Answer:
The correct answers are
option(1), option (4),option (5)
Step-by-step explanation:
One to one: Every image has exactly one unique pre-image in domain.
(1)
f(a) = b, f(b)=a, f(c)=c, f(d)=d
b has a pre-image in domain i.e a
a has a pre-image in domain i.e b
c has a pre-image in domain i.e c
d has a pre-image in domain i.e d
Here all elements have a unique pre- image in domain.
Therefore it is one to one.
(2)
f(a) = b, f(b)=a, f(c)=c, f(d)=d
b has a pre-image in domain i.e a
a has a pre-image in domain i.e b
c has a pre-image in domain i.e c
d has a pre-image in domain i.e d
Here all elements have a unique pre- image in domain.
Therefore it is one to one.
(3)
f(a) = b, f(b)=b, f(c)=d, f(d)=c
b has a pre-image in domain i.e a
b has a pre-image in domain i.e b
d has a pre-image in domain i.e c
c has a pre-image in domain i.e d
b has two pre image.
Here all elements have not a unique pre- image in domain.
Therefore it is not a one to one mapping.
(4)
f(a) = b, f(b)=b, f(c)=d, f(d)=c
b has a pre-image in domain i.e a
b has a pre-image in domain i.e b
d has a pre-image in domain i.e c
c has a pre-image in domain i.e d
b has two pre image.
Here all elements have not a unique pre- image in domain.
Therefore it is not a one to one mapping.
(5)
f(a) = d, f(b)=b, f(c)=d, f(d)=c
d has a pre-image in domain i.e a
b has a pre-image in domain i.e b
d has a pre-image in domain i.e c
c has a pre-image in domain i.e d
Here all elements have a unique pre- image in domain.
Therefore it is a one to one mapping.
(6)
f(a)=d, f(b)=b,f(c)=c,f(d)=d
d has a pre-image in domain i.e a
b has a pre-image in domain i.e b
c has a pre-image in domain i.e c
d has a pre-image in domain i.e d
Here all elements have a unique pre- image in domain.
Therefore it is a one to one mapping.
f(a) = b, f(b) = a, f(c) = c, f(d) = d and f(a) = d, f(b) = b, f(c) = c, f(d) = d are one-to-one function and f(a) = b, f(b) = b, f(c) = d, f(d) = c is not one-to-one function. Then the correct statements are 1, 4, and 5.
What is a function?The function is an expression, rule, or law that defines the relationship between one variable to another variable. Functions are ubiquitous in mathematics and are essential for formulating physical relationships.
One-to-one - every image has exactly one unique pre-image in the domain.
1. f(a) = b, f(b) = a, f(c) = c, f(d) = d
b has a pre image in a domian that is a
a has a pre image in a domian that is b
c has a pre image in a domian that is c
d has a pre image in a domian that is d
Here all elements have a unique pre-image in a domain.
Therefore it is one-to-one.
2. f(a) = b, f(b) = a, f(c) = c, f(d) = d
b has a pre image in a domian that is a
a has a pre image in a domian that is b
c has a pre image in a domian that is c
d has a pre image in a domian that is d
Here all elements have a unique pre-image in a domain.
Therefore it is one-to-one.
3. f(a) = b, f(b) = b, f(c) = d, f(d) = c
b has a pre image in a domian that is a
b has a pre image in a domian that is b
d has a pre image in a domian that is c
c has a pre image in a domian that is d
Here all elements do have not a unique pre-image in a domain.
Therefore it is not one-to-one.
4. f(a) = b, f(b) = b, f(c) = d, f(d) = c
b has a pre image in a domian that is a
b has a pre image in a domian that is b
d has a pre image in a domian that is c
c has a pre image in a domian that is d
Here all elements do have not a unique pre-image in a domain.
Therefore it is not one-to-one.
5. f(a) = d, f(b) = b, f(c) = c, f(d) = d
d has a pre image in a domian that is a
b has a pre image in a domian that is b
c has a pre image in a domian that is c
d has a pre image in a domian that is d
Here all elements have a unique pre-image in a domain.
Therefore it is one-to-one.
6. f(a) = d, f(b) = b, f(c) = c, f(d) = d
d has a pre image in a domian that is a
b has a pre image in a domian that is b
c has a pre image in a domian that is c
d has a pre image in a domian that is d
Here all elements have a unique pre-image in a domain.
Therefore it is one-to-one.
More about the function link is given below.
https://brainly.com/question/5245372
g Suppose that when a certain lake is stocked with fish, the birth and death rates ˇ and ı are both inversely propor- (a) Show that ????1 p????2 P.t/D 2ktC P0 ; where k is a constant. (b) If P0 D 100 and after 6 months there are 169 fish in the lake, how many will there be after 1 year?
Answer:
k=1
P(12)=256
Step-by-step explanation:
The model for population involving birth and death is given as:
[tex]\frac{dP}{dt}=(b-d)P[/tex]
where b=birth rate and d=death rate.
If the birth and death rate are inversely proportional to [tex]\sqrt{P}[/tex]
[tex]b=\frac{A}{\sqrt{P} } \\d=\frac{B}{\sqrt{P} }[/tex] where A and B are constants of variation.
Substituting b and d into our model
[tex]\frac{dP}{dt}=(\frac{A}{\sqrt{P} }-\frac{B}{\sqrt{P} })P\\\frac{dP}{dt}=\frac{k}{\sqrt{P} }P\\[/tex] where A-B=k, another constant
Simplifying using indices
[tex]\frac{dP}{dt}={k}P^{1-\frac{1}{2} }\\\frac{dP}{dt}={k}P^\frac{1}{2} \\[/tex]
Next, we Separate Variables and Integrate both sides
[tex]\frac{dP}{\sqrt{P} }={k}dt\[/tex]
[tex]\int\frac{dP}{\sqrt{P} }=\int{k}dt\[/tex]
[tex]2P^{1/2} =kt+C[/tex] where C is the constant of integration
[tex]P(t) =(\frac{kt}{2} +C)^2[/tex]
When t=0, P(t)=[tex]P_0[/tex], C=[tex]\sqrt{P_0}[/tex]
[tex]P(t) =(\frac{kt}{2} +\sqrt{P_0})^2[/tex] as required.
(b)If [tex]P_0[/tex]=100, t=6 months, P(t)=169
[tex]P(t) =(\frac{kt}{2} +\sqrt{P_0})^2[/tex]
[tex]169 =(\frac{6k}{2} +\sqrt{100})^2\\\sqrt{169} =3k+100\\13=3k+10\\3k=13-10=3\\k=1[/tex]
Since we have found the constant k, we can then calculate the population after 1 year. Note that we use 12 months since we used month earlier to get k.
[tex]P(t) =(\frac{kt}{2} +\sqrt{P_0})^2[/tex]
[tex]P(12) =(\frac{1X12}{2} +\sqrt{100})^2\\=(6+10)^2=256[/tex]
Therefore the population after a year is 256.
The population of fish after 1 year would be 285.
Let P be the population of fish at any time t months.
We are given:
Initial population: [tex]P_0[/tex] = 100 fishPopulation after 6 months: P(6) = 169 fishWe need to determine the population after 12 months.
The logistic growth model can be expressed as:
dP/dt = rP(1 - P/K)
where r is the intrinsic growth rate and K is the carrying capacity of the environment. Here, since the birth and death rates are inversely proportional to √P, we can infer that the growth rate might be proportional to √P, leading us to use a simplified logistic growth equation.
Given P(6) = 169, let’s find the intrinsic growth rate , r. From population changes between 0 and 6 months:
Let’s assume the logistic growth function in the form:
P(t) = [tex]P_0 e^{rt}[/tex]
We solve for r:
169 = [tex]100 e^{6r}[/tex]
1.69 = [tex]e^{6r}[/tex]
Taking natural logarithms:
ln(1.69) = 6r
0.524 = 6r
r = 0.0873/month
Now, to find P(12):
P(12) = [tex]100 e^{0.0873*12}[/tex]
= [tex]100 e^{1.048}[/tex]
= 100 * 2.853
= 285.3 fish
So, the population after 1 year is approximately 285 fish.
"According to contractarian logic, we should be willing to make concessions to others if they agree to make comparable and reciprocal concessions, with the overall result being that everyone gets a desired benefit with an acceptable minimum of sacrifice on each side. Please identify, and briefly analyze, a situation or scenario that illustrates this principle. "
Answer:In a couple with a newborn baby at home, to take turns on feeding the baby at night.
Step-by-step explanation: Here both parents are willing to sacrifice a few minutes, if not hours of sleeptime with the promise to be allowed to rest the next time the baby needs to be fed. There is no certainty in how long it will take for the baby to go back to sleep or how long it will be for the baby to be awake again, but the chances are the same for both parents, so they both agree to take care of the child one at a time with the promise to be in turns, this is an example of contractarian logic.
Final answer:
Contractarian logic is exemplified in international trade agreements where nations mutually lower tariffs under the expectation of reciprocal actions, demonstrating the principle of reciprocity and mutual advantage.
Explanation:
According to contractarian logic, we should be willing to make concessions to others if they agree to make comparable and reciprocal concessions, leading to a situation where everyone benefits with a minimal level of sacrifice. A classic example illustrating this principle is international trade agreements. Nations often agree to lower tariffs and grant each other favorable trade terms under the condition that the other nation reciprocates. These agreements are founded on the expectation that both sides will adhere to the agreements, benefiting both by expanding their markets and reducing costs for consumers. Here, the benefit is mutual economic growth, and the sacrifice might involve foregoing the protection of certain domestic industries in the interest of broader gains. This scenario mirrors the foundational ideas of social contract theory where rational, self-interested agents come together to agree on a set of rules or actions that benefit all parties involved, thereby demonstrating the principle of reciprocity and mutual advantage that is central to contractarian logic.
From past experience, a company has found that in carton of transistors: 92% contain no defective transistors 3% contain one defective transistor, 3% contain two defective transistors, and 2% contain three defective transistors. Calculate the mean and variance for the defective transistors. Mean = Variance = (Please round answers to 4 decimal places.)
Answer:
[tex] E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500[/tex]
In order to find the variance we need to find first the second moment given by:
[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i)[/tex]
And replacing we got:
[tex] E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300[/tex]
The variance is calculated with this formula:
[tex] Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075[/tex]
And the standard deviation is just the square root of the variance and we got:
[tex] Sd(X) = \sqrt{0.3075}= 0.5545[/tex]
Step-by-step explanation:
Previous concepts
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
Solution to the problem
LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X
X 0 1 2 3
P(X) 0.92 0.03 0.03 0.02
We can calculate the expected value with the following formula:
[tex] E(X) = \sum_{i=1}^n X_i P(X_i)[/tex]
And replacing we got:
[tex] E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500[/tex]
In order to find the variance we need to find first the second moment given by:
[tex] E(X^2) = \sum_{i=1}^n X^2_i P(X_i)[/tex]
And replacing we got:
[tex] E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300[/tex]
The variance is calculated with this formula:
[tex] Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075[/tex]
And the standard deviation is just the square root of the variance and we got:
[tex] Sd(X) = \sqrt{0.3075}= 0.5545[/tex]
The required value of mean 0.330 and standard deviation 0.5545 for the defective transistors.
Given that,
A company has found that in carton of transistors: 92% contain no defective transistors,
3% contain one defective transistor, 3% contain two defective transistors,
and 2% contain three defective transistors.
We have to find,
Calculate the mean and variance for the defective transistors. Mean = Variance.
According to the question,
Let, X the random variable who represent the number of defective transistors.
The following probability distribution for X,
X 0 1 2 3
P(X) 0.92 0.03 0.03 0.02
To calculate the expected value with the following formula:
[tex]E(X) = \sum^{n}_{i=1} X_i. P.(X_i)[/tex]
On substitute all the values in the formula,
[tex]E(X) = 0\times0.92 + 1\times0.03+2\times0.03 + 3\times0.02\\\\E(X) = 0.15[/tex]
To find the variance first the second moment given by:
[tex]E(X^2) = \sum^{n}_{i=1} X_i^2. P.(X_i)[/tex]
On substitute all the values in the formula;
[tex]E(X) = 0^2\times0.92 + 1^2\times0.03+2^2\times0.03 + 3^2\times0.02\\\\E(X) = 0.330[/tex]
Therefore, The variance is calculated with this formula:
[tex]Var(X) = E(X)^2 - (E(X))^2 = 0.33 - (0.15)^2 = 0.3075[/tex]
And the standard deviation is just the square root of the variance,
[tex]Standard \ deviation = \sqrt{0.0375} = 0.5545[/tex]
Hence, The required value of mean 0.330 and standard deviation 0.5545 for the defective transistors.
To know more about Mean click the link given below.
https://brainly.com/question/15167067
The probability that an Oxnard University student is carrying a backpack is .70. If 10 students are observed at random, what is the probability that fewer than 7 will be carrying backpacks
Answer:
35.03% probability that fewer than 7 will be carrying backpacks
Step-by-step explanation:
For each student, there are only two possible outcomes. Either they are carrying a backpack, or they are not. The probability of a student carrying a backpack is independent from other students. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
The probability that an Oxnard University student is carrying a backpack is .70.
This means that [tex]p = 0.7[/tex]
If 10 students are observed at random, what is the probability that fewer than 7 will be carrying backpacks
This is [tex]P(X < 7)[/tex] when [tex]n = 10[/tex]. So
[tex]P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{10,0}.(0.7)^{0}.(0.3)^{10} = 0.000006[/tex]
[tex]P(X = 1) = C_{10,1}.(0.7)^{1}.(0.3)^{9} = 0.0001[/tex]
[tex]P(X = 2) = C_{10,2}.(0.7)^{2}.(0.3)^{8} = 0.0014[/tex]
[tex]P(X = 3) = C_{10,3}.(0.7)^{3}.(0.3)^{7} = 0.0090[/tex]
[tex]P(X = 4) = C_{10,4}.(0.7)^{4}.(0.3)^{6} = 0.0368[/tex]
[tex]P(X = 5) = C_{10,5}.(0.7)^{5}.(0.3)^{5} = 0.1029[/tex]
[tex]P(X = 6) = C_{10,6}.(0.7)^{6}.(0.3)^{4} = 0.2001[/tex]
[tex]P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.000006 + 0.0001 + 0.0014 + 0.0090 + 0.0368 + 0.1029 + 0.2001 = 0.3503[/tex]
35.03% probability that fewer than 7 will be carrying backpacks
Final answer:
To find the probability that fewer than 7 students will be carrying backpacks, use the binomial probability formula. The final probability is 0.9143, or 91.43%.
Explanation:
To find the probability that fewer than 7 students will be carrying backpacks, we can use the binomial probability formula. In this case, the probability of success (carrying a backpack) is 0.70. The number of trials is 10. We want to find the probability of getting fewer than 7 successes.
We can calculate this by finding the sum of the probabilities of getting 0, 1, 2, 3, 4, 5, and 6 successes, using the binomial probability formula for each value. Then, we subtract this sum from 1 to get the probability of getting fewer than 7 successes.
The final probability for this scenario is 0.9143, or 91.43%.
There are two machines available for cutting corks intended for use in bottles. The first produces corks with diameters that are normally distributed with mean 3 cm and standard deviation 0.1 cm. The second machine produces corks with diameters that have a normal distribution with mean 3.04 cm and standard deviation 0.02 cm. Acceptable corks have diameters between 2.9 cm and 3.1 cm. Which machine is more likely to produce an acceptable cork? What should the acceptable range for cork diameters be (from 3 − d cm to 3 + d cm) to be 90% certain for the first machine to produce an acceptable cork?
Answer:
a) The second machine is more likely to produce an acceptable cork.
b) Acceptable range for cork diameters produced by the first machine with a 90% confidence = (2.8355, 3.1645)
Step-by-step explanation:
This is a normal distribution problem
For the first machine,
Mean = μ = 3 cm
Standard deviation = σ = 0.1 cm
And we want to find which percentage of the population falls between 2.9 cm and 3.1 cm.
P(2.9 ≤ x ≤ 3.1) = P(x ≤ 3.1) - P(x ≤ 2.9)
We first standardize this measurements.
The standardized score for any value is the value minus the mean then divided by the standard deviation.
For 2.9 cm
z = (x - μ)/σ = (2.9 - 3.0)/0.1 = - 1.00
For 3.1 cm
z = (x - μ)/σ = (3.1 - 3.0)/0.1 = 1.00
P(x ≤ 3.1) = P(z ≤ 1.00) = 0.841
P(x ≤ 2.9) = P(z ≤ -1.00) = 0.159
P(2.9 ≤ x ≤ 3.1) = P(-1.00 ≤ z ≤ 1.00) = P(z ≤ 1.00) - P(z ≤ -1.00) = 0.841 - 0.159 = 0.682 = 68.2%
This means that 68.2% of the diameter of corks produced by the first machine lies between 2.9 cm and 3.1 cm.
For the second machine,
Mean = μ = 3.04 cm
Standard deviation = σ = 0.02 cm
And we want to find which percentage of the population falls between 2.9 cm and 3.1 cm.
P(2.9 ≤ x ≤ 3.1) = P(x ≤ 3.1) - P(x ≤ 2.9)
We standardize this measurements.
The standardized score for any value is the value minus the mean then divided by the standard deviation.
For 2.9 cm
z = (x - μ)/σ = (2.9 - 3.04)/0.02 = - 7.00
For 3.1 cm
z = (x - μ)/σ = (3.1 - 3.0)/0.02 = 3.00
P(x ≤ 3.1) = P(z ≤ 3.00) = 0.999
P(x ≤ 2.9) = P(z ≤ -7.00) = 0.0
P(2.9 ≤ x ≤ 3.1) = P(-7.00 ≤ z ≤ 3.00) = P(z ≤ 3.00) - P(z ≤ -7.00) = 0.999 - 0.0 = 0.999 = 99.9%
This means that 99.9% of the diameter of corks produced by the second machine lies between 2.9 cm and 3.1 cm.
Hence, we can conclude that the second machine is more likely to produce an acceptable cork.
b) Margin of error = (z-multiplier) × (standard deviation of the population)
For 90% confidence interval, z-multiplier = 1.645 (from literature and the z-tables)
Standard deviation for first machine = 0.1
Margin of error, d = 1.645 × 0.1 = 0.1645.
The acceptable range = (mean ± margin of error)
Mean = 3
Margin of error = 0.1645
Lower limit of the acceptable range = 3 - d = 3 - 0.1645 = 2.8355
Upper limit of the acceptable range = 3 + d = 3 + 0.1645 = 3.1645
Acceptable range = (2.8355, 3.1645)
Final answer:
To determine which machine is more likely to produce acceptable corks, we examine their distribution characteristics. Machine 2 may be more reliable due to its tighter control despite a slightly higher mean. To find a 90% certain acceptable range for Machine 1, we calculate using its standard deviation and the z-score for the 90th percentile.
Explanation:
The question involves comparing two machines based on their ability to produce corks within a specified acceptable diameter range using normal distribution properties, and calculating the range for diameters to ensure a 90% certainty of producing acceptable corks for the first machine.
Comparing the Two Machines
For the first machine with a mean diameter of 3 cm and a standard deviation of 0.1 cm, and the second machine with a mean diameter of 3.04 cm and a standard deviation of 0.02 cm, the question is which machine is more likely to produce corks within the acceptable range of 2.9 cm to 3.1 cm.
Machine 1 produces corks closer to the center of the acceptable range but with a wider spread (higher standard deviation), while Machine 2 produces corks that are skewed slightly larger but with a much tighter spread around their mean (lower standard deviation). To determine which machine is more likely to produce acceptable corks, we would need to calculate the z-scores for the acceptance limits for both machines and compare the probabilities. However, intuitively, Machine 2 might be seen as more reliable due to its tighter control (lower standard deviation), assuming its mean is not too far out of the acceptable range.
Finding the Acceptable Range for 90% Certainty
To ensure 90% certainty that a cork produced by Machine 1 falls within an acceptable diameter range, we need to determine d in the range of 3 − d cm to 3 + d cm. This involves finding the z-score that corresponds to the 5th and 95th percentiles due to the symmetric nature of normal distribution, then solving for d using the properties of normal distribution and the given standard deviation of 0.1 cm.
The z-score corresponding to the 5th and 95th percentiles (for a 90% certainty) typically falls around ±1.645. Using the formula for z-score, which is (X − μ) / σ, and solving for d, we can find the acceptable range of diameters for the first machine to produce an acceptable cork with 90% certainty.
Using the fixed-time period inventory model, and given an average daily demand of 200 units, 4 days between inventory reviews, 5 days for lead time, 120 units of inventory on hand, a "z" of 1.96, and a standard deviation of demand over the review and lead time of 3 units, which of the following is the order quantity?
A. About 1,086
B.About 1,686
C. About 1,806
D. About 2,206
E. About 2,686
Answer:
Correct option: B. About 1,686.
Step-by-step explanation:
The formula to compute the order quantity (Q) is:
[tex]Q=(q_{d}\times (I+L))+(z\times\sigma_{I+L})-I_{n}[/tex]
Here
[tex]q_{d}=average\ daily\ semand=200\\I = Inventory\ review\ time=4\\L=lead\ time=5\\\sigma_{I+L}=standard\ deviation\ over\ the\ review\ and\ lead\ time=3\\I_{n}=number\ of\ units\ of\ inventory\ on\ hand=120[/tex]
Compute the order quantity as follows:
[tex]Q=(q_{d}\times (I+L))+(z\times\sigma_{I+L})-I_{n}\\=(200\times(4+5))+(1.96\times 3)-120\\=1800+5.88-120\\=1685.88\\\approx1686[/tex]
Thus, the order quantity was about 1,686.
You are certain to get a heart comma diamond comma club comma or spade when selecting cards from a shuffled deck. Express the indicated degree of likelihood as a probability value between 0 and 1 inclusive
Answer:
Probability = 1
Step-by-step explanation:
The number of each type of card described in the question from within a full deck of cards is as follows;
Hearts = 13
Clubs = 13
Diamonds = 13
Spades = 13
These add up to a total of 52 cards. Since a deck only has 52 cards, these make up all the cards in the deck.
Since the probability of taking out a card from these four suits is going to be as follows:
Probability = number of ways we can take out a corresponding suit card / total number of cards
Probability = 52 / 52
Probability = 1
Thus, we can see that the probability of taking out a card belonging to one of the four suits (heart,diamond,club,spade) is 1.
In the last homework, you analyzed the equation ˙x = cx − x 3 graphically, for c positive, zero and negative. Now perform a formal linear stability analysis, and compare your results with a graphical analysis
Answer:
Please find attached file for complete answer solution and explanation of same question.
Step-by-step explanation:
Solve 0 = (x – 4)2 – 1 by graphing the related function.
What are the solutions to the equation?
3 and 5 is the answer
AND THAT'S JUST ON PERIOD POOH!
Answer:
Therefore, the solutions of the quadratic equations are:
[tex]x=5,\:x=3[/tex]
The graph is also attached.
Step-by-step explanation:
The solution of the graph could be obtained by finding the x-intercept.
[tex]y=\left(x-\:4\right)^2-1[/tex]
Finding the x-intercept by substituting the value y = 0
so
[tex]y=\left(x-\:4\right)^2-1[/tex]
[tex]\:0\:=\:\left(x\:-\:4\right)^2\:-\:1[/tex] ∵ y = 0
[tex]\left(x-4\right)^2-1=0[/tex]
[tex]\left(x-4\right)^2-1+1=0+1[/tex]
[tex]\left(x-4\right)^2=1[/tex]
[tex]\mathrm{For\:}\left(g\left(x\right)\right)^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}[/tex]
[tex]\mathrm{Solve\:}\:x-4=\sqrt{1}[/tex]
[tex]\mathrm{Apply\:rule}\:\sqrt{1}=1[/tex]
[tex]x-4=1[/tex]
[tex]x=5[/tex]
[tex]\mathrm{Solve\:}\:x-4=-\sqrt{1}[/tex]
[tex]\mathrm{Apply\:rule}\:\sqrt{1}=1[/tex]
[tex]x-4=-1[/tex]
[tex]x=3[/tex]
So, when y = 0, then x values are 3, and 5.
Therefore, the solutions of the quadratic equations are:
[tex]x=5,\:x=3[/tex]
The graph is also attached. As the graph is a Parabola. It is visible from the graph that the values of y = 0 at x = 5 and x = 3. As the graph is a Parabola.
Answer:
3 and 5 is the answer
Step-by-step explanation:
We have a biased coin (probability of heads is equal to 1/4). Consider the following 2 step process: In the first step we flip the coin until we get a heads. Let X denote the trial on which the first heads occurs. In the second step we flip the coin X more times. Let Y be the number of heads in the second step. (a) For each non-negative integer, k, what is the probability that X = k? (b) Conditioned on the event {X = k}. What is the probability Y = 0? (c) Use the Law of Total Probability to compute the unconditional probability that Y = 0.
Answer:
See the attached pictures for answer.
Step-by-step explanation:
See the attached picture for detailed explanation.
The method of tree ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population of x values has an approximately normal distribution. 1313 1243 1271 1313 1268 1316 1275 1317 1275 (a) Use a calculator with mean and standard deviation keys to find the sample mean year x and sample standard deviation s. (Round your answers to the nearest whole number.) x
Answer:
[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
And replacing we got:
[tex]\bar X = \frac{1313+1243+1271+1313+1268+1316+1275+1317+1275}{9}=1287.89 \approx 1288[/tex]
In order to find the sample deviation we can use this formula:
[tex]s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
s= \sqrt{\frac{(1313-1287.89)^2 +(1243-1287.89)^2 +(1271-1287.89)^2 +(1313-1287.89)^2 +(1268-1287.89)^2 + (1316-1287.89)^2 +(1275-1287.89)^2 +(1317-1287.89)^2 +(1275-1287.89)^2}{9-1}}= 27.218 \approx 27
Step-by-step explanation:
For this case we have the following data given:
1313 1243 1271 1313 1268 1316 1275 1317 1275
In order to calculate the sample mean we can use the following formula:
[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
And replacing we got:
[tex]\bar X = \frac{1313+1243+1271+1313+1268+1316+1275+1317+1275}{9}=1287.89 \approx 1288[/tex]In order to find the sample deviation we can use this formula:
[tex]s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
And replacing we have:
s= \sqrt{\frac{(1313-1287.89)^2 +(1243-1287.89)^2 +(1271-1287.89)^2 +(1313-1287.89)^2 +(1268-1287.89)^2 + (1316-1287.89)^2 +(1275-1287.89)^2 +(1317-1287.89)^2 +(1275-1287.89)^2}{9-1}}= 27.218 \approx 27
Find a formula for the general term of the sequence 5 3 , − 6 9 , 7 27 , − 8 81 , 9 243 , assuming that the pattern of the first few terms continues. SOLUTION We are given that a1 = 5 3 a2 = − 6 9 a3 = 7 27 a4 = − 8 81 a5 = 9 243 .
Answer:
The formula to the sequence
5/3, -6/9, 7/27, -8/81, 9/243, ...
is
(-1)^n. (4 + n). 3^(-n)
For n = 1, 2, 3, ...
Step-by-step explanation:
The sequence is
5/3, - 6/9, 7/27, - 8/81, 9/243, ...
We notice the following
- That the numbers are alternating between - and +
- That the numerator of a number is one greater than the numerator of the preceding number. The first number being 5.
- That the denominator of a number is 3 raised to the power of (2 minus the position of the number)
Using these observations, we can write a formula for the sequence.
(-1)^n for n = 1, 2, 3, ... takes care of the alternation between + and -
(4 + n) for n = 1, 2, 3, ... takes care of the numerators 5, 6, 7, 8, ...
3^(-n) for n = 1, 2, 3, ... takes care of the denominators 3, 9, 27, 81, 243, ...
Combining these, we have the formula to be
(-1)^n. (4 + n). 3^(-n)
For n = 1, 2, 3, ...
The final formula is [tex]a_n = ((-1)^{ (n+1)} * (n + 4)) / (3^n).[/tex]
Finding the General Term of the Sequence
The sequence given is: 5/3, -6/9, 7/27, -8/81, 9/243. To find the formula for the general term (nth term) of this sequence, we need to carefully analyze the patterns in both the numerators and the denominators separately.
Numerator Analysis: The numerators of the given sequence are 5, -6, 7, -8, 9. Notice the pattern: the numerators alternate between positive and negative signs and increase by 1 each time. Thus, for the nth term, the numerator can be given by the formula:[tex](-1)^{(n+1)} * (n + 4).[/tex]Denominator Analysis: The denominators of the sequence are 3, 9, 27, 81, 243. These form a geometric sequence where each term is multiplied by 3. The nth term of this sequence can be expressed as [tex]3^n.[/tex]Combining the results from the numerator and denominator analysis, the general term of the sequence, an, is:
[tex]a_n = ((-1)^{ (n+1)} * (n + 4)) / (3^n).[/tex]
Complete Question:- Find a formula for the general term of the sequence 5 3 , − 6 9 , 7 27 , − 8 81 , 9 243 , assuming that the pattern of the first few terms continues
What is the slope of the line that passes through the points (1,7) and (-4,-8)?
Answer:
The slope of the line that passes through the points (1,7) and (-4,-8) is 3.
Step-by-step explanation:
The equation of a line has the following format.
[tex]y = ax + b[/tex]
In which a is the slope.
Passes through the point (1,7)
When [tex]x = 1, y = 7[/tex]
So
[tex]y = ax + b[/tex]
[tex]7 = a + b[/tex]
Passes through the point (-4, -8)
When [tex]x = -4, y = -8[/tex]
So
[tex]y = ax + b[/tex]
[tex]-8 = -4a + b[/tex]
We have to solve the following system
[tex]a + b = 7[/tex]
[tex]-4a + b = -8[/tex]
We want to find a.
From the first equation
[tex]b = 7 - a[/tex]
Replacing in the second equation
[tex]-4a + b = -8[/tex]
[tex]-4a + 7 - a = -8[/tex]
[tex]-5a = -15[/tex]
[tex]5a = 15[/tex]
[tex]a = \frac{15}{5}[/tex]
[tex]a = 3[/tex]
The slope of the line that passes through the points (1,7) and (-4,-8) is 3.
g On any day, the probability of rain is 0.3. The occurrence of rain on any day is independent of the occurrence of rain on any other day. Calculate the probability that, starting with tomorrow, the second day ofrain will occur within 5 days
Answer:
0.249579
Step-by-step explanation:
P(rain) = 0.3
P(no rain) = 1 - 0.3 = 0.7
The event of rain falling a second time within the next 5 days is possible in these ways
1. Rain on days 1 and 2
2. Rain on days 1 and 3; none on day 2
3. Rain on days 1 and 4; none on days 2 and 3
4. Rain on days 1 and 5; none on days 2, 3 and 4
5. Rain on days 1 and 6; none on day 2, 3, 4 and 5
[tex]P(\text{second rain within 5 days}) = 0.3^2+0.3^2\times0.7+0.3^2\times0.7^2+0.3^2\times0.7^3+0.3^2\times0.7^4 = 0.3^2(1+0.7+0.7^2+0.7^3+0.7^4)= 0.09\times2.7731=0.249579[/tex]
A stone is thrown horizontally with a speed of 15 m/s from the top of a vertical cliff at the edge of a lake. If the stone hits the water 2.0 s later, the height of the cliff is closest to
Answer:
20 m
Step-by-step explanation:
We are given that
Initial horizontal speed ,[tex]u_x=15 m/s[/tex]
Time, t=2 s
Initial vertical velocity, [tex]u_y=0[/tex]
We know that
[tex]h=u_yt+\frac{1}{2}gt^2[/tex]
Where [tex]g=-9.8m/s^2[/tex]
Using the formula
[tex]h=0(2)+\frac{1}{2}(-9.8)(2)^2[/tex]
[tex]h=-19.6 m\approx -20 m[/tex]
The negative sign indicates that the displacement of stone is in downward direction.
Hence, the height of the cliff is closest to 20 m
A blood bank asserts that a person with type O blood and a negative Rh factor (Rh?) can donate blood to any person with any blood type. Their data show that 43% of people have type O blood and 19% of people have Rh? factor; 45% of people have type O or Rh? factor.1.) Find the probability that a person has both type O blood and the Rh? factor.2.) Find the probability that a person does NOT have both type O blood and the Rh? factor.
Answer:P(O)UP(Rh)=17% while n(O)U(Rh) complement is 87%
Step-by-step explanation:
Since 43% represent those blood group O and 19% those with blood Rh,summing up gives 62%
Subtract 45% from 62%=17%
The probability that the person doesn't have either blood group is1 -17%=83%
Answer:
The probability that a person has a positive Rh factor given that he/she has type O blood is 82 percent.
There is a greater probability for a person to have a Positive Rh factor given type A blood than a person to have a positive Rh factor given type O blood.
Which of the following is the cheapest route to visit each city using the "Brute Force Method"
starting from A and ending at A.
Answer:
ACDBA, $900Step-by-step explanation:
Since there are 4 cities, there are (4-1)! = 6 possible routes. Half of those are the reverse of the other half, so there are 6/2 = 3 different possible routes. All of those are listed among the answer choices, along with their cost. All you need to do is choose the answer with the lowest cost:
ACDBA, $900
__
At $960, the other two routes are higher cost.
1. Hank is an intelligent student and usually makes good grades, provided that he can review the course material the night before the test. For tomorrow's test, Hank is faced with a small problem: His fraternity brothers are having an all-night party in which he would like to participate. Hank has three options: a1 - party all night: 2-divide the night equally between studying and partying: 3 - study all night. Tomorrow's exam can be easy (s1), moderate (S2) or tough (53), depending on the professor's unpredictable mood. Hank anticipates the following scores:
S1 S2 S3
a1 85 60 40
a2 92 85 81
a3 100 88 82
(a) Recommend a course of action for Hank based on each of the four criteria of decisions under uncertainty.
(b) Suppose that Hank is more interested in the letter grade he will get. The dividing scores for the passing letter grades A to Dare 90, 80, 70 and 60, respectively. What should the decision/s be?
Answer and Step-by-step explanation:
The answer is attached below
The expected pay-off for a₃ is maximum. Then the decision a₃ (study all night) is considered.
What are statistics?Statistics is the study of collection, analysis, interpretation, and presentation of data or to discipline to collect, summarise the data.
Hank is an intelligent student and usually makes good grades, provided that he can review the course material the night before the test.
For tomorrow's test, Hank is faced with a small problem: His fraternity brothers are having an all-night party in which he would like to participate.
Hank has three options:
a₁ - party all night
a₂ - divide the night equally between studying and partying
a₃ - study all night
Tomorrow's exam can be easy (S₁), moderate (S₂) or tough (S₃), depending on the professor's unpredictable mood. Hank anticipates the following scores:
S₁ S₂ S₃
a₁ 85 60 40
a₂ 92 85 81
a₃ 100 88 82
Decision under uncertainty
1. For maximum criterion when the exam is tough.
a₁ = 40, a₂ = 81, and a₃ = 82
Since 82 is the maximum out of the minimum. Then the optional action is a₃ (study all night).
2. For maximum criterion when the exam is easy.
a₁ = 85, a₂ = 92, and a₃ = 100
Since 82 is the maximum out of the maximum. Then the optional action is a₃ (study all night).
3. Regret criterion
First, find the regret matrix.
S₁ S₂ S₃ Max. regret
a₁ 15 28 42 42
a₂ 8 3 1 8
a₃ 0 0 0 0
From the maximum regret column, we find that the regret corresponding to the course of action is a₃ is minimum. Therefore, decision a₃ (study all night) will be considered.
4. Laplace criterion
The probability of occurrence is 1/3.
Therefore, the expected pay-off for each decision will be
E(a₁) = 61.67, E(a₂) = 86, and E(a₃) = 90
Therefore, the expected pay-off for a₃ is maximum.
Thus, decision a₃ (study all night) is considered.
More about the statistics link is given below.
https://brainly.com/question/10951564
The Supreme Court recently ruled that a police department in Florida did not violate any rights of privacy when a police helicopter flew over the backyard of a suspected drug dealer and noticed marijuana growing on his property. Many people, including groups like the Anti-Common Logic Union, felt that the suspect's right to privacy outweighed the police department's need to protect the public at large. The simple idea of sacrificing a right to serve a greater good should be allowed in certain cases. In this particular case the danger to the public wasn't extremely large; marijuana is probably less dangerous than regular beer. But anything could have been in that backyard—a load of cocaine, an illegal stockpile of weapons, or other major threats to society.
Final answer:
The question addresses the complex balance between individual privacy rights and law enforcement's authority to conduct searches, as protected and outlined by the Fourth Amendment. The Supreme Court has ruled on various cases that determine the scope of these rights, including exceptions that allow warrantless searches under specific circumstances. The ongoing evolution of privacy rights aligns with societal and technological changes, demanding constant legal reassessment.
Explanation:
Understanding Privacy Rights and Law Enforcement Searches
The case you are referring to touches on the complexities of privacy rights within the scope of law enforcement. Specifically, it deals with the interpretation of the Fourth Amendment which protects citizens from unreasonable searches and seizures. This protection extends to government actions and sets boundaries for police searches to respect individual privacy. However, there have been exceptions carved out that enable law enforcement to operate under certain circumstances without a warrant. This issue becomes even more complex with modern technology like drones, which can bypass traditional expectations of privacy.
For instance, the reasonable expectation of privacy is a key legal concept that dictates whether particular searches or seizures may be deemed reasonable without a warrant. Situations such as being visible from public airspaces or instances of exigent circumstances can fall outside the protections intended by the Fourth Amendment. Moreover, the amendment necessitates a search warrant to be obtained before conducting most searches or seizures. Nevertheless, Supreme Court rulings have established that there are scenarios where the warrant requirement is not applicable, such as when the items in question are in plain view or consent to search is given.
Privacy rights continue to evolve with societal changes and technological advancements. Courts and lawmakers constantly revisit and redefine the levels of privacy individuals can expect, balancing this against the interests of law enforcement and public safety. Matters such as the decriminalization of marijuana at state levels and exceptions to privacy within educational settings reflect the continuing dialogue and legal interpretation surrounding privacy rights and enforcement powers.
A supervisor must split 60 hours of overtime between five people. One employee must be assigned twice the number of hours as each of the other four employees. How many hours of overtime will be assigned to each employee?
Solution:
Given that,
A supervisor must split 60 hours of overtime between five people
One employee must be assigned twice the number of hours as each of the other four employees
Let "x" be the number of hours overtime per person.
One person does 2x hours of overtime
Which means,
(number of hours overtime per person)(6 person) = 60 hours
6x = 60
x = 10
Thus, 4 people do 10 hours and one person does 20 hours
At one SAT test site students taking the test for a second time volunteered to inhale supplemental oxygen for 10 minutes before the test. In fact, some received oxygen, but others (randomly assigned) were given just normal air. Test results showed that 42 of 66 students who breathed oxygen improved their SAT scores, compared to only 35 of 63 students who did not get the oxygen. Which procedure should we use to see if there is evidence that breathing extra oxygen can help test-takers think more clearly
the correct choice is:
E. 2-proportion Z-test
To determine if there is evidence that breathing extra oxygen can help test-takers think more clearly, we should use the 2-proportion Z-test.
This test is appropriate because we are comparing two proportions (the proportion of students who improved their SAT scores among those who breathed oxygen and those who did not) from two independent groups (students who received oxygen and those who did not).
Therefore, the correct choice is:
E. 2-proportion Z-test
The probable question maybe:
At one SAT test site students taking the test for a second time volunteered to inhale supplemental oxygen for 10 minutes before the test. In fact, some received oxygen but others (randomly assigned) were given just normal air. Test results showed that 42 of 66 students who breathed oxygen improved their SAT scores, compared to only 35 of 63 students who did not get the oxygen Which procedure should we use to see if there is evidence that breathing extra oxygen can help test-takers think more clearly?
A. 1-proportion 2-test
B matched pairs t-test
C 2-sample t-test
D. 1-sample t-test
E. 2-proportion Z-test
A chi-square test for independence should be used to analyze the effect of breathing extra oxygen on SAT score improvements, by comparing observed frequencies of score improvements with expected frequencies under the null hypothesis.
To determine if there is evidence that breathing extra oxygen can help test-takers think more clearly, a statistical test of significance is appropriate. In this scenario, you would typically use a chi-square test for independence to see if there is a significant association between the treatment (oxygen vs. normal air) and the outcome (improvement in SAT scores). The chi-square test compares the observed frequencies of events (here, the number of students who improved) with the frequencies we would expect to see if there were no association between the treatment and the outcome.
The procedure involves calculating a chi-square statistic, which reflects how far the observed frequencies are from the expected frequencies assuming the null hypothesis is true (no effect of breathing extra oxygen). If the resulting p-value is less than the chosen significance level (commonly 0.05), we can reject the null hypothesis and conclude that there is evidence to suggest a relationship between breathing extra oxygen and improved SAT scores.
suppose you have 3 bags containing only apples and oranges. bag a has 2 apples and 4 oranges, bag b has 8 apples and 4 oranges, and bag c has 1 apple and 3 oranges. you pick 1 fruit (at random) from each bag. a) what is the probability that you picked exactly 2 apples? b) suppose you picked 2 apples but forgot which bag they came from. what is the probability that you picked an apple from bag a?
Answer:
Step-by-step explanation:
There are three bags
Bag A
2apples and 4 oranges
P(A¹)=2/6
P(A¹)=⅓
P(O¹)=4/6
P(O¹)=⅔
Bag B
8 apples and 4 oranges
P(A²)=8/12
P(A²)=⅔
P(O²)=4/12
P(O²)=⅓
Bag C
1 apple and 3 oranges
P(A³)=¼
P(O³)=¾
Note
P(A¹) means probability of Apple in bag A
P(A²) means probability of Apple in bag B
P(A³) means probability of Apple in bag C
P(O¹) means probability of oranges in bag A.
P(O²) means probability of oranges in bag B.
P(O³) means probability of oranges in bag C.
a. The probability of picking exactly two apples can be analyzed as
Picking apple in bag A and picking apple in bag B and picking orange in bag C or picking apple in bag A and picking orange in bag B and picking apple in bag C or picking orange in bag A and picking apple in bag B and picking apple in bag C.
Then,
P(exactly two apples)=P(A¹ n A² n O³) + P(A¹ n O² n A³) + P(O¹ n A² n A³)
Since they are mutually exclusive
Then,
P(exactly two apples) =
P(A¹) P(A²)P(O³) + P(A¹)P(O²)P(A³) + P(O¹) P(A²) P(A³)
P(exactly two apples) =
(⅓×⅔×¾)+(⅓×⅓×¼)+(⅔×⅔¼)
P(exactly two apples)=1/6 +1/36 +1/9
P(exactly two apples)= 11/36
b. Probability that an apple comes from Bag A out of the two apple will be Picking apple in bag A and picking apple in bag B and picking orange in bag C or picking apple in bag A and picking orange in bag B and picking apple in bag C.
P(an apple belongs to bag A)=P(A¹ n A² n O³) + P(A¹ n O² n A³)
Since they are mutually exclusive
Then,
P(an apple belongs to bag A) =
P(A¹) P(A²)P(O³) + P(A¹)P(O²)P(A³)
P(an apple belongs to bag A) =
(⅓×⅔×¾)+(⅓×⅓×¼)
P(an apple belongs to bag A)=1/6 +1/36
P(an apple belongs to bag A)= 7/36
The University of Arkansas recently approved out of state tuition discounts for high school students from any state. The students must qualify by meeting certain standards in terms of GPA and standardized test scores. The goal of this new policy is to increase the geographic diversity of students from states beyond Arkansas and its border states. Historically, 90% of all new students came from Arkansas or a bordering state. Ginger, a student at the U of A, sampled 180 new students the following year and found that 157 of the new students came from Arkansas or a bordering state. Does Ginger’s study provide enough evidence to indicate that this new policy is effective with a level of significance 10%? What would be the correct decision?
a. Reject H0; conclude that the new policy does not increase the percentage of students from states that don’t border Arkansasb. Fail to reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas
c. Reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas
d. Fail to reject H0; conclude that the new policy does not increase the percentage of students from states that don’t border Arkansas
Answer:
[tex]z=\frac{0.872 -0.9}{\sqrt{\frac{0.9(1-0.9)}{180}}}=-1.252[/tex]
[tex]p_v =P(z<-1.252)=0.105[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% the proportion of students who came from Arkansas or a bordering state is not significantly lower than 0.9
b. Fail to reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas
Step-by-step explanation:
Data given and notation n
n=180 represent the random sample taken
X=157 represent the students who came from Arkansas or a bordering state
[tex]\hat p=\frac{157}{180}=0.872[/tex] estimated proportion of students who came from Arkansas or a bordering state
[tex]p_o=0.9[/tex] is the value that we want to test
[tex]\alpha=0.1[/tex] represent the significance level
Confidence=90% or 0.90
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the true proportion is higher or not than 0.9.:
Null hypothesis:[tex]p\geq 0.9[/tex]
Alternative hypothesis:[tex]p < 0.9[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.872 -0.9}{\sqrt{\frac{0.9(1-0.9)}{180}}}=-1.252[/tex]
Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided [tex]\alpha=0.1[/tex]. The next step would be calculate the p value for this test.
Since is a left tailed test the p value would be:
[tex]p_v =P(z<-1.252)=0.105[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% the proportion of students who came from Arkansas or a bordering state is not significantly lower than 0.9
b. Fail to reject H0; conclude that the new policy increases the percentage of students from states that don’t border Arkansas
A drawer contains 4 different pairs of gloves. Suppose we choose 3 gloves randomly, what is the probability that there is no matching pair?
Answer:
The probability that there is no matching pair is 4/7 = 0.5714286.
Step-by-step explanation:
For the first glove we have no restrictions. For the second glove, we have 6 gloves that works for us and 1 that doesnt work (the one that matches the first glove), hence we have 6 possibilities out of 7. Once we pick a good second glove, for the last glove, we only have 4 cases that doesnt match the other two pairs, out of 6 total. This means that the probability that there is no matching pair is 6/7*4/6 = 4/7.
Bob can row 13mph in still water. The total time to travel downstream and return upstream to the starting point is 2.6 hours. If the total distance downstream and back is 32 miles. Determine the speed of the river (current speed)
Answer:
"s" is the river speed 16 miles is the one-way distance
Time = distance / velocity
2.6 = [16 / (13 + s) ] + [16 / (13 -s) ]
That solves to s = 3 miles per hour
Step-by-step explanation:
Answer: the speed of the current is 3 mph
Step-by-step explanation:
Let x represent the speed of the river current.
Bob can row 13mph in still water. Assuming that while rowing upstream, he rowed against the current, this means that his total speed upstream is (13 - x) mph
Assuming that while rowing downstream, he rowed with the current, this means that his total speed downstream is (13 + x) mph.x
If the total distance downstream and back is 32 miles. Assuming the distance upstream and downstream is the same, then the distance upstream is 32/2 = 16 miles. Distance downstream is also 16 miles.
Time = Distance/speed
Time taken to row upstream is
16/(13 - x)
Time taken to row downstream is
16/(13 + x)
If total time spent is 2.6 hours, it means that
16/(13 - x) + 16/(13 + x) = 2.6
Cross multiplying, it becomes
16(13 + x) + 16(13 - x) = 2.6(13 + x)(13 - x)
= 208 + 16x + 208 - 16x = 2.6(169 - 13x + 13x - x²)
416 = 2.6(169 - x²)
416/2.6 = 169 - x²
160 = 169 - x²
x² = 169 - 160 = 9
x =√9
x = 3
You bicycle along a straight flat road with a safety light attached to one foot. Your bike moves at a speed of 10 km/hr and your foot moves in a circle of radius 24 cm centered 34 cm above the ground, making one revolution per second.
(a) Find parametric equations for x and y which describe the path traced out by the light, where y is distance (in cm) above the ground and x the horizontal distance (in cm) starting position of the center of the circle around which your foot moves. Assuming the light starts cm above the ground, at the front of its rotation.
x(t)=
y(t)=
(b) How fast (in revolutions/sec) would your foot have to be rotating if an observer standing at the side of the road sees the light moving backward?
Rotate at ? revolutions/second.
The parametric equations similar to a sinusoidal wave are x(t) = 10000t/3600 + 24cos(2πt) and y(t) = 34 + 24sin(2πt). For an observer to see the light moving backward, the foot would have to be making physical revolutions faster than the bike is moving forward, or approximately 7 revolutions/sec.
Explanation:The light attached to the foot is effectively forming a sinusoidal path as it moves along, creating a circular path while also advancing. Let's start by exploring the parametric equations.
The horizontal position (x) will be a combination of the distance traveled by the bike in time t (which is 10 km/hr * t converted to cm/sec) and the horizontal projection of the circular motion of the foot. The vertical position (y) will be a combination of the base height above the ground and the vertical projection of the circular motion of the foot.
So we have:
x(t) = 10000t/3600 + 24cos(2πt)
y(t) = 34 + 24sin(2πt)>
For your foot to appear to move backward from the perspective of an observer, the foot would have to move faster than the bicycle. This could be calculated by the ratio of bike speed to circumference of rotation. The rotation speed needs to at least meet this ratio.
Rotation speed = 10km/hr / (2π * 0.24m) = ~7 revolutions/sec
Learn more about Physics of cycling here:https://brainly.com/question/35869557
#SPJ3
Each of the ODEs below is second order in y, with y1 as a solution. Reduce the ODE from being second order in y to being first order in ????, with ???? being the only response variable appearing in the ODE. Combine like terms. Show your work.
Answer:
Step-by-step explanation:
The detailed steps and appropriate workings is as shown in the attached file.
The question asks about reducing a second-order ODE to first-order given a known solution, which is achieved using the method of reduction of order to find a new function v(y) leading to a first-order ODE.
To reduce a second-order ordinary differential equation (ODE) to a first-order ODE, given that y1 is a solution, you can apply the method of reduction of order. This involves introducing a new function v(y) such that y can be expressed as a product of the known solution y1 and this new function v(y). Essentially, you substitute y = y1*v into the original second-order ODE and differentiate as necessary to obtain an equation in terms of v and its derivatives only, effectively transforming the equation into first-order.
The steps include (i) expressing y in terms of y1 and v, (ii) differentiating this expression to find the derivatives of y, and (iii) substituting all of this into the equation that y1 obeys. The resultant first-order equation will only involve the function v and its first derivative, v'. By solving this simplified equation, you can find v and thus the second solution to the original second-order ODE.
A 2005 survey found that 7% of teenagers (ages 13 to 17) suffer from an extreme fear of spiders (arachnophobia). At a summer camp there are 10 teenagers sleeping in each tent. Assume that these 10 teenagers are independent of each other. What is the probability that at least one of them suffers from arachnophobia
Answer:
Probability that at least one of them suffers from arachnophobia is 0.5160.
Step-by-step explanation:
We are given that a 2005 survey found that 7% of teenagers (ages 13 to 17) suffer from an extreme fear of spiders (arachnophobia).
Also, At a summer camp there are 10 teenagers sleeping in each tent.
Firstly, the binomial probability is given by;
[tex]P(X=r) =\binom{n}{r}p^{r}(1-p)^{n-r} for x = 0,1,2,3,....[/tex]
where, n = number of trials(teenagers) taken = 10
r = number of successes = at least one
p = probability of success and success in our question is % of
the teenagers suffering from arachnophobia, i.e. 7%.
Let X = Number of teenagers suffering from arachnophobia
So, X ~ [tex]Binom(n= 10,p=0.07)[/tex]
So, probability that at least one of them suffers from arachnophobia
= P(X >= 1) = 1 - probability that none of them suffers from arachnophobia
= 1 - P(X = 0) = 1 - [tex]\binom{10}{0}0.07^{0}(1-0.07)^{10-0}[/tex]
= 1 - (1 * 1 * [tex]0.93^{10}[/tex] ) = 1 - 0.484 = 0.5160 .
Therefore, Probability that at least one of them suffers from arachnophobia is 0.5160 .
Maya is planning a bridal shower for her best friend. At the party, she wants to serve 3 beverages, 3 appetizers, and 3 desserts, but she does not have time to cook. She can choose from 12 bottled drinks, 12 frozen appetizers, and 10 prepared desserts at the supermarket. How many different ways can Maya pick the food and drinks to serve at the bridal shower
Using the combinations formula C(n, r), the number of ways Maya can choose the food and drinks for the bridal shower is computed as: C(12,3) for drinks * C(12,3) for appetizers * C(10,3) for desserts.
Explanation:The subject of this question is combinations in Mathematics. Maya has choices of 12 bottled drinks, 12 frozen appetizers, and 10 prepared desserts. She wants to pick 3 of each, and the order of selection does not matter. We can use the combination formula to calculate the number of ways she can do this:
For the bottled drinks, the number of combinations can be calculated as C(12,3).For the frozen appetizers, the number of combinations can be calculated as C(12,3).For the prepared desserts, the number of combinations can be calculated as C(10,3).So the total number of different ways Maya can pick the food and drinks to serve at the bridal shower is C(12,3) * C(12,3) * C(10,3).
Learn more about Combinations here:https://brainly.com/question/24703398
#SPJ3
A random sample of 8 recent college graduates found that starting salaries for architects in New York City had a mean of $42,653 and a standard deviation of $9,114. There are no outliers in the sample data set. Construct a 95% confidence interval for the average starting salary of all architects in the city.
A. (35222.41, 50083.59)
B. (34506.12, 50799.88)
C. (36337.32, 48968.68)
D. (35032.29, 50273.71)
Answer:
C. (36337.32, 48968.68)
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]
Now, find M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 1.96*\frac{9114}{\sqrt{8}} = 6315.68[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 42653 - 6315.68 = 36337.32.
The upper end of the interval is the sample mean added to M. So it is 42653 + 6315.68 = 48968.68.
So the correct answer is:
C. (36337.32, 48968.68)
The correct option is D.
[tex](35032.29,50273.71)[/tex]
Probability Sampling:Probability sampling is described as a sampling method in which the person or researcher chooses samples from a larger population using a method based on the theory of probability. For the participant, it is necessary to choose a random selection.
Note that margin of Error [tex]E=\frac{t\alpha }{2}\ast \frac{s}{\sqrt{n}} \\[/tex]
Lower Bound [tex]X=\frac{-t\alpha }{2}\ast \frac{s}{\sqrt{n}} \\[/tex]
Upper Bound [tex]X=\frac{+t\alpha }{2}\ast \frac{s}{\sqrt{n}}[/tex]
Where,
[tex]\frac{\alpha }{2}=\frac{\left ( 1-confidence \ level \right )}{2}=0.025\\\frac{t\alpha }{2}=critical \ t \ for \ the \ confidence \ interval=2.364624252[/tex]
[tex]S[/tex]=sample standard deviation[tex]=9114[/tex]
[tex]n[/tex]=sample size[tex]=8[/tex]
[tex]df=n-1=7[/tex]
Thus, the Margin of Error[tex]E=7619.49468[/tex]
Lower bound[tex]=35033.50532[/tex]
Upper bound[tex]=50272.49468[/tex][
Thus, the confidence interval is[tex](35033.50532 \ , 50272.49468 )[/tex]
Learn more about the topic Probability Sampling: https://brainly.com/question/22241230