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%.
A manufacturing process outputs parts having a normal distribution with a mean of 30 cm and standard deviation of 2 cm. From a production sample of 80 parts, what proportion of the sample can be expected to fall between 28 and 32 cm
Answer:
68% of the sample can be expected to fall between 28 and 32 cm
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 30
Standard deviation = 2
What proportion of the sample can be expected to fall between 28 and 32 cm
28 = 30 - 2
28 is one standard deviation below the mean
32 = 30 + 2
32 is one standard deviation above the mean.
By the Empirical Rule, 68% of the sample can be expected to fall between 28 and 32 cm
Based on the provided information about the characteristic roots and the right hand side function g(t), determine the appropriate form of a particular solution to be used with the undetermined coefficient method.
(a) r1=-2i; r2=2i g(t)=2sin(2t) + 3cos(2t)
(b) r1=r2=0; r3=1 g(t)= t^2 +2t + 3
Answer:
Yp = t[Asin(2t) + Acos(2t)]
Yp = t²[At² + Bt + C]
Step-by-step explanation:
The term "multiplicity" means when a given equation has a root at a given point is the multiplicity of that root.
(a) r1=-2i; r2=2i g(t)=2sin(2t) + 3cos(2t)
As you can notice the multiplicity of this equation is 1 since the roots r1 = 2i and r2 = 2i appear for only once.
The form of a particular solution will be
Yp = t[Asin(2t) + Acos(2t)]
where t is for multiplicity 1
(b) r1=r2=0; r3=1 g(t)= t² +2t + 3
As you can notice the multiplicity of this equation is 2 since the roots r1 = r2 = 0 appears 2 times.
The form of a particular solution will be
Yp = t²[At² + Bt + C]
where t² is for multiplicity 2
Suppose there is currently a tax of $50 per ticket on airline tickets. Sellers of airline tickets are required to pay the tax to the government. If the tax is reduced from $50 per ticket to $30 per ticket, then thea. demand curve will shift upward by $20, and the price paid by buyers will decrease by less than $20.b. demand curve will shift upward by $20, and the price paid by buyers will decrease by $20.c. supply curve will shift downward by $20, and the effective price received by sellers will increase by less than $20.d. supply curve will shift downward by $20, and the effective price received by sellers will increase by $20.
Answer:
the demand curve will shift upward by $20 and the price paid by buyers will decrease by $20.
Step-by-step explanation:
the reduction in the fixed amount of tax from $50 t0 $30 will bring about reduction of $20 in the price of ticket. the reduction in the price of the ticket, other factors held constant, will brings about change in the demand curve.
The correct answer is D, as supply curve will shift downward by $20, and the effective price received by sellers will increase by $20.
This is so because by reducing the tax, the value that each airline will effectively increase. In this way, as they obtain greater income for each ticket, the airlines will in turn increase the offer of tickets.
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PLEASE HELP 69 POINTS, BRAINLIEST, 5 STARS, AND THANKS.
Answer:
a) 125 scoops b) 31 scoops
Step-by-step explanation:
The sink = 4000/3 pi
a) The volume of a Cone is 1/3 (pi)(h)(r)^2, meaning...
(1/3)(pi)(8)(2^2)=32/3 pi, and...
(4000pi/3)/(32pi/3) =
4000/32 =
125
b) The volume of a Cone is 1/3 (pi)(h)(r)^2, meaning...
(1/3)(pi)(8)(4^2) = 128pi/3, and...
(4000pi/3)/(128pi/3) =
4000/128 =
31.25, which is approx. 31
Answer:
a) 125
b) 31
Step-by-step explanation:
a) radius = 2
Volume of cup = ⅓×pi×r²×h
= 32pi/3
No. of scoops = 4000pi/3 ÷ 32pi/3
= 125
b) radius = 4
Volume of cup = ⅓×pi×r²×h
= 128pi/3
No. of scoops = 4000pi/3 ÷ 128pi/3
= 31.25
Nearest whole number: 31
Consider a sample with a mean of 500 and a standard deviation of 100. What are the z-scores for the following data values: 560, 650, 500, 450, and 300? z-score for 560 z-score for 650 z-score for 500 z-score for 450 z-score for 300
The z-scores for the given data values are approximately:
560: 0.6
650: 1.5
500: 0
450: -0.5
300: -2
Given the mean (μ) of 500 and the standard deviation (σ) of 100, we can calculate the z-scores for the provided data values:
The z-score (also known as the standard score) measures how many standard deviations a data point is away from the mean. It is calculated using the formula:
z = (x - μ) / σ
Where:
x is the data value
μ is the mean of the sample
σ is the standard deviation of the sample
For x = 560:
z = (560 - 500) / 100 = 0.6
For x = 650:
z = (650 - 500) / 100 = 1.5
For x = 500:
z = (500 - 500) / 100 = 0
For x = 450:
z = (450 - 500) / 100 = -0.5
For x = 300:
z = (300 - 500) / 100 = -2
So, the z-scores for the given data values are approximately:
560: 0.6
650: 1.5
500: 0
450: -0.5
300: -2
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The z-scores for the data values 560, 650, 500, 450, and 300 in a sample with a mean of 500 and a standard deviation of 100 are 0.6, 1.5, 0, -0.5, and -2 respectively. They are computed using the formula z = (X - μ) / σ.
Explanation:This question refers to the concept of z-scores in statistics, which is a part of Mathematics. A z-score indicates how many standard deviations a given data point is from the mean. The formula to calculate the z-score is: z = (X - μ) / σ, where X is the data point, μ is the mean, and σ is the standard deviation.
The given sample has a mean (μ) of 500 and a standard deviation (σ) of 100. Let's calculate the z-scores:
For X = 560: z = (560 - 500) / 100 = 0.6 For X = 650: z = (650 - 500) / 100 = 1.5 For X = 500: z = (500 - 500) / 100 = 0 For X = 450: z = (450 - 500) / 100 = -0.5 For X = 300: z = (300 - 500) / 100 = -2
So the z-scores for the data values 560, 650, 500, 450, and 300 are 0.6, 1.5, 0, -0.5, and -2 respectively.
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A number between 1 and 15
is chosen at random. What
is the probability that the
number chosen will be a
multiple of 5? please help just learning this will mark brainest got people answering it maki g up junk to get the answer then needing help
The probability that the chosen will be a multiple of 5 between 1 and 15 is [tex]\frac{2}{13}[/tex].
Solution:
Given data:
Number between 1 and 15 is 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 and 14.
Total numbers between 1 and 15 = 13
N(S) = 13
Multiple of 5 between 1 and 15 = 5, 10
Number of multiples of 5 between 1 and 15 = 2
N(A) = 2
Probability of multiple of 5 between 1 and 15:
[tex]$P(A)=\frac{N(A)}{N(S)}[/tex]
[tex]$P(A)=\frac{2}{13}[/tex]
The probability that the chosen will be a multiple of 5 between 1 and 15 is [tex]\frac{2}{13}[/tex].
Final answer:
To determine the probability of selecting a multiple of 5 between 1 and 15, count the multiples (5 and 10) and divide by the total number (15), resulting in a probability of 2/15.
Explanation:
The question is related to the concept of probability in mathematics. To find the probability that a randomly chosen number between 1 and 15 is a multiple of 5, you simply count how many multiples of 5 are in that range and divide by the total number of possibilities (15). The multiples of 5 between 1 and 15 are 5 and 10, so there are 2 favorable outcomes. Hence, the probability is 2 divided by 15, which equals 2/15.
Baseball's World Series is a maximum of seven games, with the winner being the first team to win four games. Assume that the Atlanta Braves and the Minnesota Twins are playing in the World Series and that the first two games are to be played in Atlanta, the next three games at the Twins' ballpark, and the last two games, if necessary, back in Atlanta. Taking into account the projected starting pitchers for each game and the home field advantage, the probabilities of Atlanta winning each game are as follows: Game 1 2 3 4 5 6 7 Probability of Win 0.4 0.55 0.42 0.56 0.55 0.39 0.52 Set up a spreadsheet simulation model for which whether Atlanta wins or loses each game is a random variable. What is the probability that the Atlanta Braves win the World Series? If required, round your answer to two decimal places. What is the average number of games played regardless of winner? If required, round your answer to one decimal place.
Answer:
For part ;a)
probability that the Atlanta Braves win the World Series=0.40
For part b ;b)
average number of games played regardless of winner =5.8
Step-by-step explanation:
using the rand() function for the simulation of the probabilities and compare with given probabilites if less than above then Atlanta wins otherwise loses
a)
probability that the Atlanta Braves win the World Series=0.40
b)
average number of games played regardless of winner =5.8
Consider the population of four juvenile condors. Their weights in pounds are : 4, 5, 7, 12 (a) Let x be the weight of a juvenile condor. Write the possible unique values for x: (NOTE: Separate each value in the list with a comma.) . (b) Find the mean of the population: (c) Let x¯ be the average weight from a sample of two juvenile condors. List all possible outcomes for x¯. (If a value occurs twice, make sure to list it twice.) This is the sampling distribution for samples of size 2: (NOTE: Separate each value in the list with a comma.) . (d) Find the mean of the sampling distribution: Note: You can earn partial credit on this problem.
Answer:
a) 4, 5, 7, 12
b) 7
c) 4.5, 6.5, 8, 6, 8.5, 9.5
d) 7.167
Step-by-step explanation:
We are given the following in the question:
4, 5, 7, 12
a) unique values for x
4, 5, 7, 12
b) mean of the population
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]\mu =\displaystyle\frac{28}{4} = 7[/tex]
c) sampling distribution for samples of size 2
Sample size, n = 2
Possible samples of size 2 are (4,5),(4,7),(4,12),(5,7),(5,12),(7,12)
Sample means are:
[tex]\bar{x_1} = \dfrac{4+5}{2} = 4.5\\\\\bar{x_2} = \dfrac{4+7}{2} = 6.5\\\\\bar{x_3} = \dfrac{4+12}{2} = 8\\\\\bar{x_4} = \dfrac{5+7}{2} = 6\\\\\bar{x_5} = \dfrac{5+12}{2} = 8.5\\\\\bar{x_6} = \dfrac{7+12}{2} = 9.5[/tex]
Thus, the list is 4.5, 6.5, 8, 6, 8.5, 9.5
d) mean of the sampling distribution
[tex]\bar{x} = \dfrac{4.5 + 6.5 + 8+ 6 + 8.5+ 9.5}{6} = \dfrac{43}{6} = 7.167[/tex]
The unique values for x are 4, 5, 7, and 12 pounds. The population mean is 7 pounds, and the sampling distribution for the average weight from a sample of two juvenile condors has possible outcomes of 4.5, 5.5, 6, 8, 8.5, and 9.5 pounds, with a mean of 7 pounds.
Explanation:To answer the student's questions regarding juvenile condor weights:
Possible unique values for x: The unique weights of the juvenile condors are 4, 5, 7, 12 pounds.Mean of the population: The mean (μ) is the average of all the values. To calculate the mean, add all the weights and divide by the number of condors: (4 + 5 + 7 + 12) / 4 = 28 / 4 = 7 pounds.Possible outcomes for ¯x (the average weight from a sample of two juvenile condors): To list all possible outcomes for ¯x, calculate the mean of all possible pairs of condor weights:Since some pairs can be chosen in two ways (e.g., condor 1 and 2, or condor 2 and 1), record the mean for both occurrences. However, in this example, no average is repeated because all condor weights are unique.Mean of the sampling distribution: We can find this by averaging the possible outcomes for ¯x: (4.5 + 5.5 + 8 + 6 + 8.5 + 9.5) / 6 = 42 / 6 = 7 pounds, which is identical to the population mean.
Suppose Gabe, an elementary school student, has just finished dinner with his mother, Judy. Eyeing the nearby cookie jar, Gabe asks his mother if he can have a cookie for dessert. She tells Gabe that she needs to check his backpack to make sure that he finished his homework. Gabe cannot remember where he left his backpack, but he knows for sure that he did not complete his homework and will not be allowed to eat a cookie. Gabe believes his only option is to quickly steal a cookie while his mother is out of the room. Judy then leaves the room to look for Gabe's backpack. Assume that Judy could return at any time in the next 90 seconds with equal probability. For the first 30 seconds, Gabe sheepishly wonders if he will get caught trying to grab a nearby cookie. After waiting and not seeing his mother, Gabe decides that he needs a cookie and begins to take one from the jar. Assuming it takes Gabe 15 seconds to grab a cookie from the jar and devour it without a trace, what is the probability that his mother returns in time to catch Gabe stealing a cookie? Please round your answer to the nearest two decimal places.
Answer:
the probability that his mother would return on time to catch Gabe stealing is given by:[tex]\frac{15}{60}=0.25[/tex].
Step-by-step explanation:
probability is given by; the required outcome over the number of possible outcome.
Step1; we have to subtract the first initial 30 secs that Gabe spent wondering if he will be caught. i.e 90 sec-30 secs= 60 Sec.
Step2: we divide the number of seconds that he can finish a cookie without being caught over the number of seconds left before his mother returns. that is why we have [tex]\frac{15}{60} = 0.25[/tex]
The probability that his mother returns in time to catch Gabe stealing a cookie is 0.25.
Calculation of the probability:Judy could return at any time in the next 90 seconds with equal probability. For the first 30 seconds, Gabe sheepishly wonders if he will get caught trying to grab a nearby cookie. And, it takes Gabe 15 seconds to grab a cookie from the jar.
Now the probability is
[tex]= 15 \div (90 - 30)\\\\= 15 \div 60[/tex]
= 0.25
Hence, we can conclude that The probability that his mother returns in time to catch Gabe stealing a cookie is 0.25.
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Determine which of the following numbers could not be probabilities, and why? Select all that apply. A. The number 175% could not be a probability because it is larger than 100%. B. The number 5.91 could not be a probability because it is larger than 1. C. The number 0.66 could not be a probability because it is smaller than 1. D. The number 0.002 could not be a probability because it is extremely small. E. The number negative 110% could not be a probability because it is negative.
Answer:
A Not a probability B.Probability C.not a probability d.Probability E propability F.not a propability
Step-by-step explanation:
A propability of any event is a number between zero and one
A)175% converted 1.75 so lies outside zero and one not a propability
B)5.91 lies outside the zero and one rage not a propability
C)0.66 lies in the zero and one rage therefore a propability
D)0.002 Lies within the zero and one range therefore a propability
E) -110% converted equals -1.1 so does not lies in this range not a propability
The following numbers could not be probabilities:
A. The number 175% could not be a probability because it is larger than 100%.
B. The number 5.91 could not be a probability because it is larger than 1.
E. The number negative 110% could not be a probability because it is negative.
Probabilities must be between 0 and 1, inclusive. This is because a probability represents the likelihood of an event occurring, and there is no event that can be more likely than certain (1) or less likely than impossible (0).
The numbers 0.66 and 0.002 are valid probabilities.
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A shipping company handles containers in three different sizes: (1) 27 ft3 (3 × 3 × 3), (2) 125 ft3, and (3) 512 ft3. Let Xi (i = 1, 2, 3) denote the number of type i containers shipped during a given week. With μi = E(Xi) and σi2 = V(Xi), suppose that the mean values and standard deviations are as follows: μ1 = 220 μ2 = 250 μ3 = 120 σ1 = 9 σ2 = 13 σ3 = 8 (a) Assuming that X1, X2, X3 are independent, calculate the expected value and variance of the total volume shipped. [Hint: Volume = 27X1 + 125X2 + 512X3.] expected value ft3 variance ft6 (b) Would your calculations necessarily be correct if the Xi's were not independent? Explain. Both the expected value and the variance would be correct. The expected value would be correct, but the variance would not be correct. Neither the expected value nor the variance would be correct. The expected value would not be correct, but the variance would be correct.
a. Expectation is linear, so that
[tex]E[27X_1+125X_2+512X_3]=27E[X_1]+125E[X_2]+512E[X_3]=98,630[/tex]
The variance of a sum of independent random variables is equal to a weighted sum of the variances, with weights equal to the squares of the coefficients of the [tex]X_i[/tex]:
[tex]V[27X_1+125X_2+512X_3]=27^2V[X_1]+125^2V[X_2]+512^2V[X_3]=2,306,838[/tex]
b. If the [tex]X_i[/tex] were dependent on one another, we would have the same expectation, but now the variance of a sum of random variables becomes the sum of their covariances:
[tex]\displaystyle\sum_{i=1}^3V[\alpha_iX_i]=\sum_{i=1}^3\sum_{j=1}^3\mathrm{Cov}[X_i,X_j]=\sum_{i=1}^3{\alpha_i}^2V[X_i]+2\sum_{i\neq j}\alpha_i\alpha_j\mathrm{Cov}[X_i,X_j][/tex]
where
[tex]\mathrm{Cov}[X_i,X_j]=E[(X_i-E[X_i])(X_j-E[X_j])]=E[X_iX_j]-E[X_i]E[X_j][/tex]
When we assumed independence, we were granted [tex]E[X_iX_j]=E[X_i]E[X_j][/tex], but this may not be the case if the variables are dependent.
The expected value of the total volume shipped is 98630 ft3. The variance of the total volume shipped is 17423640 ft6. If the variables were not independent, only the expected value would remain correct.
Explanation:To calculate the expected value of the total volume shipped, we use the linearity of the expectation:
E(Total Volume) = E(27X1 + 125X2 + 512X3)
E(Total Volume) = 27E(X1) + 125E(X2) + 512E(X3)
E(Total Volume) = 27(μ1) + 125(μ2) + 512(μ3)
E(Total Volume) = 27(220) + 125(250) + 512(120)
E(Total Volume) = 5940 + 31250 + 61440
E(Total Volume) = 98630 ft3
To calculate the variance of the total volume shipped, given that X1, X2, X3 are independent:
V(Total Volume) = V(27X1 + 125X2 + 512X3)
V(Total Volume) = 272V(X1) + 1252V(X2) + 5122V(X3)
V(Total Volume) = 729(σ12) + 15625(σ22) + 262144(σ32)
V(Total Volume) = 729(92) + 15625(132) + 262144(82)
V(Total Volume) = 729(81) + 15625(169) + 262144(64)
V(Total Volume) = 59049 + 2639375 + 16777216
V(Total Volume) = 17423640 ft6
If the Xi's were not independent, the expected value would still be correct but the variance would not be correct because the calculation of variance for non-independent variables includes covariance terms that are not present for independent variables.
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What is the greatest common factor of 10x^2y^2 – 8xy?
(^=exponent)
Answer:
2xy
Step-by-step explanation:
10x²y² - 8xy
2xy(5xy - 4)
The greatest common factor (GCF) of the expressions 10x^2y^2 and 8xy is 2xy. The GCF for the numbers is 2, and for the variables x and y it's x and y, respectively.
Explanation:The subject is the greatest common factor (GCF), applied to algebraic expressions. The GCF is the largest expression that can be multiplied by another expression to get the original expression. For the expressions 10x^2y^2 and 8xy, the numbers are 10 and 8. Their GCF is 2. The term x appears in both expressions, and the lowest power of x is 1. The term y also appears in both expressions, with the lowest power being 1. So, 2x and y are the GCFs for the numbers and the variables respectively. Combining them, we get 2xy, which is the GCF of 10x^2y^2 and 8xy.
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Prove the superposition principle for nonhomogeneous equations. Suppose that y1 is a solution to Ly1 = f(x) and y2 is a solution to Ly2 = g(x) (same linear operator L). Show that y = y1 + y2 solves Ly = f(x) + g(x). Differential Equation.
Answer:
Step-by-step explanation:
Given Data
Suppose That [tex]y_{1}[/tex] is a solution of L[tex]y_{1}[/tex] = F(x)
and [tex]y_{2}[/tex] is a solution of [tex]Ly_{2}[/tex] = g(x)
L is liner operator
∴ [tex]L(y_{1}+y_{2} )[/tex] = L[tex]y_{1}[/tex] +[tex]Ly_{2}[/tex]
L(y) = F(x) + g(x)
[tex]y_{1}+y_{2}[/tex] is the solution to L(y) = F(x) + g(x)
E.g the liner operator be L = [tex]\frac{d}{dx}[/tex]
[tex]\frac{d}{dx}[/tex] [tex]y_{1}[/tex] = f(x)
[tex]\frac{d}{dx}[/tex] [tex]y_{2}[/tex] = g(x)
[tex]\frac{d}{dx}[/tex] [tex](y_{1}+y_{2} )[/tex] = [tex]\frac{d}{dx}[/tex] [tex]y_{1}[/tex] + [tex]\frac{d}{dx}[/tex] [tex]y_{2}[/tex] = f(x) + g(x)
Final Answer:
y = y1 + y2 solves the equation Ly = f(x) + g(x).
Explanation:
Certainly! To prove the superposition principle for nonhomogeneous linear differential equations, we will utilize the properties of linear operators and the given solutions for the differential equations.
Let's define L as a linear differential operator. Being linear implies that for any functions u(x) and v(x), and any constants a and b, the operator satisfies the following properties:
1. L(u + v) = L(u) + L(v) (additivity)
2. L(au) = aL(u) (homogeneity)
Given there are two functions y1(x) and y2(x) that are solutions to the nonhomogeneous linear differential equations:
Ly1 = f(x)
Ly2 = g(x)
We need to prove that if you take a linear combination of y1 and y2, the result y = y1 + y2 will also be a solution to the combined nonhomogeneous equation Ly = f(x) + g(x).
Here's how we do it:
Consider a linear combination of y1 and y2, denoted as y = y1 + y2. Apply the linear operator L to both sides of this equation:
L(y) = L(y1 + y2)
Since L is a linear operator, we can apply the additivity property:
L(y) = L(y1) + L(y2)
Now, we know that y1 and y2 are solutions to their respective nonhomogeneous equations, so we can substitute f(x) for L(y1) and g(x) for L(y2):
L(y) = f(x) + g(x)
Thus, we have shown that y = y1 + y2 solves the equation Ly = f(x) + g(x).
This is the proof of the superposition principle for nonhomogeneous linear differential equations. It shows that solutions to such equations can be added together to obtain a new solution corresponding to the sum of the nonhomogeneous parts.
Please help..........
bro I hope someone can help you like an expert or something...
The length of time needed to complete a certain test is normally distributed with mean 35 minutes and standard deviation 15 minutes. Find the probability that it will take between 31 and 40 minutes to complete the test.
Answer:
Probability = 0.23572 .
Step-by-step explanation:
We are given that the length of time needed to complete a certain test is normally distributed with mean 35 minutes and standard deviation 15 minutes.
Let X = length of time needed to complete a certain test
Since, X ~ N([tex]\mu,\sigma^{2}[/tex])
The z probability is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1) where, [tex]\mu[/tex] = 35 and [tex]\sigma[/tex] = 15
So, P(31 < X < 40) = P(X < 40) - P(X <= 31)
P(X < 40) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{40-35}{15}[/tex] ) = P(Z < 0.33) = 0.62930
P(X <= 31) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{31-35}{15}[/tex] ) = P(Z < -0.27) = 1 - P(Z <= 0.27)
= 1 - 0.60642 = 0.39358
Therefore, P(31 < X < 40) = 0.62930 - 0.39358 = 0.23572 .
To find the probability that it takes between 31 and 40 minutes to complete the test when the test completion times are normally distributed with a mean (μ) of 35 minutes and a standard deviation (σ) of 15 minutes, we can use the properties of the normal distribution.
Firstly, we need to standardize our times (31 and 40 minutes) to find the corresponding z-scores. The z-score is a measure of how many standard deviations an element is from the mean. We use the following formula to calculate the z-score:
\[ z = \fraction{(X - \mu)}{\sigma} \]
where:
- \( X \) is the value for which we are finding the z-score
- \( \mu \) is the mean
- \( σ \) is the standard deviation
We will calculate the z-scores for both 31 minutes and 40 minutes.
For \( X = 31 \):
\[ z_{31} = \fraction{(31 - 35)}{15} = \fraction{-4}{15} \approx -0.267 \]
For \( X = 40 \):
\[ z_{40} = \fraction{(40 - 35)}{15} = \fraction{5}{15} \approx 0.333 \]
Next, we look up these z-scores in the standard normal distribution (z-distribution) table, which will give us the area to the left of each z-score.
If we don't have the z-distribution table available, we might use statistical software or a calculator with a normal distribution function. But let's proceed as if we are using the z-table.
Let's say our z-table gives us the following areas:
Area to the left of \( z_{31} \approx -0.267 \): Approximately 0.3944 (please note that actual values will depend on the specific z-table or calculator you are using).
Area to the left of \( z_{40} \approx 0.333 \): Approximately 0.6304.
Now we want the area between these two z-scores, which represents the probability that the test completion time is between 31 and 40 minutes. To find this, we can subtract the area for \( z_{31} \) from the area for \( z_{40} \):
Probability \( P(31 < X < 40) = P(z_{40}) - P(z_{31}) \)
Substituting the values,
\[ P(31 < X < 40) = 0.6304 - 0.3944 = 0.2360 \]
So the probability that it will take between 31 and 40 minutes to complete the test is approximately 0.2360, or 23.60%.
A sampling method is ________ dependent quantitative independent qualitative when an individual selected for one sample does not dictate which individual is to be in the second sample.
Answer:
A sampling method is independent qualitative when an individual selected for one sample does not dictate which individual is to be in the second sample.
Step-by-step explanation:
individuals selected for one sample do not dictate which individuals are to be in the second sample, while dependent quantitative involves individuals selected to be in one sample are used to determine the individuals in the second sample.
Final answer:
An independent sampling method allows for selection of individuals for one sample without influencing the selection for another, maintaining the integrity of quantitative research.
Explanation:
A sampling method is considered independent when an individual selected for one sample does not dictate which individual is to be in the second sample. This type of sampling strategy ensures that the selection of participants for one sample does not influence the selection for another, which is crucial for maintaining the validity and reliability of research findings, particularly in quantitative research.
In research, particularly in studies involving statistical analysis, it is essential to use appropriate sampling methods to ensure that the results are representative of the wider population. Probability sampling methods are often employed in quantitative research as they provide a means to make generalizations from the sample to the population. Types of probability samples include simple random samples, stratified samples, and cluster samples.
In Quebec, 90 percent of the population subscribes to the Roman Catholic religion. In a random sample of eight Quebecois, find the probability that the sample contains at least five Roman Catholics.
Answer:
Probability that the sample contains at least five Roman Catholics = 0.995 .
Step-by-step explanation:
We are given that In Quebec, 90 percent of the population subscribes to the Roman Catholic religion.
The Binomial distribution probability is given by;
P(X = r) = [tex]\binom{n}{r}p^{r}(1-p)^{n-r}[/tex] for x = 0,1,2,3,.......
Here, n = number of trials which is 8 in our case
r = no. of success which is at least 5 in our case
p = probability of success which is probability of Roman Catholic of
0.90 in our case
So, P(X >= 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)
= [tex]\binom{8}{5}0.9^{5}(1-0.9)^{8-5} + \binom{8}{6}0.9^{6}(1-0.9)^{8-6} + \binom{8}{7}0.9^{7}(1-0.9)^{8-7} + \binom{8}{8}0.9^{8}(1-0.9)^{8-8}[/tex]
= 56 * [tex]0.9^{5} * (0.1)^{3}[/tex] + 28 * [tex]0.9^{6} * (0.1)^{2}[/tex] + 8 * [tex]0.9^{7} * (0.1)^{1}[/tex] + 1 * [tex]0.9^{8}[/tex]
= 0.995
Therefore, probability that the sample contains at least five Roman Catholics is 0.995.
The probability that a random sample of eight Quebecois contains at least five Roman Catholics is approximately 0.9950 or 99.50%.
Given:
- The probability of success (subscribing to the Roman Catholic religion) p = 0.9 .
- The number of trials (sample size) n = 8 .
- We want to find the probability of getting at least 5 successes.
The binomial probability formula is:
[tex]\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \][/tex]
Where:
- [tex]\( \binom{n}{k} \)[/tex] is the binomial coefficient [tex]\(\frac{n!}{k!(n-k)!} \)[/tex],
- k is the number of successes,
- p is the probability of success,
- (1-p) is the probability of failure.
We need to find [tex]\( P(X \geq 5) \)[/tex], which is the sum of the probabilities of getting exactly 5, 6, 7, and 8 successes.
[tex]\[ P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) \][/tex]
Calculating each term individually:
1.
[tex]\( P(X = 5) \):\[ P(X = 5) = \binom{8}{5} (0.9)^5 (0.1)^3 \]\[ \binom{8}{5} = \frac{8!}{5!3!} = 56 \]\[ P(X = 5) = 56 \times (0.9)^5 \times (0.1)^3 \]\[ P(X = 5) = 56 \times 0.59049 \times 0.001 \]\[ P(X = 5) = 56 \times 0.00059049 \]\[ P(X = 5) \approx 0.0331 \][/tex]
2. P(X = 6) :
[tex]\[ P(X = 6) = \binom{8}{6} (0.9)^6 (0.1)^2 \]\[ \binom{8}{6} = \frac{8!}{6!2!} = 28 \]\[ P(X = 6) = 28 \times (0.9)^6 \times (0.1)^2 \]\[ P(X = 6) = 28 \times 0.531441 \times 0.01 \]\[ P(X = 6) = 28 \times 0.00531441 \]\[ P(X = 6) \approx 0.1488 \][/tex]
3. P(X = 7) :
[tex]\[ P(X = 7) = \binom{8}{7} (0.9)^7 (0.1)^1 \]\[ \binom{8}{7} = \frac{8!}{7!1!} = 8 \]\[ P(X = 7) = 8 \times (0.9)^7 \times 0.1 \]\[ P(X = 7) = 8 \times 0.4782969 \times 0.1 \]\[ P(X = 7) = 8 \times 0.04782969 \]\[ P(X = 7) \approx 0.3826 \][/tex]
4. P(X = 8) :
[tex]\[ P(X = 8) = \binom{8}{8} (0.9)^8 (0.1)^0 \]\[ \binom{8}{8} = 1 \]\[ P(X = 8) = 1 \times (0.9)^8 \times 1 \]\[ P(X = 8) = (0.9)^8 \]\[ P(X = 8) = 0.43046721 \]\[ P(X = 8) \approx 0.4305 \][/tex]
Now sum these probabilities:
[tex]\[ P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) \]\[ P(X \geq 5) \approx 0.0331 + 0.1488 + 0.3826 + 0.4305 \]\[ P(X \geq 5) \approx 0.9950 \][/tex]
So, the probability that the sample contains at least five Roman Catholics is approximately 0.9950 or 99.50%.
In a sample of 100 steel canisters, the mean wall thickness was 8.1 mm with a standard deviation of 0.5 mm. Someone says that the mean thickness is less than 8.2 mm. With what level of confidence can this statement be made? (Express the final answer as a percent and round to two decimal places.)
Answer:
This statement can be made with a level of confidence of 97.72%.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 8.1 mm
Standard Deviation, σ = 0.5 mm
Sample size, n = 100
We are given that the distribution of thickness is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
Standard error due to sampling:
[tex]=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.5}{\sqrt{100}} = 0.05[/tex]
P(mean thickness is less than 8.2 mm)
P(x < 8.2)
[tex]P( x < 8.2)\\\\ = P( z < \displaystyle\frac{8.2 - 8.1}{0.05})\\\\ = P(z < 2)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x < 8.2) =0.9772 = 97.72\%[/tex]
This statement can be made with a level of confidence of 97.72%.
The z-score calculation reveals that the mean thickness is less than 8.2 mm with approximately 97.50% confidence.
Explanation:To determine the level of confidence that the mean thickness is less than 8.2 mm, we can calculate the z-score of the sample mean and then find the corresponding percentile in the standard normal distribution. The z-score is calculated by the formula:
z = (x - μ) / (σ / √n)
where x is the sample mean, μ is the population mean, σ is the standard deviation, and n is the sample size.
Using the given information, x = 8.1 mm, μ = 8.2 mm (the mean to test against), σ = 0.5 mm, and n = 100, we get:
z = (8.1 mm - 8.2 mm) / (0.5 mm / √100) = -0.1 / 0.05 = -2
The z-score corresponds to a percentile in the standard normal distribution, which can be found using a z-table or statistical software. A z-score of -2 corresponds to approximately the 2.5th percentile, which means that the sample mean is less than 8.2 mm with about 97.5% confidence.
Therefore, the statement that the mean thickness is less than 8.2 mm can be made with approximately 97.50% confidence.
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What length is the shortest path from A to Cin the graph below?
Answer: It's C.5
Step-by-step explanation:
Dandelions are studied for their effects on crop production and lawn growth. In one region, the mean number of dandelions per square meter was found to be 2. We are interested in the number of dandelions in this region. (a) Find the probability of no dandelions in a randomly selected area of 1 square meter in this region. (Round your answer to four decimal places.) (b) Find the probability of at least one dandelion in a randomly selected area of 1 square meter in this region. (Round your answer to four decimal places.)
Answer:
(a) The probability that there are no dandelions in a randomly selected area of 1 square meter in this region is 0.1353.
(b) The probability that there are at least one dandelion in a randomly selected area of 1 square meter in this region is 0.8647.
Step-by-step explanation:
Let X = number of dandelions per square meter.
The average number of dandelions per square meter is, λ = 2.
The random variable X follows a Poisson distribution with parameter λ = 2.
The probability mass function of X is:
[tex]P(X=x)=\frac{e^{-2}2^{x}}{x!};\ x=0,1,2,3...[/tex]
(a)
Compute the value of P (X = 0) as follows:
[tex]P(X=0)=\frac{e^{-2}2^{0}}{0!}=\frac{0.13534\times 1}{1}=0.1353[/tex]
Thus, the probability that there are no dandelions in a randomly selected area of 1 square meter in this region is 0.1353.
(b)
Compute the value of P (X ≥ 1) as follows:
P (X ≥ 1) = 1 - P (X < 1)
= 1 - P (X = 0)
[tex]=1-0.1353\\=0.8647[/tex]
Thus, the probability that there are at least one dandelion in a randomly selected area of 1 square meter in this region is 0.8647.
To find the probability of no dandelions, use the Poisson distribution formula with the mean value of 2. To find the probability of at least one dandelion, use the complement rule.
Explanation:To find the probability of no dandelions in a randomly selected area of 1 square meter in this region, we need to use the Poisson distribution. The mean number of dandelions per square meter is 2. The probability of no dandelions is given by the formula: P(X=0) = e^(-2) * (2^0 / 0!). Calculating this gives us approximately 0.1353.
To find the probability of at least one dandelion, we can use the complement rule. The probability of at least one dandelion is equal to 1 minus the probability of no dandelions. So, P(X >= 1) = 1 - P(X=0). Calculating this gives us approximately 0.8647.
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Based on outcomes since 1967, the probability a team from the National League wins the baseball World Series in any given year is 0.44. Based on outcomes since 1967, the probability a team from the National Football conference wins the Superbowl in any given year is 0.53. Assume whether a team from the National League wins the World Series in a year is independent of whether a team from the National Football Conference wins the Superbowl in the same year. What is the probability a team from the National League wins the World Series in a year if a team from the National Football Conference wins the Superbowl in the same year
Answer:
Probability a team from national football conference wins the same year will be equal to 0.53
Step-by-step explanation:
Probability that a National League team wins baseball world series = 0.44
Probability that a National Football conference team wins the Superbowl = 0.53
The two events are independent, since the sports played are different (baseball and football). This is a logical assumption to make.
Since these events are independent, the occurrence of one event will not change the probability of the other event. This means that it does not matter whether the national league team won or lost the world series. The probability of the football team winning the Superbowl will remain the same, which is 0.53.
A student needs 11 more classes to graduate. If she has met the prerequisites for all the classes, how many possible schedules for next semester could she make if she plans to take 3 classes?
Using the combination formula, the student can make 165 different possible schedules for next semester if she is planning to take three classes from the eleven available that she has met the prerequisites for.
Explanation:This problem is a combination problem where students need to select 3 classes from 11 available classes to plan her schedule. The combination formula is used when the order of election does not matter. Here, the formula to find a combination is written as C(n, r) = n! / [r!(n - r)!].
In this particular case, the student has 11 classes (n=11) and plans to take 3 classes (r=3). So, the combination becomes C(11,3) = 11! / [3!(11-3)!] = (11*10*9) / (3*2*1) = 165.
This means, the student can make 165 different possible schedules for next semester assuming she is planning to take three classes from the eleven available classes.
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The student could make 165 different schedules.
The formula for combinations is given by:
[tex]\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \][/tex]
where n! (n factorial) is the product of all positive integers up to n, and r is the number of items to choose.
Given that the student has 11 classes left and she plans to take 3 classes next semester, we substitute n = 11 and r = 3 into the combination formula:
[tex]\[ \binom{11}{3} = \frac{11!}{3!(11-3)!} \][/tex]
Now, we simplify the factorial expressions:
[tex]\[ \binom{11}{3} = \frac{11!}{3! \times 8!} \][/tex]
We can cancel out the common terms in the numerator and the denominator:
[tex]\[ \binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} \][/tex]
Now, we perform the arithmetic:
[tex]\[ \binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = \frac{990}{6} = 165 \][/tex]
Thus, the student has 165 different possible schedules for next semester if she is to take 3 classes out of the remaining 11.
The number of accidents per week at a hazardous intersection varies with mean 2.2 and standard deviation 1.1. This distribution takes only whole-number values, so it is certainly not Normal. (a) Let x be the mean number of accidents per week at the intersection during a year (52 weeks). What is the approximate distribution of x according to the central limit theorem
Answer:
Step-by-step explanation:
Hello!
The definition of the Central Limi Theorem states that:
Be a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.
As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.
X[bar]≈N(μ;σ²/n)
If the variable of interest is X: the number of accidents per week at a hazardous intersection.
There is no information about the distribution of this variable, but a sample of n= 52 weeks was taken, and since the sample is large enough you can approximate the distribution of the sample mean to normal. With population mean μ= 2.2 and standard deviation σ/√n= 1.1/√52= 0.15
I hope it helps!
Gabe Amodeo, a nuclear physicist, needs 60 liters of a 60% acid solution. He currently has a 40% solution and a 70% solution. How many liters of each does he need
to make the needed 60 liters of 60% acid solution?
Answer:
20 liters of the 40% solution
60-20=40 liters of the 70% solution
Step-by-step explanation:
40%x+70%(60-x)=60%(60)
0.4x+0.7(60-x)=0.6(60)
0.4x+42-0.7x=36
-0.3x+42=36
-0.3x=-42+36
-0.3x=-6
0.3x=6
3x=60
x=60/3
x=20 liters of the 40% solution
60-20=40 liters of the 70% solution
check: 0.4*20=8
0.7*40=28
8+28=36 and 0.6*60=36 also
A normally distributed population has mean 57,800 and standard deviation 750. Find the probability that a single randomly selected element X of the population is between 57,000 and 58,000. Find the mean and standard deviation of X - for samples of size 100. Find the probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000.
Answer:
(a) Probability that a single randomly selected element X of the population is between 57,000 and 58,000 = 0.46411
(b) Probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = 0.99621
Step-by-step explanation:
We are given that a normally distributed population has mean 57,800 and standard deviation 75, i.e.; [tex]\mu[/tex] = 57,800 and [tex]\sigma[/tex] = 750.
Let X = randomly selected element of the population
The z probability is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
(a) So, P(57,000 <= X <= 58,000) = P(X <= 58,000) - P(X < 57,000)
P(X <= 58,000) = P( [tex]\frac{X-\mu}{\sigma}[/tex] <= [tex]\frac{58000-57800}{750}[/tex] ) = P(Z <= 0.27) = 0.60642
P(X < 57000) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{57000-57800}{750}[/tex] ) = P(Z < -1.07) = 1 - P(Z <= 1.07)
= 1 - 0.85769 = 0.14231
Therefore, P(31 < X < 40) = 0.60642 - 0.14231 = 0.46411 .
(b) Now, we are given sample of size, n = 100
So, Mean of X, X bar = 57,800 same as before
But standard deviation of X, s = [tex]\frac{\sigma}{\sqrt{n} }[/tex] = [tex]\frac{750}{\sqrt{100} }[/tex] = 75
The z probability is given by;
Z = [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
Now, probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000 = P(57,000 < X bar < 58,000)
P(57,000 <= X bar <= 58,000) = P(X bar <= 58,000) - P(X bar < 57,000)
P(X bar <= 58,000) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] <= [tex]\frac{58000-57800}{\frac{750}{\sqrt{100} } }[/tex] ) = P(Z <= 2.67) = 0.99621
P(X < 57000) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{57000-57800}{\frac{750}{\sqrt{100} } }[/tex] ) = P(Z < -10.67) = P(Z > 10.67)
This probability is that much small that it is very close to 0
Therefore, P(57,000 < X bar < 58,000) = 0.99621 - 0 = 0.99621 .
Using the normal distribution and the central limit theorem, it is found that:
There is a 0.4641 = 46.41% probability that a single randomly selected element X of the population is between 57,000 and 58,000.For samples of size 100, the mean is of 57800 and the standard deviation is 75.There is a 0.9965 = 99.65% probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000.In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
It 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, which is the percentile of X.By the Central Limit Theorem, for sampling distributions of samples of size n, the mean is [tex]\mu[/tex] and the standard deviation is [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].In this problem:
Mean of 57800, thus [tex]\mu = 57800[/tex].Standard deviation of 750, thus [tex]\sigma = 750[/tex].The probability that a single randomly selected element X of the population is between 57,000 and 58,000 is the p-value of Z when X = 58000 subtracted by the p-value of Z when X = 57000, thus:
X = 58000:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{58000 - 57800}{750}[/tex]
[tex]Z = 0.27[/tex]
[tex]Z = 0.27[/tex] has a p-value of 0.6064.
X = 57000:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{57000 - 57800}{750}[/tex]
[tex]Z = -1.07[/tex]
[tex]Z = -1.07[/tex] has a p-value of 0.1423.
0.6064 - 0.1423 = 0.4641.
0.4641 = 46.41% probability that a single randomly selected element X of the population is between 57,000 and 58,000.
For samples of size 100, [tex]n = 100[/tex], and then:
[tex]s = \frac{750}{\sqrt{100}} = 75[/tex]
For samples of size 100, the mean is of 57800 and the standard deviation is 75.
Then, the probability is:
X = 58000:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{58000 - 57800}{75}[/tex]
[tex]Z = 2.7[/tex]
[tex]Z = 2.7[/tex] has a p-value of 0.9965.
X = 57000:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{57000 - 57800}{75}[/tex]
[tex]Z = -10.7[/tex]
[tex]Z = -10.7[/tex] has a p-value of 0.
0.9965 - 0 = 0.9965.
0.9965 = 99.65% probability that the mean of a sample of size 100 drawn from this population is between 57,000 and 58,000.
A similar problem is given at https://brainly.com/question/24663213
Multiply 2 1/2 x 3 2/3
Answer: 9.16666666667
Step-by-step explanation:
The function f(x,y)=2x + 2y has an absolute maximum value and absolute minimum value subject to the constraint 9x^2 - 9xy + 9y^2 =25. Use Lagrange multipliers to find these values.
Answer:
the minimum is located in x = -5/3 , y= -5/3
Step-by-step explanation:
for the function
f(x,y)=2x + 2y
we define the function g(x)=9x² - 9xy + 9y² - 25 ( for g(x)=0 we get the constrain)
then using Lagrange multipliers f(x) is maximum when
fx-λgx(x)=0 → 2 - λ (9*2x - 9*y)=0 →
fy-λgy(x)=0 → 2 - λ (9*2y - 9*x)=0
g(x) =0 → 9x² - 9xy + 9y² - 25 = 0
subtracting the second equation to the first we get:
2 - λ (9*2y - 9*x) - (2 - λ (9*2x - 9*y))=0
- 18*y + 9*x + 18*x - 9*y = 0
27*y = 27 x → x=y
thus
9x² - 9xy + 9y² - 25 = 0
9x² - 9x² + 9x² - 25 = 0
9x² = 25
x = ±5/3
thus
y = ±5/3
for x=5/3 and y=5/3 → f(x)= 20/3 (maximum) , while for x = -5/3 , y= -5/3 → f(x)= -20/3 (minimum)
finally evaluating the function in the boundary , we know because of the symmetry of f and g with respect to x and y that the maximum and minimum are located in x=y
thus the minimum is located in x = -5/3 , y= -5/3
Solve for x: 2 over x - 2 + 7 over x^2 - 4 = 5 over x
A. x = -4/3 and x = −5
B. x = -4/3 and x = 5
C. x = 4/3 and x = −5
D. x = 4/3 and x = 5
Answer:
B. x = -4/3 and x = 5
Step-by-step explanation:
2 / (x − 2) + 7 / (x² − 4) = 5 / x
2 / (x − 2) + 7 / ((x + 2) (x − 2)) = 5 / x
Multiply both sides by x − 2:
2 + 7 / (x + 2) = 5 (x − 2) / x
Multiply both sides by x + 2:
2 (x + 2) + 7 = 5 (x − 2) (x + 2) / x
Multiply both sides by x:
2x (x + 2) + 7x = 5 (x − 2) (x + 2)
Simplify:
2x² + 4x + 7x = 5 (x² − 4)
2x² + 4x + 7x = 5x² − 20
0 = 3x² − 11x − 20
Factor:
0 = (3x + 4) (x − 5)
x = -4/3 or x = 5
In an experiment, A,B, C, andD are events with probabilitiesP[A UB] = 5/8,P[A] =3/8,
P[C ∩D] = 1/3, andP[C] =1/2. Furthermore, Aand B are disjoint, whileC and D areindependent.
(a) Find P[A∩ B],P[B],P[A ∩Bc], andP[A UBc].
(b) Are A andB independent?
(c) FindP[D],P[C ∩Dc],P[Cc∩ Dc],andP[C|D].
(d) Find P[CU D] andP[C UDc].
(e) Are C andDcindependent?
Answer:
Step-by-step explanation:
Hello!
You have 4 events A, B, C and D
With probabilities:
P(A∪B)= 5/8
P(A)= 3/8
P(C∩D)= 1/3
P(C)= 1/2
A and B are disjoint events, this means that there are no shared elements between then and their intersection is void, symbolically A∩B= ∅, in consequence, these events are mutually exclusive.
C and D are independent events, this means that the occurrence of one of them does not affect the probability of occurrence of the other one in two consecutive repetitions.
a.
i. P(A∩B)= 0
⇒ Since A and B are disjoint events, the probability of their intersection is zero.
ii. A and B are mutually exclusive events, this means that P(A∪B)= P(A)+P(B)
⇒ From this expression, you can clear the probability of b as P(B)= P(A∪B)-P(A)= 5/8-3/8= 1/4
iii. If Bc is the complementary event of B, its probability would be P(Bc)= 1 - P(B)= 1 - 1/4= 3/4. If the events A and B are mutually exclusive and disjoint, it is logical to believe that so will be the events A and Bc, so their intersection will also be void:
P(A∩Bc)= 0
vi.P(A∪Bc)= P(A) + P(Bc)= 3/8+3/4= 9/8
b.
If A and B are independent then the probability of A is equal to the probability of A given B, symbolically:
P(A)= P(A/B)
[tex]P(A/B)= \frac{P(AnB)}{P(B)}= \frac{0}{1/4}= 0[/tex]
P(A)= 3/8
P(A) ≠ P(A/B) ⇒ A and B are not independent.
c.
i. P(D) ⇒ Considering C and D are two independent events, then we know that P(C∩D)= P(C)*P(D)
Then you can clear the probability of D as:
P(D)= P(C∩D)/P(C)= (1/3)/(1/2)= 2/3
ii. If Dc is the complementary event of D, then its probability is P(Dc)= 1 - P(D) = 1 - 2/3= 1/3
P(C∩Dc)= P(C)*P(Dc)= (1/2)*(1/3)= 1/6
iii. Now Cc is the complementary event of C, its probability is P(Cc)= 1 - P(C)= 1 - 1/2= 1/2
P(Cc∩Dc)= P(Cc)*P(Dc)= (1/2)*(1/3)= 1/6
vi. and e.
[tex]P(C/D)= \frac{P(CnD)}{P(D)} = \frac{1/3}{2/3} = 1/2[/tex]
P(C)=1/2
As you can see the P(C)=P(C/D) ⇒ This fact proves that the events C and D are independent.
d.
i. P(C∪D)= P(C) + P(D) - P(C∩D)= 1/2 + 2/3 - 1/3= 5/6
ii. P(C∪Dc)= P(C) + P(Dc) - P(C∩Dc)= 1/2 + 1/3 - 1/6= 2/3
I hope it helps!
Consider the following. (If an answer does not exist, enter DNE.) f '(x) = x2 + x − 20 (a) Find the open intervals on which f '(x) is increasing or decreasing.
Answer:
f'(x) is increasing in the open interval (-1/2, +∞), and it is decreasing in the interval (-∞, -1/2).
Step-by-step explanation:
f'(x) is a quadratic function with positive main coefficient, then it will be decreasing until the x-coordinate of its vertex and it will be increasing from there onwards. The x-coordinate of the vertex is given by the equation -b/2a = -1/2. Hence
- f'(x) is incresing in the interval (-1/2, +∞)
- f'(x) is decreasing in the interval (-∞, -1/2)
Final answer:
To find intervals where f'(x) is increasing or decreasing, locate the critical points by solving f'(x) = 0 and test the signs of f'(x) around these points. This process determines where the original function f(x) is increasing or decreasing.
Explanation:
To determine the open intervals in which the derivative of a function f(x), denoted as f'(x), is increasing or decreasing, one must:
Find all solutions of f'(x) = 0 within the interval [a, b]. These points are known as critical or stationary points.Examine the sign of f'(x) at points other than the critical points to determine the intervals where f'(x) is > 0 (indicating increasing behavior of f(x)) and where f'(x) is < 0 (indicating decreasing behavior).Identify the changes in the sign of f'(x) at the critical points to confirm if those points are local maxima or minima. The function f(x) will be evaluated at each of these stations to determine local extrema.In the scenario one would solve the equation for f'(x) = 0 to find the critical points. Once these are found, the signs of f'(x) to the left and right of these points can be tested to see where the function is increasing or always decreasing.