture is $4.20, so the total cost is:
2.40x + 9.90y = 4.20 * 30
Simplifying the second equation, we get:
2.40x + 9.90y = 126
We can solve this system of equations using substitution, elimination, or a calculator. The solution is x = 24 lbs and y = 6 lbs. Therefore, you will need 24 lbs of the first type of chocolate and 6 lbs of the second type of chocolate to obtain the desired mixture.
A branch of a certain bank has six ATMs. Let X represent the number of machines in use at a particular time of day. The cdf of X is as follows: F(x) = 0 x < 0 0.08 0 ≤ x < 1 0.20 1 ≤ x < 2 0.40 2 ≤ x < 3 0.62 3 ≤ x < 4 0.83 4 ≤ x < 5 0.99 5 ≤ x < 6 1 6 ≤ x
Final answer:
The question pertains to calculating probabilities from a given discrete Cumulative Distribution Function (CDF) for the number of ATMs in use. By utilizing the step function provided in the CDF, probabilities for specific intervals or exact values can be determined.
Explanation:
The student's question involves understanding the Cumulative Distribution Function (CDF) for a discrete random variable, specifically the number of ATMs in use at a particular time.
The CDF is used to find the probability that the random variable X, representing the number of ATMs used, is less than or equal to a certain value x.
In this scenario, we have a step function showing the probabilities associated with different intervals of x. The function increases as x increases, which is typical of a CDF, as it accumulates the probabilities for all values up to x.
To calculate probabilities for specific intervals from the given CDF, one would subtract the CDF value just before the interval from the CDF value at the end of the interval. For example, to find the probability that exactly four ATMs are in use (P(X=4)), one would calculate P(X<5) - P(X<4), which is 0.83 - 0.62 = 0.21.
[6 points] Prove Bayes’ Theorem. Briefly explain why it is useful for machine learning problems, i.e., by converting posterior probability to likelihood and prior probability.
Answer: Mathematically Bayes’ theorem is defined as
P(A\B)=P(B\A) ×P(A)
P(B)
Bayes theorem is defined as where A and B are events, P(A|B) is the conditional probability that event A occurs given that event B has already occurred (P(B|A) has the same meaning but with the roles of A and B reversed) and P(A) and P(B) are the marginal probabilities of event A and event B occurring respectively.
Step-by-step explanation: for example, picking a card from a pack of traditional playing cards. There are 52 cards in the pack, 26 of them are red and 26 are black. What is the probability of the card being a 4 given that we know the card is red?
To convert this into the math symbols that we see above we can say that event A is the event that the card picked is a 4 and event B is the card being red. Hence, P(A|B) in the equation above is P(4|red) in our example, and this is what we want to calculate. We previously worked out that this probability is equal to 1/13 (there 26 red cards and 2 of those are 4's) but let’s calculate this using Bayes’ theorem.
We need to find the probabilities for the terms on the right-hand side. They are:
P(B|A) = P(red|4) = 1/2
P(A) = P(4) = 4/52 = 1/13
P(B) = P(red) = 1/2
When we substitute these numbers into the equation for Bayes’ theorem above we get 1/13, which is the answer that we were expecting.
Bayes’ theorem, featuring components of likelihood and prior probability, is fundamental to probability theory and statistics. It's a tool that allows us to update our previous assumptions based on new data, making it essential in machine learning.
Explanation:Bayes’ theorem is a fundamental theorem in probability theory and statistics, named after Thomas Bayes. It formalizes the process of updating probabilities based on new data. The theorem can be derived from the definition of conditional probability, P(A|B), which is the probability of event A given that event B has occurred.
Bayes' theorem is generally represented by the formula:
P(A|B) = (P(B|A) * P(A)) / P(B)
Bayes’ theorem has two main components:
Likelihood: P(B|A) shows the likelihood of the data under a particular hypothesis.Prior Probability: P(A) represents our prior belief in the hypothesis before seeing any data.In the realm of Machine Learning, Bayes' theorem allows us to update our previous assumptions (priors) with data to get updated, more accurate results (posterior). The prior and posterior are probability distributions over the same event space, but Bayesian inference allows us to adjust our initial assumptions in light of new evidence, making it a dynamic and flexible method.
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A dishwasher has a mean life of 12 years with an estimated standard deviation of 1.25 years ("Appliance life expectancy," 2013). Assume the life of a dishwasher is normally distributed. Find the number of years that the bottom 25% of dishwasher would last.
Answer:
[tex]z=-0.674<\frac{a-12}{1.25}[/tex]
And if we solve for a we got
[tex]a=12 -0.674*1.25=11.157[/tex]
So the value of height that separates the bottom 25% of data from the top 75% is 11.157.
Step-by-step explanation:
Previous concepts
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".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the mean life of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(12,1.25)[/tex]
Where [tex]\mu=12[/tex] and [tex]\sigma=1.25[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.75[/tex] (a)
[tex]P(X<a)=0.25[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.25[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.25[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-0.674<\frac{a-12}{1.25}[/tex]
And if we solve for a we got
[tex]a=12 -0.674*1.25=11.157[/tex]
So the value of height that separates the bottom 25% of data from the top 75% is 11.157.
The lifespan of the dishwashers in the bottom 25% is approximately 11.16 years, found by calculating the corresponding value to the 25th percentile z-score from the normal distribution of the appliance's lifespan.
Explanation:The problem is asking for the life of the dishwasher that falls in the bottom 25% of the normally distributed lifespan range. In other words, we need to find the dishwasher lifespan that is the cutoff for the bottom quartile. For a normal distribution, the 25th percentile (the cutoff for the bottom quartile) is commonly found by using a z-score, which is a measure of how many standard deviations a value is from the mean.
The z-score for the 25th percentile is typically -0.675. We can find the lifespan corresponding to this z-score using the formula: lifespan = mean + z*sd. Here, 'mean' is the average lifespan, 'z' is the z-score, and 'sd' is the standard deviation.
Therefore, substituting the known values into the formula, we get: lifespan = 12 + (-0.675)*1.25 = 11.15625 years.
Therefore, the number of years the bottom 25% of dishwashers would last is approximately 11.16 years.
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The American Community Survey is an ongoing survey that provides data every year to give communities the current information they need to plan investments and services. The 2010 American Community Survey estimates that 14.6% of Americans live below the poverty line, 20.7% speak a language other than English (foreign language) at home, and 4.2% fall into both categories.
a. Draw a Venn diagram summarizing the variables and their associated probabilities
b. What percent of Americans live below the poverty line and only speak English at home?
c. What percent of Americans live below the poverty line or speak a foreign language at home?
Answer:
b. 10.4
c. 26.9
Step-by-step explanation:
Let the universal set U = 100% which is the total no of people in the American community
Let A = 14.6% which is the total no of people living below poverty line
Let B = 20.7% which is the total no of people speaking foreign Language
C = 4.2% no of people who both speak foreign language and live below poverty line
X = no of people who neither live below poverty line nor speak foreign language
P (A) = 14.6%
P (B) = 20.7%
P (C) = P (A ∩ B) = 4.2%
P (A – C) = P (A ∩ U) = 14.6 – 4.2 = 10.4%
P (B – C) = P (B ∩ U) = 20.7 – 4.2 = 16.5%
P (X) = P (A ᴜ B) c =100 – (10.4 + 4.2 + 16.5) = 68.9%
a. The venn diagram is as shown above
b. Percent of Americans who live below poverty line and Speak English at home(minus foreign lang speakers living below poverty line) that is A only
= A – C
= 14.6 – 4.2
= 10.4%
c. Percentage of Americans Living below poverty line or Speaking foreign language
= A only + B only
A only = A – C ( People living below poverty line only)
= 14.6 -4.2
= 10.4%
B only = B – C ( people speaking foreign languages only)
= 20.7 – 4.2
= 16.5%
Hence
A only + B only = 10.4 + 16.5 = 26.9%
In 1906 Kennelly developed a simple formula for predicting an upper limit on the fastest time that humans could ever run distances from 100 yards to 10 miles. His formula is given by t equals . 0588 s Superscript 1.125 where s is the distance in meters and t is the time to run that distance in seconds. a. Find Kennelly's estimate for the fastest a human could possibly run 1606 meters. tequals nothing seconds (Round to the nearest thousandth as needed.) b. Find StartFraction dt Over ds EndFraction when sequals20 and interpret your answer. StartFraction dt Over ds EndFraction almost equals nothing sec/m (Round to the nearest thousandth as needed.) When the distance is 20 meters, this rate gives the number of seconds per meter ▼ by which the fastest possible time is increasing. by which the fastest possible time is decreasing. that the fastest human could possibly run.
Answer / Step-by-step explanation:
(1) Given t = 0.0588s ¹.¹²⁵
where s is the distance and t is the time to run that distance.
The second portion asks us to find the derivative of the equation when our s value is equal to 20 and interpret.
(2) First, we try to convert the unit from miles to meters
Therefore, 1 mile = 1609 meters
Then,
t = 0.0588 ( 1609 ) ¹.¹²⁵
=238 . 09
This gives us the instantaneous rate of change of seconds between every 20 meters ran.
The last portion asks us to compare this estimate to current world records. And have they been surpassed?
As of today, the fastest official record for a standard mile is held by a man from Morocco named Hichan El Guerrouj. The time was recorded at 3.43 minutes in Rome, Italy on July 7th, 1999.
Now, keep in mind that this is almost a full minute slower than the estimated time. However, how do these projections hold up against Usain Bolt, the man that is considered the fastest man in the world ?
Although, Usain Bolt does run long distances, he holds records in nearly every sprinting event that he has ever competed in.
Hence, Kennelly's estimate for the fastest mile is 238.09
(3) Now, noting that since dt / ds = 0.0588 ( 1.25 ) s ⁰.¹²⁵
Then,
dt / ds I 100 = 0.0588 (1.25) (20) ⁰.¹²⁵
= 0.1176
Two cars are driving towards an intersection from perpendicular directions. The first car's velocity is 101010 meters per second and the second car's velocity is 666 meters per second. At a certain instant, the first car is 444 meters from the intersection and the second car is 333 meters from the intersection. What is the rate of change of the distance between the cars at that instant (in meters per second)
The rate of change of the distance between the two cars at that instant is 100344 meters per second.
Here, we have,
To find the rate of change of the distance between the two cars at a certain instant, we can use the concept of relative velocity.
The relative velocity between the two cars is the difference between their velocities.
Since the cars are moving towards each other, the relative velocity will be the sum of their individual velocities.
Let's denote the distance between the two cars at the certain instant as
d (in meters). The rate of change of this distance d with respect to time (t) is the derivative of d with respect to t (i.e., d/dt).
At the given instant, the first car's velocity (V₁) is 101010 meters per second, and the second car's velocity (V₂) is 666 meters per second.
The distance traveled by the first car (d₁) at the given instant is 444 meters, and the distance traveled by the second car (d₂) is 333 meters.
The distance between the two cars at the given instant (d) is the difference between the distances traveled by the first and second cars:
d=d₁−d₂
Now, let's find the rate of change of d with respect to time (t):
d/dt = d₁/dt - d₂/dt
Since the velocities of the cars are constants (not changing with time), the derivatives of d₁ and d₂ with respect to time are simply their velocities:
d/dt =V₁−V₂
Now, substitute the given values for V₁ and V₂:
d/dt=101010meters per second−666meters per second
d/dt =100344meters per second
Therefore, the rate of change of the distance between the two cars at that instant is 100344 meters per second.
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The rate of change of the distance between the cars at the given instant is 1210 m/s. We used the Pythagorean theorem to express the distance between the two cars at any time 't' and then differentiated this distance with respect to 't' to get the relative speed of the cars.
Explanation:To calculate the rate of change of the distance between the two cars at a certain instance, we can use the Pythagorean theorem. Consider the two cars as points moving along perpendicular paths. At any time 't', car 1 is '1010t' m away from the intersection and car 2 is '666t' m away from the intersection. The distance 'd' between them is √[(1010t)² + (666t)²] which simplifies to d = 1210t.
The rate of change of the distance with respect to time, i.e., the derivative of 'd' with respect to 't' (dd/dt) would be the relative speed of the two cars at any instant. This could be found by differentiating 'd' with respect to 't'. Thus, at the moment when car 1 is 444 meters away from the crossing and car 2 is 333 meters away from the crossing, the rate of change of the distance between the cars (the relative velocity) at that instant would be d/dt = 1210 m/s.
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(1 point) Suppose we will flip a fair coin 100 times. (a) What does 35 heads correspond to on the standard scale? (enter exact answer) (b) What does z=2.4 on the standard scale correspond to on the number of heads scale? (enter exact answer)
Answer:
(a) 35 heads correspond to -3 on the standard scale.
(b) z = 2.4 corresponds to 62 heads on the number of heads scale.
Step-by-step explanation:
(a) If we flip a fair coin once, probability of getting head = 0.5
If we flip a fair coin 100 times, mean number of heads = 100(0.5) = 50
If there are N draws with a P probability of success, the standard deviation (SD) is given as:
[tex]SD = \sqrt{(N)(P)(1 - P)}[/tex]
Here, the probability of getting a head (P) is 0.5 while the number of draws (N) is 100. So,
[tex]SD = \sqrt{(100)(0.5)(1-0.5)}[/tex]
SD = 5
The standard scale value is: (35 - 50) / 5 = -3
Hence, 35 heads correspond to -3 on the standard scale.
(b) The standard scale value is 2.4 and we need to find the number of heads.
(X - 50) / 5 = 2.4
X - 50 = 12
X = 62
Hence, z = 2.4 on the standard scale corresponds to 62 on the number of heads scale.
The 35 heads corresponds to -3 on the standard scale, and the number of heads is 62
The head on the standard scale
The given parameter is:
n = 100
In a flip of a coin, the probability of a head is:
p = 1/2
So, the mean of the distribution is:
[tex]\bar x = np[/tex]
[tex]\bar x = 100 * 1/2[/tex]
[tex]\bar x = 50[/tex]
The standard deviation is:
[tex]\sigma = \sqrt{np(1 - p)}[/tex]
So, we have:
[tex]\sigma = \sqrt{100 * 1/2 * (1 - 1/2)}[/tex]
[tex]\sigma = 5[/tex]
The corresponding value on the standard scale is then calculated as:
[tex]z= \frac{x - \mu}{\sigma}[/tex]
For 35 heads, we have:
[tex]z= \frac{35 - 50}{5}[/tex]
[tex]z = -3[/tex]
Hence, the 35 heads corresponds to -3 on the standard scale
(b) The number of heads
We have:
[tex]z= \frac{x - \mu}{\sigma}[/tex]
When z = 2.4, the equation becomes
[tex]2.4= \frac{x - 50}{5}[/tex]
Multiply both sides by 5
[tex]12= x - 50[/tex]
Add 50 to both sides
[tex]x = 62[/tex]
Hence, the number of heads is 62
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The temperature at a point (x,y,z) is given by T(x,y,z)=200e^-x^2-3y^-9z^2, where T measured in degrees Celsius and x,y,z in meters. Find the rate of change of temperature at the point P(2, -1, 2) in the direction toward the point (3, -3, 3)
Answer:
Therefore the rate change of temperature at the point P(2,-1,2) in the direction toward the point (3,-3,3) is [tex]-\frac{5200\sqrt6}{3}e^{-43}[/tex] °C/m.
Step-by-step explanation:
Given that, the temperature at a point (x,y,z) is
[tex]T(x,y,z)=200e^{-x^2-3y^2-9z^2}[/tex].
Rate change of temperature at the point P(2,-1,2) in the direction toward the point Q (3,-3,3) is [tex]D_uT(2,-1,2)[/tex]
[tex]T_x= 200(-2x)e^{-x^2-3y^2-9z^2}[/tex][tex]=-400x200.2xe^{-x^2-3y^2-9z^2}[/tex]
[tex]T_y = 200.(-6y)e^{-x^2-3y^2-9z^2}[/tex][tex]=-1200ye^{-x^2-3y^2-9z^2}[/tex]
[tex]T_z=200(-18z)e^{-x^2-3y^2-9z^2}[/tex][tex]=-3600ze^{-x^2-3y^2-9z^2}[/tex]
The gradient of the temperature
[tex]\bigtriangledown T(x,y,z)= (-400x200.2xe^{-x^2-3y^2-9z^2},-1200ye^{-x^2-3y^2-9z^2},-3600ze^{-x^2-3y^2-9z^2})[/tex] [tex]=-400e^{-x^2-3y^2-9z^2}(x,3y,9z)[/tex]
[tex]\bigtriangledown T(2,-1,2) = -400e^{-2^2-3(-1)^2-9.2^2}(2,3.(-1),9.2)[/tex]
[tex]=-400 e^{-43}(2,-3,18)[/tex]
V=[tex]\overrightarrow {PQ}= \vec{Q}-\vec{P}[/tex]=(3,-3,3)-(2,-1,2)=(1,-2,1)
The unit vector of V is [tex]\frac{(1,-2,1)}{\sqrt{1^2+(-2)^2+1^2}}[/tex]
[tex]=\frac{1}{\sqrt6}(1,-2,1)[/tex]
Therefore,
[tex]D_uT(2,-1,2) = \bigtriangledown T(2,-1,2).\frac{1}{\sqrt6}(1,-2,1)[/tex]
[tex]=-400 e^{-43}(2,-3,18).\frac{1}{\sqrt6}(1,-2,1)[/tex]
[tex]=-\frac{400}{\sqrt6}e^{-43}(2.1+(-3).(-2)+18.1)[/tex]
[tex]=-\frac{10400}{\sqrt6}e^{-43}[/tex]
[tex]=-\frac{5200\sqrt6}{3}e^{-43}[/tex] °C/m
Therefore the rate change of temperature at the point P(2,-1,2) in the direction toward the point (3,-3,3) is [tex]-\frac{5200\sqrt6}{3}e^{-43}[/tex] °C/m.
A 16-inch candle is lit and burns at a constant rate of 0.8 inches per hour. Let t represent the number of hours since the candle was lit, and suppose f is a function such that f(t) represents the remaining length of the candle (in inches) t hours after it was lit. a. Write a function formula for f. f(t)- 16-0.8t Previewb. What is the domain of f relative to this context? c. What is the range of f relative to this context?
Answer:
See the explanation.Step-by-step explanation:
a.
The initial length of the candle is 16 inch. It also given that, it burns with a constant rate of 0.8 inch per hour.
After one hour since the candle was lit, the length of the candle will be (16 - 0.8) = 15.2 inch.
After two hour since the candle was lit, the length of the candle will be (15.2 - 0.8) = 14.4 inch. The length of the candle after two hours can also be represented by {16 - 2(0.8)}.
Hence, the length of the candle after t hours when it was lit can be represented by the function, [tex]f(t) = 16 - 0.8t[/tex]. [tex]f(t) = 0[/tex] at t = 20.
b.
The domain of the function is 0 to 20.
c.
The range is 0 to 16.
The function formula for the given context is f(t) = 16 - 0.8*t. The domain of this function is from 0 to 20 hours, and the range is from 0 to 16 inches.
Explanation:The function you want to define for the scenario presented represents the remaining length, in inches, of the candle after being lit for a certain number of hours (t). Given that the initial length of the candle is 16 inches and burns at the constant rate of 0.8 inches per hour, the function will follow a linear format: f(t) = 16 - 0.8*t.
The domain of this function, referring to the permissible values for t in this context, would be any value greater than or equal to 0 and less than or equal to 20 (since it would take 20 hours for the candle to completely burn down). Therefore, the domain of this function is [0, 20].
The range of this function will be the possible lengths the candle could have, depending on the time it has burned. As the candle was 16 inches long initially and reduces to 0 after burning for 20 hours, the range lies between 0 and 16 (inclusive). Thus, the range of this function is [0, 16].
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What is the solution
2 – 3y=1
x – 2y = 6
Answer: x = - 16
y = - 11
Step-by-step explanation:
The given system of linear equations is expressed as
2x - 3y = 1- - - - - - - - - - - - -1
x - 2y = 6 - - - - - - - - - - - -2
We would eliminate x by multiplying equation 1 by 1 and equation 2 by 2. It becomes
2x - 3y = 1- - - - - - - - - - -3
2x - 4y = 12- - - - - - - - - - - -4
Subtracting equation 4 from equation 3, it becomes
y = - 11
Substituting y = - 11 into equation 2, it becomes
x - 2 × - 11 = 6
x + 22 = 6
Subtracting 22 from the left hand side and the right hand side of the equation, it becomes
x + 22 - 22 = 6 - 22
x = - 16
According to Scarborough Research, more than 85% of working adults commute by car. Of all U.S. cities, Washington, D.C., and New York City have the longest commute times. A sample of 25 commuters in the Washington, D.C., area yielded the sample mean commute time of 27.97 minutes and sample standard deviation of 10.04 minutes. Construct and interpret a 99% confidence interval for the mean commute time of all commuters in Washington D.C. area.
Answer:
The 99% confidence interval for the mean commute time of all commuters in Washington D.C. area is (22.35, 33.59).
Step-by-step explanation:
The (1 - α) % confidence interval for population mean (μ) is:
[tex]CI=\bar x\pm z_{\alpha /2}\frac{\sigma}{\sqrt{n}}[/tex]
Here the population standard deviation (σ) is not provided. So the confidence interval would be computed using the t-distribution.
The (1 - α) % confidence interval for population mean (μ) using the t-distribution is:
[tex]CI=\bar x\pm t_{\alpha /2,(n-1)}\frac{s}{\sqrt{n}}[/tex]
Given:
[tex]\bar x=27.97\\s=10.04\\n=25\\t_{\alpha /2, (n-1)}=t_{0.01/2, (25-1)}=t_{0.005, 24}=2.797[/tex]
*Use the t-table for the critical value.
Compute the 99% confidence interval as follows:
[tex]CI=27.97\pm 2.797\times\frac{10.04}{\sqrt{25}}\\=27.97\pm5.616\\=(22.354, 33.586)\\\approx(22.35, 33.59)[/tex]
Thus, the 99% confidence interval for the mean commute time of all commuters in Washington D.C. area is (22.35, 33.59).
You are collecting a sample of 60 data points from a population that you know to follow an exponential distribution. It is not appropriate to use the results of the Central Limit Theorem (CLT) to calculate the confidence intervals for the mean.
Answer:
For the exponential distribution:
[tex] \mu = \frac{1}{\lambda}[/tex]
[tex] \sigma^2 = \frac{1}{\lambda^2}[/tex]
We know that the exponential distribution is skewed but the sample mean for this case using a sample size of 60 would be approximately normal, so then we can conclude that if we have a sample size like this one and an exponential distribution we can approximate the sample mean to the noemal distribution and indeed use the Central Limit theorem.
[tex] \bar X \sim N(\mu_{\bar X} , \frac{\sigma}{\sqrt{n}})[/tex]
[tex] \mu_{\bar X} = \bar X[/tex]
[tex]\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}[/tex]
Step-by-step explanation:
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
For this case we have a large sample size n =60 >30
The exponential distribution is the probability distribution that describes the time between events in a Poisson process.
For the exponential distribution:
[tex] \mu = \frac{1}{\lambda}[/tex]
[tex] \sigma^2 = \frac{1}{\lambda^2}[/tex]
We know that the exponential distribution is skewed but the sample mean for this case using a sample size of 60 would be approximately normal, so then we can conclude that if we have a sample size like this one and an exponential distribution we can approximate the sample mean to the noemal distribution and indeed use the Central Limit theorem.
[tex] \bar X \sim N(\mu_{\bar X} , \frac{\sigma}{\sqrt{n}})[/tex]
[tex] \mu_{\bar X} = \bar X[/tex]
[tex]\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}[/tex]
Which of the following when added to 4a^2+9 will result in a perfect square for all integer values of a?
(A) 0
(B) 3a
(C) 6a
(D) 9a
(E) 12a
Answer:
(D) 9a
Step-by-step explanation:
Given
4a^2 + 9
Add 9a to the expression
4a^2 + 9 + 9a
This can be written as
(2a)^2 + (3)^2 + (3)^2 a
Answer:
E. 12a
Step-by-step explanation:
TBH no idea why thats correct.
A password is required to be 12 to 16 characters in length. Characters can be digits (0-9), upper or lower-case letters (A-Z, a-z) or special characters. There are 10 permitted special characters. There is an additional rule that not all characters can be letters (i.e. there has to be at least one digit or one special character.) How many permitted passwords are there? Give your answer in un-evaluated/un-simpli ed form and explain it fully.
Answer:
Step-by-step explanation:
You can find your answer in attached document.
The number of passwords is the sum from i=12 to 16 of 66^i minus the sum from i=12 to 16 of 52^i, representing all combinations of characters from length 12 to 16, excluding all-letter combinations.
Explanation:The subject of your question pertains to combinatorics, specifically about counting password possibilities. Given that we have 66 different characters (10 digits, 52 letters lower and upper-case and 10 special characters), we have 66^12 to 66^16 possibilities for the length. However, we need to exclude the passwords composed strictly of letters. There are 52 letter characters, so we have 52^12 to 52^16 such passwords. Hence, the number of permitted passwords, in un-evaluated/un-simplified form, is the sum from i=12 to 16 of 66^i minus the sum from i=12 to 16 of 52^i
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Write an equation (-2, 10), (10, -14)
[tex]\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{10})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{-14}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-14}-\stackrel{y1}{10}}}{\underset{run} {\underset{x_2}{10}-\underset{x_1}{(-2)}}}\implies \cfrac{-24}{10+2}\implies \cfrac{-24}{12}\implies -2[/tex]
[tex]\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{10}=\stackrel{m}{-2}[x-\stackrel{x_1}{(-2)}]\implies y-10=-2(x+2) \\\\\\ y-10=-2x-4\implies y=-2x+6[/tex]
A company is interested in evaluating its current inspection procedure on large shipments of identical items. The procedure is to take a sample of 5 items and pass the shipment if no more than 1 item is found to be defective. It is known that items are defective at a 10% rate overall. a. What is the probability that the inspection procedure will pass the shipment? b. What is the expected number of defectives in this process of inspecting 5 items? c. If items are defective at a 20% rate overall, what is the probability that you will find 4 defectives in a sample of 5?
Answer:
(a) The probability that the inspection procedure will pass the shipment is 0.9185.
(b) The expected number of defectives in this process of inspecting 5 items is 0.50.
(c) The probability that there will be 4 defectives in a sample of 5 is 0.0016.
Step-by-step explanation:
Let X = number of defective items.
The probability of selecting a defective item is, p = 0.10.
A sample of n = 5 items is selected at random.
The random variable X follows a Binomial distribution with parameters n = 5 and p = 0.10.
The probability mass function of X is:
[tex]P(X=x)={5\choose x}0.10^{x}(1-0.10)^{5-x};\ x=0,1,2,3....[/tex]
It is provided that the shipment will pass if there are no more than 1 defective items in the selected 5 units.
(a)
Compute the probability that the inspection procedure will pass the shipment as follows:
P (X ≤ 1) = P (X = 0) + P (X = 1)
[tex]={5\choose 0}0.10^{0}(1-0.10)^{5-0}+{5\choose 1}0.10^{1}(1-0.10)^{5-1}\\=(1\times 1\times 0.59049) + (5\times 0.10\times 0.6561)\\=0.59049+0.32805\\=0.91854\\\approx0.9185[/tex]
Thus, the probability that the inspection procedure will pass the shipment is 0.9185.
(b)
The expected value of a Binomial distribution is:
[tex]E(X)=np[/tex]
Compute the expected number of defectives in this process of inspecting 5 items as follows:
[tex]E(X)=5\times0.10=0.50[/tex]
Thus, the expected number of defectives in this process of inspecting 5 items is 0.50.
(c)
The probability of finding a defective is now changed to 0.20.
Compute the probability that there will be 4 defectives in a sample of 5 as follows;
[tex]P(X=4)={5\choose 4}0.20^{4}(1-0.20)^{5-4}\\=5\times0.0016\times0.20\\=0.0016[/tex]
Thus, the probability that there will be 4 defectives in a sample of 5 is 0.0016.
Answer:
(a) Probability that the inspection procedure will pass the shipment = 0.91854
(b) E(X) = [tex]5 \times 0.1[/tex] = 0.5
(c) Probability of 4 defectives in a sample of 5 is 0.0064
Step-by-step explanation:
We are given that a company is interested in evaluating its current inspection procedure on large shipments of identical items. The procedure is to take a sample of 5 items and pass the shipment if no more than 1 item is found to be defective. It is known that items are defective at a 10% rate overall.
The above situation can be represented through Binomial distribution;
[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]
where, n = number of trials (samples) taken = 5 items
r = number of success
p = probability of success which in our question is % of items that are
defective, i.e., 10%
LET X = Number of defective items
Also, it is given that a sample of 5 items is taken,
So, it means X ~ [tex]Binom(n=5,p=0.10)[/tex]
(a) Probability that the inspection procedure will pass the shipment is given by the fact that if no more than 1 item is found to be defective, then only shipment is passed, i.e.;
P(X [tex]\leq[/tex] 1) = P(X = 0) + P(X = 1)
= [tex]\binom{5}{0}0.1^{0} (1-0.1)^{5-0} + \binom{5}{1}0.1^{1} (1-0.1)^{5-1}[/tex]
= [tex]1 \times 1 \times 0.9^{5} +5 \times 0.1 \times 0.9^{4}[/tex]
= 0.59049 + 0.32805 = 0.91854
(b) The expected number of defectives in this process of inspecting 5 items is given by = E(X) or Mean of X
So, mean of binomial distribution is, E(X) = [tex]n \times p[/tex]
So, E(X) = [tex]5 \times 0.1[/tex] = 0.5
Therefore, the expected number of defectives in this process of inspecting 5 items is 1 (after rounding to nearest integer).
(c) Now, the probability of success which is % of items that are defective is changed to 20% rate overall, i.e., p = 0.20 now.
So, probability that you will find 4 defectives in a sample of 5 = P(X = 4)
P(X = 4) = [tex]\binom{5}{4}0.2^{4} (1-0.2)^{5-4}[/tex]
= [tex]5 \times 0.2^{4} \times 0.8^{1}[/tex] = 0.0064
In remodeling a kitchen, a builder decides to place a splash-guard behind the sink consisting of eight 6-inch-square ceramic tiles decorated with different botanical herbs. The tiles will be installed in a custom-made wooden panel. The tile supplier has 12 different herb designs to choose from, and the builder selects 8 of these 12 at random. Suppose the order in which the tiles are arranged on the splash-guard does not matter. Fill in the blanks. (Give your answers to four decimal places.)
Two of the 12 herb tiles contain a blue tint that matches the kitchen color scheme. The probability that these 2 tiles will be included in the splash-guard is __.
The family actually grows 5 of the 12 herbs in a backyard garden. The probability that all 5 of these will be included on the splash-guard is _
Answer:
(a) 0.4242
(b) 0.0707
Step-by-step explanation:
The total number of ways of selecting 8 herbs from 12 is
[tex]\binom{12}{8}=495[/tex]
(a) If 2 herbs are selected, then there are 8 - 2 = 6 herbs to be selected from 12 - 8 = 10. The number of ways of the selection is then
[tex]\binom{10}{6}= 210[/tex]
Note that this is the number of ways that both are included. We would have multiplied by 2! if any of them were to be included.
The probability = [tex]\dfrac{210}{495}=0.4242[/tex]
(b) If 5 herbs are selected, then there are 8 - 5 = 3 herbs to be selected from 12 - 5 = 7. The number of ways of the selection is then
[tex]\binom{7}{3}= 35[/tex]
This is the number of ways that both are included. We would have multiplied by 5! if any of them were to be included. In that case, our probability will exceed 1; this implies that certainly, at least, one of them is included.
The probability = [tex]\dfrac{35}{495}=0.0707[/tex]
The probability that the 2 tiles with a blue tint will be included in the splash-guard is 0.0455. The probability that all 5 of the herbs grown in the backyard garden will be included on the splash-guard is 0.0058.
The probability that the 2 tiles with a blue tint will be included in the splash-guard is 0.0455.
The probability that all 5 of the herbs grown in the backyard garden will be included on the splash-guard is 0.0058.
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Two fair six-sided dice are tossed independently. Let M 5 the maximum of the two tosses (so M(1,5) 5 5, M(3,3) 5 3, etc.). a. What is the pmf of M? [Hint: First determine p(1),
Answer with Step-by-step explanation:
S={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),..,(2,6),(3,1),(3,2),...,(3,6),(4,1),..,(4,6),(5,1),(5,2),(5,3),...,(5,6),(6,1),(6,2),..,(6,6)}
Total number of cases=36
M=Maximum of the two tosses
M=1,
(1,1)
Therefore, [tex]P(1)=\frac{1}{36}[/tex]
Using the formula, P(E)=[tex]\frac{favorable\;cases}{Total\;number\;of\;cases}[/tex]
M=2
(1,2),(2,1),(2,2)
Favorable cases=3
[tex]P(2)=\frac{3}{36}=\frac{1}{12}[/tex]
M=3
(1,3),(3,1),(3,2),(2,3),(3,3)
[tex]P(3)=\frac{5}{36}[/tex]
M=4
(1,4),(4,1),(2,4),(3,4),(4,2)(4,3),(4,4)
[tex]P(4)=\frac{7}{36}[/tex]
M=5
(1,5),(2,5),(3,5),(4,5),(5,1),(5,2),(5,3),(5,4),(5,5)
[tex]P(5)=\frac{9}{36}=\frac{1}{4}[/tex]
M=6
(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
[tex]P(6)=\frac{11}{36}[/tex]
Please hurry i need help
Examine this expanded form.
1
(a)(a)(a)(a)(a)(a)
Which is the expression in exponential form?
6-a
6a
a-6
6a-1
Answer:
yes its C
stop scrolling !!
Step-by-step explanation:
2020 edg
In between the given options expression 6⁻ᵃ in the exponential form, So Option A is correct
What are exponential functions?An exponential function is a mathematical function which we write as a
f(x) = aˣ, where a is constant and x is variable term. The most commonly used exponential function is eˣ , where e is constant having value 2.7182
Given that,
The expressions,
6⁻ᵃ
6a
a-6
6a-1
Exponential form is a mathematical representation where a number (the base) is raised to a power (the exponent).
The exponent tells us how many times the base is multiplied by itself. For example, in the expression "6a", 6 is the base and "a" is the exponent.
This means that the expression can be written as 6 * 6 * 6 * ... * 6,
where there are "a" number of 6's being multiplied together.
The expression "6⁻ᵃ" is in exponential form.
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Set up but do not solve for the appropriate particular solution yp for the differential equation y′′+4y=5xcos(2x) using the Method of Undetermined Coefficients (primes indicate derivatives with respect to x). In your answer, give undetermined coefficients as A, B, etc.
Answer:
So, solution of the differential equation is
[tex]y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\[/tex]
Step-by-step explanation:
We have the given differential equation: y′′+4y=5xcos(2x)
We use the Method of Undetermined Coefficients.
We first solve the homogeneous differential equation y′′+4y=0.
[tex]y''+4y=0\\\\r^2+4=0\\\\r=\pm2i\\\\[/tex]
It is a homogeneous solution:
[tex]y_h(t)=c_1e^{-2i t}+c_2e^{2i t}[/tex]
Now, we finding a particular solution.
[tex]y_p(t)=A5x\cos 2x\\\\y'_p(t)=A5\cos 2x-A10x\sin 2x\\\\y''_p(t)=-A20\sin 2x-A20x\cos 2x\\\\\\\implies y''+4y=5x\cos 2x\\\\-A20\sin 2x-A20x\cos 2x+4\cdot A5x\cos 2x=5x\cos 2x\\\\-A20\sin 2x=5x\cos 2x\\\\A=-\frac{x}{4} \cot 2x\\[/tex]
we get
[tex]y_p(t)=A5\cos 2x\\\\y_p(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x\\\\\\y(t)=y_p(t)+y_h(t)\\\\y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\[/tex]
So, solution of the differential equation is
[tex]y(t)=-\frac{5x^2}{4}\cot 2x\cdot \cos 2x+c_1e^{-2it}+c_2e^{2it}\\[/tex]
To set up the particular solution for the differential equation using the Method of Undetermined Coefficients, we propose a function similar to the nonhomogeneous term with unknown coefficients. The form is a combination of polynomial and trigonometric terms, reflecting the equation's right-hand side.
Explanation:To set up the particular solution yp for the differential equation y″+4y=5xcos(2x) using the Method of Undetermined Coefficients, first identify the form of the nonhomogeneous term 5xcos(2x). Since it's a product of a polynomial and a trigonometric function, we anticipate a particular solution that mirrors this form, but with unknown coefficients to solve for.
The lowest degree polynomial in the nonhomogeneous term is x, so we begin with Ax and will also need to account for the derivative of cosine, which is sine. Therefore, our yp will be of the form:
yp = x(Acos(2x) + Bsin(2x)) + Cx²cos(2x) + Dx²sin(2x).
Here, A, B, C, and D are the undetermined coefficients that will be found by differentiating this assumed form and substituting back into the original differential equation.
Suppose a struggling student who is currently taking pre-statistics and not passing (60%) wants to predict his introductory statistics course grade. Should the regression line be use to make this prediction?
Answer:
No
Step-by-step explanation:
Because the model was developed using only pre-statistics course grades between 70% and 95% so it is risky to assume that linear trend will continue far beyond that span of values.
A large sample of tires from cabs driven within a city have an average tire tread depth of 0.25cm at the end of the winter. If the city’s average winter tire tread depth is 2.2cm, with a standard deviation of 0.33cm at the end of the same winter, what is the probability cab tires depths would be shallower than 0.25cm.
Answer:
0% probability cab tires depths would be shallower than 0.25cm.
Step-by-step explanation:
Problems of normally distributed samples can be 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.
In this problem, we have that:
[tex]\mu = 2.2, \sigma = 0.33[/tex]
What is the probability cab tires depths would be shallower than 0.25cm.
This probability is the pvalue of Z when X = 0.25. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.25 - 2.2}{0.33}[/tex]
[tex]Z = -5.9[/tex]
[tex]Z = -5.9[/tex] has a pvalue of 0.
0% probability cab tires depths would be shallower than 0.25cm.
Final answer:
The Z-score for a tire tread depth of 0.25 cm when compared to the city's average tread depth is -5.91, which suggests an extremely low probability, nearly zero, of finding a tire tread depth shallower than 0.25 cm.
Explanation:
To determine the probability that the tire tread depth for cab tires is shallower than 0.25 cm, we will use the standard normal distribution and the process of finding a Z-score. We are given the population mean (μ) to be 2.2 cm and the population standard deviation (σ) to be 0.33 cm.
The Z-score is calculated as:
Z = (X - μ) / σ
Where:
X is the value we are comparing to the mean (0.25 cm in this case)μ is the mean tire tread depth (2.2 cm)σ is the standard deviation of the tire tread depth (0.33 cm)So, the Z-score for 0.25 cm is:
Z = (0.25 - 2.2) / 0.33 = -5.91
A Z-score of -5.91 is significantly beyond the typical range for a standard normal distribution table, which indicates that the probability of having a tire tread depth of less than 0.25 cm is extremely low, essentially approaching zero.