It is concluded that the alternative hypothesis H1: σ² < 100 is true.
The variance is the square of the standard deviation of a sample of observations. In order to test whether a given variance of the population is equal to a given value, we make use of the chi-square distribution.
Thus, let X be a random variable that has a normal distribution with mean μ and variance σ². The formula to calculate chi-square distribution is as follows:
chi-square (x²) = (n-1) * S² / σ²Where n = sample size, S² = sample variance, and σ² = population variance.
Now, let's perform a hypothesis test with the given data:
Null hypothesis:H0: σ² = 100
Alternative hypothesis:
H1: σ² < 100
The value of the test statistic is:chi-square (x²) = (n-1) * S² / σ²= (25-1) * 131.29 / 100= 33.82
The degrees of freedom (df) for the test is
:df = n - 1= 25 - 1= 24
The critical value for chi-square distribution at df = 24 and α = 0.01 is 9.7097.
Since the calculated test statistic (33.82) is greater than the critical value (9.7097), we reject the null hypothesis and conclude that there is evidence to suggest that the variance of the miles per gallon (mpg) is less than 100.
Therefore, it is concluded that the alternative hypothesis H1: σ² < 100 is true.
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determine whether the sequence =7 sin(11 6)11 6 converges or diverges. if it converges, find the limit.
The given sequence, (7 sin(nπ/6))/(nπ/6), converges to zero as n approaches infinity.
To determine whether the sequence converges or diverges, we can analyze the behavior of the terms as n approaches infinity.
Let's rewrite the sequence as (7 sin(πn/6))/(πn/6).
As n approaches infinity, the term πn/6 also approaches infinity. We know that the function sin(x) oscillates between -1 and 1 as x varies, but when x becomes very large, sin(x) approaches zero.
Since the numerator of the sequence is a bounded function (sin(πn/6) is bounded between -1 and 1), and the denominator (πn/6) grows infinitely, the entire sequence tends to zero.
Therefore, the given sequence converges to zero as n approaches infinity.
In summary, the sequence (7 sin(11π/6))/(11π/6) converges to zero as n approaches infinity.
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What is the probability of the event when we randomly select a permutation of the 26 lowercase letters of the English alphabet where a immediately precedes m, which immediately precedes z, in the permutation?
24!/26!
24/26
24/26!
1/26!
1/26
it is not 1/26
Therefore, the probability of randomly selecting a permutation with the desired arrangement is 24!/26!.
Since we want the letters "a", "m", and "z" to appear in the specified order in the permutation, we can treat them as a single unit. So we have 24 remaining letters to arrange along with the unit "amz".
The total number of permutations of the 26 letters is 26!.
Since "a", "m", and "z" are treated as a single unit, the total number of permutations with "a" immediately preceding "m" and "m" immediately preceding "z" is 24!.
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Axline Computers manufactures personal computers at two plants, one in Texas and the other in Hawall. The Texas plant has 50 employees; the Hawall plant has 20. A random sample of 10 employees is to be asked to fill out a benefits questionnaire. Round your answers to four decimal places.. a. What is the probability that none of the employees in the sample work at the plant in Hawaii? b. What is the probability that 1 of the employees in the sample works at the plant in Hawail? c. What is the probability that 2 or more of the employees in the sample work at the plant in Hawaii? d. What is the probability that 9 of the employees in the sample work at the plant in Texas?
a. Probability that none of the employees in the sample work at the plant in Hawaii: 0.0385
b. Probability that 1 of the employees in the sample works at the plant in Hawaii: 0.3823
c. Probability that 2 or more of the employees in the sample work at the plant in Hawaii: 0.5792
d. Probability that 9 of the employees in the sample work at the plant in Texas: 0.2707
a. To find the probability that none of the employees in the sample work at the plant in Hawaii, we need to calculate the probability of selecting all employees from the Texas plant.
The probability of selecting an employee from the Texas plant is (number of employees in Texas plant)/(total number of employees) = 50/70 = 0.7143.
Since we are sampling without replacement, the probability of selecting all employees from the Texas plant is:
P(All employees from Texas) = [tex](0.7143)^{10}[/tex] ≈ 0.0385.
Therefore, the probability that none of the employees in the sample work at the plant in Hawaii is approximately 0.0385.
b. To find the probability that 1 of the employees in the sample works at the plant in Hawaii, we need to calculate the probability of selecting exactly 1 employee from the Hawaii plant.
The probability of selecting an employee from the Hawaii plant is (number of employees in Hawaii plant)/(total number of employees) = 20/70 = 0.2857.
The probability of selecting exactly 1 employee from the Hawaii plant is given by the binomial probability formula:
P(1 employee from Hawaii) = [tex]C(10, 1) * (0.2857)^1 * (1 - 0.2857)^{10-1}[/tex] ≈ 0.3823.
Therefore, the probability that 1 of the employees in the sample works at the plant in Hawaii is approximately 0.3823.
c. To find the probability that 2 or more of the employees in the sample work at the plant in Hawaii, we need to calculate the complementary probability of selecting 0 or 1 employee from the Hawaii plant.
P(2 or more employees from Hawaii) = 1 - P(0 employees from Hawaii) - P(1 employee from Hawaii)
P(2 or more employees from Hawaii) = 1 - 0.0385 - 0.3823 ≈ 0.5792.
Therefore, the probability that 2 or more of the employees in the sample work at the plant in Hawaii is approximately 0.5792.
d. To find the probability that 9 of the employees in the sample work at the plant in Texas, we need to calculate the probability of selecting exactly 9 employees from the Texas plant.
The probability of selecting an employee from the Texas plant is 0.7143 (as calculated in part a).
The probability of selecting exactly 9 employees from the Texas plant is given by the binomial probability formula:
P(9 employees from Texas) = [tex]C(10, 9) * (0.7143)^9 * (1 - 0.7143)^{10-9}[/tex] ≈ 0.2707.
Therefore, the probability that 9 of the employees in the sample work at the plant in Texas is approximately 0.2707.
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Compute the probability that the sum of X and Y exceeds 1.
Let (X, Y) be random variables with joint density Jxy xy if 0≤x≤ 2, 0 ≤ y ≤ 1 fx,y(2,y) = = 0 otherwise
The probability that the sum of X and Y exceeds 1, with the specified joint density function, is 0. In terms of probability, this implies that the event of X + Y exceeding 1 is not possible based on the given distribution.
To compute the probability that the sum of X and Y exceeds 1, we need to calculate the integral of the joint density function over the region where X + Y > 1.
We have the joint density function:
f(x, y) = xy if 0 ≤ x ≤ 2, 0 ≤ y ≤ 1
f(x, y) = 0 otherwise
We want to find P(X + Y > 1), which can be expressed as the double integral over the region where X + Y > 1.
P(X + Y > 1) = ∫∫R f(x, y) dxdy
To determine the region R, we can set up the inequalities for X + Y > 1:
X + Y > 1
Y > 1 - X
Since the domain of x is from 0 to 2 and the domain of y is from 0 to 1, we have the following limits for integration:
0 ≤ x ≤ 2
1 - x ≤ y ≤ 1
Now, we can set up the integral:
P(X + Y > 1) = ∫∫R f(x, y) dxdy
= ∫0^2 ∫1-x¹ xy dydx
Evaluating this integral:
P(X + Y > 1) = ∫0² [x(y^2/2)]|1-x¹ dx
= ∫0² [x/2 - x^3/2] dx
= [(x^2/4 - x^4/8)]|0²
= (2/4 - 2^4/8) - (0/4 - 0^4/8)
= (1/2 - 16/8) - (0 - 0)
= (1/2 - 2) - 0
= -3/2
Therefore, the probability that the sum of X and Y exceeds 1 is -3/2. However, probabilities must be non-negative values between 0 and 1, so in this case, the probability is 0.
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between which pair of decimals should 4/7 be placed on a number line
o 0.3 and 0.4
o 0.4 and 0.5
o 0.5 and 0.6
o 0.6 and 0.7
To determine the pair of decimals between which 4/7 should be placed on a number line, we will convert 4/7 into a decimal.
We can do that by dividing 4 by 7 using a calculator or by long division method: `4 ÷ 7 = 0.5714...`.Hence, 4/7 as a decimal is 0.5714. To determine the pair of decimals between which 0.5714 should be placed on a number line, we can examine the given options.
Notice that option B is the most suitable. The number line below illustrates the correct position of 4/7 between 0.4 and 0.5:. Therefore, between the pair of decimals 0.4 and 0.5 should 4/7 be placed on a number line.
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0.5 and 0.6 are pair of decimals where 4/7 be placed on a number line.
To determine between which pair of decimals 4/7 should be placed on a number line, we need to find the approximate decimal value of 4/7.
Dividing 4 by 7, we get:
4/7
= 0.571428571...
Rounding this decimal to the nearest hundredth, we have:
=0.57
Since 0.57 is greater than 0.5 and less than 0.6, the correct pair of decimals between which 4/7 should be placed on a number line is 0.5 and 0.6.
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T is a linear transformation from R2 into R2. Show that T is invertible and find a formula for T-1. T (x1, x2) = (2x1 - 8x2, -2x1 + 7x2)
To show that the linear transformation T is invertible, we need to demonstrate that it is both injective (one-to-one) and surjective (onto).
Injectivity:
For T to be injective, we need to show that if T(x1, x2) = T(y1, y2), then (x1, x2) = (y1, y2). Let's assume that T(x1, x2) = T(y1, y2). This implies that:
(2x1 - 8x2, -2x1 + 7x2) = (2y1 - 8y2, -2y1 + 7y2).
From this, we obtain the following system of equations:
2x1 - 8x2 = 2y1 - 8y2 ---- (1)
-2x1 + 7x2 = -2y1 + 7y2 ---- (2)
To show that (x1, x2) = (y1, y2), we need to demonstrate that equations (1) and (2) hold. Let's manipulate these equations:
Equation (1) multiplied by 7:
14x1 - 56x2 = 14y1 - 56y2 ---- (3)
Equation (2) multiplied by 8:
-16x1 + 56x2 = -16y1 + 56y2 ---- (4)
Adding equations (3) and (4) together:
-2x1 = -2y1 ---- (5)
From equation (5), we can conclude that x1 = y1. Substituting this back into equation (1), we have:
2x1 - 8x2 = 2x1 - 8y2.
Simplifying this equation, we find that -8x2 = -8y2, which implies x2 = y2.
Therefore, we have shown that if T(x1, x2) = T(y1, y2), then (x1, x2) = (y1, y2), proving that T is injective.
Surjectivity:
To show that T is surjective, we need to demonstrate that for any vector (a, b) in R^2, there exists a vector (x1, x2) such that T(x1, x2) = (a, b).
Let's solve the following system of equations for x1 and x2:
2x1 - 8x2 = a ---- (6)
-2x1 + 7x2 = b ---- (7)
To solve this system, we can multiply equation (6) by 7 and equation (7) by 8, and then add them together:
14x1 - 56x2 + (-16x1 + 56x2) = 7a + 8b
-2x1 = 7a + 8b
Dividing both sides of the equation by -2:
x1 = (-7a - 8b)/2
Now, substitute x1 back into equation (6):
2((-7a - 8b)/2) - 8x2 = a
-7a - 8b - 8x2 = a
-8b - 8x2 = 8a
-8(x2 + a) = 8a - 8b
x2 + a = b - a
x2 = b - 2a
So, we have found the values of x1 and x2 in terms of a and b. Therefore, for any vector (a, b) in R^2, we can find a vector (x1, x2) such that T(x1, x2) = (a, b). This demonstrates that T is surjective.
Since T is both injective and surjective, it is invertible.
To find the formula for T^(-1), we need to determine the inverse transformation that maps vectors (a, b) back to (x1, x2).
We have found x1 = (-7a - 8b)/2 and x2 = b - 2a. Therefore, the inverse transformation T^(-1) is given by:
T^(-1)(a, b) = ((-7a - 8b)/2, b - 2a)
This formula represents the inverse of the linear transformation T.
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A student researcher was surprised to learn that the 2017 NCAA
Student-Athlete Substance Use Survey supported that college
athletes make healthier decisions in many areas than their peers in
the gener
A student researcher was surprised to learn that the 2017 NCAA Student-Athlete Substance Use Survey supported that college athletes make healthier decisions in many areas than their peers in the general population.
The 2017 NCAA Student-Athlete Substance Use Survey revealed interesting findings regarding the health behaviors of college athletes compared to their peers in the general population. Contrary to the researcher's initial expectations, the survey indicated that college athletes tended to make healthier decisions across various areas.
One key area where college athletes demonstrated healthier behaviors was substance use. The survey found that college athletes were less likely to engage in substance abuse compared to their non-athlete counterparts. This included lower rates of alcohol consumption, smoking, and illicit drug use among college athletes. These findings suggest that participating in collegiate sports may contribute to a lower likelihood of engaging in risky behaviors related to substance use.
Furthermore, the survey highlighted that college athletes were more likely to prioritize their overall health and well-being. They reported higher rates of engaging in regular physical activity and maintaining a balanced diet. This dedication to physical fitness and healthy eating habits may be attributed to the rigorous training and athletic demands placed on college athletes. Their commitment to their sport often translates into a conscious effort to maintain optimal health.
Additionally, the survey revealed that college athletes were more likely to prioritize their academic success. They reported higher rates of attending classes, completing assignments, and achieving better academic performance compared to non-athletes. This emphasis on academic success can be attributed to the unique demands placed on college athletes, who must balance their rigorous training schedules with their academic responsibilities. The discipline and time management skills required for their athletic pursuits often spill over into their academic lives, resulting in a greater commitment to their studies.
Overall, the 2017 NCAA Student-Athlete Substance Use Survey provided empirical evidence that college athletes tend to make healthier decisions in various areas compared to their peers in the general population. These findings underscore the positive impact of collegiate sports on the overall well-being of student-athletes. By promoting healthier behaviors and instilling values such as discipline and commitment, college athletics contribute to the development of well-rounded individuals who prioritize their physical and mental health, as well as their academic success.
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A student researcher was surprised to learn that the 2017 NCAA Student-Athlete Substance Use Survey supported that college athletes make healthier decisions in many areas than their peers in the general student body. He collected data of his own, focusing exclusively on male student-athletes to see if such habits vary based on one’s sport. He asked 93 male student-athletes whether they had engaged in binge-drinking in the last month (> 5 drinks in a single sitting). Data are provided in the table below.
Lacrosse
Hockey
Swimming
Row Totals
Yes – Binge
20
17
15
52
No – did not binge
16
15
10
41
Column totals
36
32
25
93
the rate of change of y with respect to x is one-half times the value of y. find an equation for y, given that when x = 0. you get:
The equation for y given that the rate of change of y with respect to x is one-half times the value of y is y = 2e^(x/2), where x is any real number.
Given that the rate of change of y with respect to x is one-half times the value of y and that the value of x is 0, find the equation for y.To solve this problem, we need to integrate both sides. [tex]dy/dx = (1/2)y, d/dy [ ln |y| ] = 1/2 dx + C[/tex], where C is a constant of integration.
If we now assume that[tex]y > 0, ln y = x/2 + C, y = e^(x/2 + C) = e^C * e^(x/2[/tex]).But we don't know the value of the constant, C, yet. To determine the value of C, we need to use the initial condition given by the question, namely that when[tex]x = 0, y = 2.C = ln 2, y = 2e^(x/2).[/tex]Therefore, the equation for y when x = 0 is y = 2.
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Assign probabilities to each outcome in the following 2
situations. Your answers may be 1 sentence each, as opposed to a
series of numbers. (a) A random experiment with five equally likely
outcomes. R
(a) In a random experiment with five equally likely outcomes, each outcome has a probability of 1/5 or 0.2.
In this situation, since there are five equally likely outcomes, each outcome has the same chance of occurring. Therefore, the probability of each outcome is equal and can be calculated by dividing 1 by the total number of outcomes. In this case, the total number of outcomes is five. Hence, the probability of each outcome is 1/5 or 0.2.
By assigning equal probabilities to each outcome, we assume that there is no preference or bias toward any specific outcome. This assumption is based on the principle of equally likely outcomes, which states that in certain situations where all outcomes are equally likely, the probability of each outcome is the same.
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describe how to translate the graph of y=sqrt x to obtain the graph of y=sqrt x+20
Answer:
The parent funcion is:
For this case we have two possible cases:
Case 1:
If the new function is:
We have the following transformation:
Horizontal translations:
Suppose that h> 0
To graph y = f (x-h), move the graph of h units to the right.
Answer:
shift right 15 units
Case 2:
If the function is:
We have the following transformation:
Vertical translations:
Suppose that k> 0
To graph y = f (x) -k, move the graph of k units down.
Answer:
shift down 15 units
Step-by-step explanation:
Answer:
To translate the graph of
�
=
�
y=
x
to obtain the graph of
�
=
�
+
20
y=
x
+20, you need to shift the entire graph vertically upwards by 20 units.
Step-by-step explanation:
The probability mass function of a discrete random variable X is given by the following table: X 1 2 3 4 5 6 P(X) 1/36 3/36 5/36 7/36 9/36 11/36 36/36-1 Find 1- Cumulative distribution function. 2- Dr
1- The cumulative distribution function (CDF) for the given probability mass function (PMF) is as follows:
X | 1 2 3 4 5 6
P(X)| 1/36 3/36 5/36 7/36 9/36 11/36
CDF | 1/36 4/36 9/36 16/36 25/36 36/36
2- The probability of the random variable X being greater than or equal to a certain value can be calculated using the CDF. The complementary probability, denoted as DR (the probability of X being less than a certain value), is calculated by subtracting the CDF value from 1. The DR values for each X are as follows:
X | 1 2 3 4 5 6
DR | 35/36 32/36 27/36 20/36 11/36 0/36
1- To calculate the cumulative distribution function (CDF), we need to sum up the probabilities of X being less than or equal to a certain value. Starting with X = 1, the CDF is 1/36 since it is the only value in the PMF. For X = 2, we add P(X=1) and P(X=2) to get 4/36, and so on until we reach X = 6.
2- The complementary probability, DR (the probability of X being less than a certain value), can be calculated by subtracting the CDF value from 1. For X = 1, DR is 1 - 1/36 = 35/36. For X = 2, DR is 1 - 4/36 = 32/36, and so on until we reach X = 6, where DR is 1 - 36/36 = 0/36.
The cumulative distribution function (CDF) for the given probability mass function (PMF) is calculated by summing up the probabilities of X being less than or equal to a certain value. The complementary probability, denoted as DR, represents the probability of X being less than a certain value. By subtracting the CDF from 1, we can find the DR values for each X.
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suppose a soup can has a height of 6 inches and a radius of 2 inches. in terms of π, how much material is needed to make each can?
The amount of material needed to make a can can be calculated by finding the surface area of the can. In this case, we have a soup can with a height of 6 inches and a radius of 2 inches.
To calculate the surface area, we need to find the area of the circular top and bottom, as well as the area of the curved side. The area of each circular top or bottom is given by the formula A = πr^2, where r is the radius. So, the total area of the circular tops and bottoms is 2π(2^2) = 8π.
The area of the curved side can be found using the formula for the lateral surface area of a cylinder, which is given by A = 2πrh, where r is the radius and h is the height. In this case, the curved side of the can forms a rectangle when it is unrolled, so the height of the rectangle is the same as the height of the can, which is 6 inches. Therefore, the area of the curved side is 2π(2)(6) = 24π.
To find the total amount of material needed, we add the areas of the circular tops and bottoms to the area of the curved side. So, the total surface area of the can is 8π + 24π = 32π square inches.
Therefore, in terms of π, the amount of material needed to make each can is 32π square inches.
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A five-digit identification card is made. Find the probability that the card will contain the digits 0,1 , 2,3 , and 4 in any order.
The probability of a five-digit identification card containing the digits 0,1,2,3 and 4 in any order is 1 or 100%.
Given a five-digit identification card is made. We have to find the probability that the card will contain the digits 0,1,2,3, and 4 in any order.
So, we need to find the total number of possible ways of arranging the digits 0,1,2,3 and 4 in a 5-digit number. We can do this by calculating the number of permutations of these digits using the formula for permutation is:
P(n, r) = n! / (n - r)!
Here, n = 5 (the total number of digits) and r = 5 (the number of digits we want to arrange).
So, the total number of possible 5-digit numbers that can be made using the digits 0,1,2,3 and 4 is:P(5, 5) = 5! / (5 - 5)! = 5! / 0! = 5! = 120
Now, we need to find the number of 5-digit numbers that contain the digits 0,1,2,3 and 4 in any order. We can do this by counting the number of permutations of these digits using the formula for permutation is:P(n, r) = n! / (n - r)!Here, n = 5 (the total number of digits) and r = 5 (the number of digits we want to arrange).So, the number of 5-digit numbers that contain the digits 0,1,2,3 and 4 in any order is:P(5, 5) = 5! / (5 - 5)! = 5! / 0! = 5! = 120
Therefore, the probability of a five-digit identification card containing the digits 0,1,2,3 and 4 in any order is:Number of 5-digit numbers that contain the digits 0,1,2,3 and 4 in any order / Total number of possible 5-digit numbers= 120 / 120 = 1 or 100%
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Robin had been separated from her husband Rob for only three weeks when she was killed in a car accident. She died intestate. Rob had moved out but they had not yet started to work on the separation agreement. She was 49 and her two children were 17 and 20. Who inherits her $40,000 estate? Both children No one - since she didn't have a will, the government will take it. Rob The 20-year old child Question 50 (1 point) Which of the following statements is true for all provinces and territories?
The correct answer is: No one - since she didn't have a will, the government will take it.
When a person dies without a will, it is known as dying intestate. In such cases, the distribution of the deceased person's estate is determined by the laws of intestacy in the jurisdiction where the person resided.
In most jurisdictions, the laws of intestacy prioritize the distribution of the estate to the closest relatives, such as a spouse and children. However, since Robin and Rob were separated and had not yet finalized their separation agreement, it is unlikely that Rob would be considered the spouse entitled to inherit her estate.
As for the children, the laws of intestacy typically distribute the estate among the children equally. However, the fact that Robin's children are both minors (17 and 20 years old) may complicate the distribution. In some jurisdictions, a legal guardian or trustee may be appointed to manage the inherited assets on behalf of the minors until they reach the age of majority.
It is important to note that the specific laws of intestacy can vary between provinces and territories in Canada. Therefore, it is always recommended to consult with a legal professional to understand the exact distribution of the estate in a particular jurisdiction.
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If a pair of fair six-sided dice are tossed, what is the probability that the sum is even OR greater than 77 a. 0.667 b. 0.25 c. 0.833 d. 0.583 Week 1 Assignment Broom Dave has several golf balls in his golf bag. Seven of them are brand A, 9 are brand 8, and 2 are brand C. He reaches into the bag and randomly selects one golf ball, then he selects a second one without replacing the first one. What is the probability that the first one is a brand A golf ball and the second one is a brand C golf ball? a. 28,288 b. 0.071 € 0.0432 d. 0.0458 Week 1 Assignment 2 Betale If events A and B are mutually exclusive, P(A or B) 0.5, and P(B) 0.3; then what is Page 25 Back to top + MacBook Pro
Hence, option (a) is correct.Option (a) is correct: 0.667. When we roll a pair of fair six-sided dice, we have a total of 36 possible outcomes. And the probability of getting a certain number on dice can be calculated by dividing the number of ways that number can be rolled by the total number of possible outcomes.
For instance, we can get a total of 11 in two different ways; by rolling a 5 on the first die and a 6 on the second die or by rolling a 6 on the first die and a 5 on the second die. Hence, the probability of rolling an 11 is 2/36 = 1/18.Solution:The sample space when rolling a pair of fair dice is 36. The following are all the possible ways the dice can be rolled and the corresponding sums:(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)The probability of rolling an even number with one die is 3/6 (or 1/2), and the probability of rolling an odd number with one die is 3/6 (or 1/2). Thus, the probability of rolling an even number with two dice is (1/2) * (1/2) = 1/4, and the probability of rolling an odd number with two dice is (1/2) * (1/2) = 1/4. The probability of rolling a sum greater than 7 is 15/36. We can use this to calculate the probability of rolling a sum greater than 7 and even as follows: The probability of rolling a sum greater than 7 and even = the probability of rolling a sum greater than 7 + the probability of rolling an even number - the probability of rolling a sum greater than 13 = 15/36 + 1/4 - 0 = 19/36. So, the probability of rolling a sum that is even or greater than 7 is the sum of the probability of rolling an even number and the probability of rolling a sum greater than 7 and even: 1/4 + 19/36 = 0.69 (rounded to two decimal places).Hence, option (a) is correct.Option (a) is correct: 0.667.
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For the following set of scores, calculate the mean, median, and
mode: 4.9; 3.9; 1.7; 4.8; 1.7; 5.3; 6.8; 9.9; 2.9; 1.7; 8.4. (Round
answer to the nearest two decimal places) Mean :
Median;
Mode:
The mean ≈ 4.55, the median is 4.8, and the mode is 1.7 for the given set of scores.
To find the mean, median, and mode of the given set of scores:
Scores: 4.9; 3.9; 1.7; 4.8; 1.7; 5.3; 6.8; 9.9; 2.9; 1.7; 8.4
Mean: To calculate the mean, sum up all the scores and divide by the total number of scores:
Mean = (4.9 + 3.9 + 1.7 + 4.8 + 1.7 + 5.3 + 6.8 + 9.9 + 2.9 + 1.7 + 8.4) / 11
Mean = 50.0 / 11
Mean ≈ 4.55 (rounded to two decimal places)
Median: To find the median, we first need to arrange the scores in ascending order:
1.7, 1.7, 1.7, 2.9, 3.9, 4.8, 4.9, 5.3, 6.8, 8.4, 9.9
Since we have an odd number of scores (11), the median is the middle value, which is the sixth score:
Median = 4.8
Mode: The mode is the most frequently occurring score in the data set. In this case, the score 1.7 appears three times, which is more than any other score:
Mode = 1.7
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Determine all the singular points of the given differential equation. (t? - t - 30)x" + (t + 5)x' - (t - 6)x = 0 The singular points are all t < -5 and t = 6. The singular points are all t > 6 and t = -5. The singular points are t = 6,-5. The singular points are all t > -5. The singular points are all t < 6. There are no singular points. Determine all the singular points of the given differential equation. In(x – 6)/' + sin(6x)y - ey=0 The singular points are all I < 6 and x = 7 The singular points are all x > 6 The singular points are all x > 7 and x = 6 There are no singular points The singular points are all x < 6 The singular points are x = 6 and x = 7
The singular points of a differential equation are the points where the coefficients of the highest and/or second-highest order derivative are zero.
These singular points usually play a vital role in the analysis of the behavior of solutions around them.
Now, let's solve the given differential equations one by one:
1. The given differential equation is `(t² - t - 30)x'' + (t + 5)x' - (t - 6)x = 0`.
We can write the equation in the form of a polynomial as follows: p(t)x'' + q(t)x' + r(t)x = 0,
`where `p(t) = t² - t - 30`, `q(t) = t + 5`, and `r(t) = -(t - 6)`.
The singular points are the values of `t` that make `p(t) = 0`.We can factorize `p(t)` as follows: `p(t) = (t - 6)(t + 5)`.
Therefore, the singular points are `t = 6` and `t = -5`.
So, the answer is "The singular points are t = 6,-5.
2. The given differential equation is `ln(x – 6) y' + sin(6x)y - ey = 0`.
We can write the equation in the form of a polynomial as follows: `p(x)y' + q(x)y = r(x)`where `p(x) = ln(x - 6)`, `q(x) = sin(6x)`, and `r(x) = e^(y)`.
The singular points are the values of `x` that make `p(x) = 0`.For `ln(x - 6) = 0`, we get `x = 7`.
So, the singular point is `x = 7`.
Therefore, the answer is "The singular points are x = 7."
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determine whether the series is absolutely convergent, conditionally convergent, or divergent. [infinity] n2 9 9n2 5 n n = 1 absolutely convergent conditionally convergent divergent
The series given below is absolutely convergent:`∑_(n=1)^∞▒1/n^2`
Consider the given data,
The given series is a p-series and the general term of this series is given by `an = 1/n^2`.
Now,Let's test for the convergence of the series using p-test for convergence:`∑_(n=1)^∞▒1/n^p`
The series is absolutely convergent if `p>1`.Therefore, for `p=2`, the given series is convergent.
Since the series is absolutely convergent, it is also convergent. So, the correct option is "Absolutely convergent".
In other words, if the series ∑(|a_n|) converges, where a_n is the nth term of the original series, then the original series is absolutely convergent.
The required answer for the given question is,
Therefore, the series given below is absolutely convergent:`∑_(n=1)^∞▒1/n^2`
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Let G be a finite group and p a prime. A theorem of Cauchy says that if p divides the order of G, then G contains an element of order p. Prove this in two parts. (a) Prove it when G is abelian. (b) Use the class equation to prove it when G is nonabelian.
Let G be a finite group and p a prime. A theorem of Cauchy says that if p divides the order of G, then G contains an element of order p. Prove this in two parts. (a) Prove it when G is abelian. (b) Use the class equation to prove it when G is nonabelian.Proof of Cauchy's Theorem Let G be a finite group and p be a prime number such that p divides the order of G. Let's assume that G is abelian first.
So, we want to show that G contains an element of order p. We will proceed by induction on the order of G. If the order of G is 1, then G contains only the identity element. It is of order p, which means that the statement is true. If the order of G is greater than 1, then we can pick an element g in G which is not the identity element. We will consider two cases: Case 1: The order of g is divisible by p. In this case, we are done since g is an element of order p. Case 2: The order of g is not divisible by p.
In this case, we consider the group H generated by g. Since H is a subgroup of G, the order of H divides the order of G. Also, the order of H is greater than 1 since it contains g. Therefore, p divides the order of H. By induction, there exists an element h in H such that the order of h is p. Since h is in H, it can be written as a power of g. Hence, g^(m*p) = h^m = e, where e is the identity element of G. This means that the order of g is at most p. But we know that the order of g is not divisible by p. Therefore, the order of g is p itself. So, G contains an element of order p if G is abelian.
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Answer the following: (10 points) a. Find the area to the right of z= -1 for the standard normal distribution. b. First year college graduates are known to have normally distributed annual salaries wi
The area to the right of z = -1 for the standard normal distribution is approximately 0.8413.
a. To find the area to the right of z = -1 for the standard normal distribution, we need to calculate the cumulative probability using the standard normal distribution table or a statistical calculator.
From the standard normal distribution table, the area to the left of z = -1 is 0.1587. Since we want the area to the right of z = -1, we subtract the left area from 1:
Area to the right of z = -1 = 1 - 0.1587 = 0.8413
Therefore, the area to the right of z = -1 for the standard normal distribution is approximately 0.8413.
b. To answer this question, we would need additional information about the mean and standard deviation of the annual salaries for first-year college graduates. Without this information, we cannot calculate specific probabilities or make any statistical inferences.
If we are provided with the mean (μ) and standard deviation (σ) of the annual salaries for first-year college graduates, we could use the properties of the normal distribution to calculate probabilities or make statistical conclusions. Please provide the necessary information, and I would be happy to assist you further.
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Find the exact value of the expressions cos(a + b), sin(a + b) and tan(a + b) under the following conditions: 15 sin(a)= 77' a lies in quadrant I, and sin(B) 24 25' Blies in quadrant II.
We are given that [tex]15 sin(a) = 77[/tex] and a lies in quadrant I. Therefore, we need to find the value of sin(a) as follows: [tex]sin(a) = 77/15[/tex]Now, we are given that sin(B) = 24/25 and B lies in quadrant II.
Therefore, we can find cos(B) and tan(B) as follows: [tex]cos(B) = -√(1 - sin²(B)) = -√(1 - (24/25)²) = -7/25tan(B) = sin(B)/cos(B) = (24/25) / (-7/25) = -24/7[/tex]Using the trigonometric sum identities, we can write: [tex]cos(a + B) = cos(a)cos(B) - sin(a)sin(B)sin(a + B) = sin(a)cos(B) + cos(a)sin(B)tan(a + B) = (tan(a) + tan(B))/(1 - tan(a)tan(B))[/tex]We already know that [tex]sin(a) = 77/15[/tex] and [tex]sin(B) = 24/25[/tex].
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In the university course Data 363, three undergraduates grades
are 79, 68, and 86. According to this data, the following answers
would be:
i) Sample mean
ii) Sample variance
iii) Sample standard devia
i) Sample mean: 77.67
ii) Sample variance: 63.26
iii) Sample standard deviation: 7.95
What are the sample mean, variance and standard deviation?Given the grades: 79, 68, and 86.
Sample mean:
Sample Mean = (Sum of all grades) / (Number of grades)
Sample Mean = (79 + 68 + 86) / 3
Sample Mean = 233 / 3
Sample Mean = 77.67
Sample variance:
Sample Variance = (Sum of (Grade - Sample Mean)^2) / (Number of grades - 1)
Sample Variance = [tex]((79 - 77.67)^2 + (68 - 77.67)^2 + (86 - 77.67)^2) / (3 - 1)[/tex]
Sample Variance = 164.6667 / 2
Sample Variance = 82.33335
Sample Variance = 82.33
Sample standard deviation:
Sample Standard Deviation = [tex]\sqrt{Sample Variance}[/tex]
Sample Standard Deviation = [tex]\sqrt{63.26}[/tex]
Sample Standard Deviation = 7.95361553006
Sample Standard Deviation = 7.95.
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i. The sample mean is 77.67
ii. The sample variance is 82.35
iii. The sample standard deviation is 9.1
What is the sample mean?To find the sample mean, sample variance, and sample standard deviation for the given data, follow these steps:
i) Sample mean:
To find the sample mean, add up all the values and divide the sum by the total number of values (in this case, 3).
Sample mean = (79 + 68 + 86) / 3 = 233 / 3 = 77.67
ii) Sample variance:
To find the sample variance, calculate the squared difference between each value and the sample mean, sum up those squared differences, and divide by the total number of values minus 1.
Step 1: Calculate the squared difference for each value:
(79 - 77.67)² = 1.77
(68 - 77.67)² = 93.51
(86 - 77.67)² = 69.4
Step 2: Sum up the squared differences:
1.77 + 93.51 + 69.4 = 164.7
Step 3: Divide by the total number of values minus 1:
164.7 / (3 - 1) = 82.35
Sample variance = 82.35
iii) Sample standard deviation:
To find the sample standard deviation, take the square root of the sample variance.
Sample standard deviation = √82.35 = 9.1
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In a study of marble color preference, Lucinda Georgette Who surveyed a simple random sample of 400 Whos from Whoville Heights, and found that 250 of them support a constitutional amendment making red the official marble color on alternate Tuesdays. A 95% confidence interval for the percentage of all Whoville Heights Whos who support this amendment is given by... O... (60.1%, 64.9%) (59.4%, 65.6%) *** O (55.5%, 69.5%) *** O... (57.7%, 67.3%) () The Tand Corporation surveys a simple random sample of 87 households from a large metropolitan area (with millions of households). The sample mean monthly disposable household income is $4560, with a standard deviation of $3100. A 90%-confidence interval for the mean disposable household income in the entire metropolitan area is given by... O... ($4236, $4884) O...A confidence interval for the population mean can't be found from this data, because the income distribution is clearly not normal - it is obviously skewed right. O... ($3898, $5222) O... ($4007, $5113)
The correct answer is: (60.1%, 64.9%) and ($4236, $4884). The standard error of the mean can be calculated as the standard deviation of the sample divided by the square root of the sample size, or $3100/sqrt(87) = $332.
For the first question about marble color preference, we have a sample size of 400 and 250 people in the sample support the amendment making red the official marble color. The sample proportion is 250/400 = 0.625. Using this information, we can calculate the standard error of the sample proportion as sqrt(0.625*(1-0.625)/400) = 0.0309.
To find a 95% confidence interval for the true proportion of all Whoville Heights Whos who support the amendment, we can use the formula:
sample proportion +/- z*standard error
where z is the critical value from the standard normal distribution corresponding to a 95% confidence level, which is approximately 1.96. Plugging in the values, we get:
0.625 +/- 1.96*0.0309
which gives us the interval (0.594, 0.656), or (59.4%, 65.6%).
For the second question about household income, we have a sample size of 87 and a sample mean of $4560 with a standard deviation of $3100. Since the sample size is relatively large, we can use a t-distribution with degrees of freedom equal to n-1 = 86 to construct a confidence interval for the population mean. A 90% confidence interval can be calculated using the formula:
sample mean +/- t*standard error
where t is the critical value from the t-distribution with 86 degrees of freedom corresponding to a 90% confidence level, which is approximately 1.67.
The standard error of the mean can be calculated as the standard deviation of the sample divided by the square root of the sample size, or $3100/sqrt(87) = $332.
Plugging in the values, we get:
$4560 +/- 1.67*$332
which gives us the interval ($4236, $4884).
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5) In a poll, 925 females and 920 males were asked "If you could get a free car which maker would you chose: Toyota, Honda, or Chevy?" Their responses are presented in the table below. Honda Chevy Toy
The probability of selecting a male that has Honda is 0.1491
Calculating the probability of selecting a male that has HondaFrom the question, we have the following parameters that can be used in our computation:
The table of values
Where we have
Male and Honda = 290
Total = 925 + 920
Total = 1845
Using the above as a guide, we have the following:
P(Male and Honda) = Male and Honda/Total
So, we have
P(Male and Honda) = 290/1945
Evaluate
P(Male and Honda) = 0.1491
Hence, the probability is 0.1491
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Question
In a poll, 925 females and 920 males were asked "If you could get a free car which maker would you chose: Toyota, Honda, or Chevy?" Their responses are presented in the table below.
Toyota Honda Chevy
Female 320 349 256
Male 325 290 305
Calculate the probability of selecting a male that has Honda
Use a known Maclaurin series to obtain a Maclaurin series for the given function. f(x) = sin (pi x/2) Find the associated radius of convergence R.
The Maclaurin series for [tex]\(f(x) = \sin\left(\frac{\pi x}{2}\right)\)[/tex] is given by:
[tex]\[\sin\left(\frac{\pi x}{2}\right) = \frac{\pi}{2} \left(x - \frac{\left(\pi^2 x^3\right)}{2^3 \cdot 3!} + \frac{\left(\pi^4 x^5\right)}{2^5 \cdot 5!} - \frac{\left(\pi^6 x^7\right)}{2^7 \cdot 7!} + \ldots\right).\][/tex]
The radius of convergence, [tex]\(R\)[/tex] , for this series is infinite since the series converges for all real values of [tex]\(x\).[/tex]
Therefore, the Maclaurin series for [tex]\(f(x) = \sin\left(\frac{\pi x}{2}\right)\)[/tex] is:
[tex]\[\sin\left(\frac{\pi x}{2}\right) = \frac{\pi}{2} \left(x - \frac{\left(\pi^2 x^3\right)}{2^3 \cdot 3!} + \frac{\left(\pi^4 x^5\right)}{2^5 \cdot 5!} - \frac{\left(\pi^6 x^7\right)}{2^7 \cdot 7!} + \ldots\right)\][/tex]
with an associated radius of convergence [tex]\(R = \infty\).[/tex]
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In Problems 55-62, write each function in terms of unit step functions. Find the Laplace transform of the given function 0 =t< 1 57. f(t) = {8 12 1 Jo, 0 =t < 30/2 58. f(t) = ( sint, t = 30/2
The Laplace transform of the given function is,
L{f(t)} = (8/s) - 4e^{-3s/2}/s - 6e^{-2s}/s
Given function is f(t) = {8 12 1 Jo, 0 ≤ t < 3/2, 3/2 ≤ t < 2, 2 ≤ t < ∞ respectively.
We have to find Laplace transform of the given function.
For first interval 0 ≤ t < 3/2,
f(t) = 8u(t) - 8u(t-3/2)
For second interval 3/2 ≤ t < 2,
f(t) = 12u(t-3/2) - 12u(t-2)
For third interval 2 ≤ t < ∞,
f(t) = Jo(u(t-2))
Hence, we can write the Laplace transform of the given function as,
L{f(t)} = L{8u(t) - 8u(t-3/2)} + L{12u(t-3/2) - 12u(t-2)} + L{Jo(u(t-2))}
Where, L is Laplace transform.
Let's calculate each Laplace transform stepwise,
1. L{8u(t) - 8u(t-3/2)}L{8u(t)} = 8/L{u(t)}L{u(t)}
= 1/sL{u(t-3/2)}
= e^{-3s/2}/s
Therefore,
L{8u(t) - 8u(t-3/2)} = 8[1/s - e^{-3s/2}/s]
2. L{12u(t-3/2) - 12u(t-2)}L{12u(t-3/2)}
= 12e^{-3s/2}/sL{12u(t-2)}
= 12e^{-2s}/s
Therefore,
L{12u(t-3/2) - 12u(t-2)} = 12[e^{-3s/2}/s - e^{-2s}/s]
3. L{Jo(u(t-2))}L{Jo(u(t-2))} = ∫_{0}^{∞}δ(t-2)e^{-st}dtL{Jo(u(t-2))}
= e^{-2s}
Hence, the Laplace transform of the given function is,
L{f(t)} = 8[1/s - e^{-3s/2}/s] + 12[e^{-3s/2}/s - e^{-2s}/s] + e^{-2s}
= (8/s) - 4e^{-3s/2}/s - 6e^{-2s}/s
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determine whether the planes are parallel, perpendicular, or neither. 9x 36y − 27z = 1, −12x 24y 28z = 0
Therefore, the given planes are neither parallel nor perpendicular.
Given planes are 9x+36y−27z=1 and −12x+24y+28z=0.
Let's compare the coefficients of x,y, and z in both planes to check whether the planes are parallel, perpendicular or neither.
We know that, two planes are parallel if and only if the normal vectors are parallel.
Two planes are perpendicular if the dot product of their normal vectors is zero.
Let's write the given planes in the vector form by equating the coefficients of x, y, and z.9x+36y−27z=1 => (9, 36, -27) . (x, y, z) = 1−12x+24y+28z=0 => (-12, 24, 28) . (x, y, z) = 0
Now let's find the dot product of the normal vectors in both planes to determine whether the planes are parallel or perpendicular(9, 36, -27) . (-12, 24, 28) = -432 - 648 + (-756) = -1836
The dot product is not zero, so the planes are not perpendicular.
Since the normal vectors are not parallel (one is not a scalar multiple of the other), the planes are not parallel.
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please answer all questions
Question 3) Let say for question 2 if we measure the anxiety score before and after intervention for male and female students. (part a 8 points and part b 7 points total 15 points) a. What statistical
The significance level, also known as alpha, is the probability of rejecting the null hypothesis when it is actually true. The common significance level is 0.05, which indicates that there is a 5% probability of rejecting the null hypothesis when it is true
.What is the p-value?
The p-value is the probability of observing a difference as large as or larger than the one observed, assuming that the null hypothesis is true. It is compared to the significance level to determine if the null hypothesis should be rejected or not.
What is the interpretation of the p-value?
A p-value of less than the significance level (0.05) indicates that there is a significant difference between the means of the two groups. A p-value of greater than the significance level suggests that there is no significant difference between the means of the two groups.
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The correct answer is option 3: Kruskal-Wallis test.
The correct answer is option 3: Two-sample t-test with the difference of after and before anxiety score.
a. The appropriate statistical test to compare the differences in anxiety scores before and after intervention for male and female students would be:
Kruskal-Wallis test
The Kruskal-Wallis test is a non-parametric test used to compare the medians of three or more independent groups.
In this case, we have two independent groups (males and females), and we want to determine if there are any differences in the anxiety score changes between these groups after the intervention.
b. If you have a larger sample size, you can use the following parametric test to analyze the differences in anxiety scores before and after intervention:
Two-sample t-test with the difference of after and before anxiety scores.
The two-sample t-test is appropriate when comparing the means of two independent groups. In this case, you can calculate the difference between the after and before anxiety scores for each individual, and then perform a two-sample t-test to determine if there is a significant difference in the mean difference between males and females.
However, it's important to note that the t-test assumes normality of the data and equality of variances between the groups. If these assumptions are violated, alternative non-parametric tests, such as permutation tests or bootstrapping, may be more appropriate.
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5. Suppose the following is true for all students who completed STA 2023 during the past Academic year: C: F: Student was a Freshman Student earned a "C" grade P(F) = 0.25 P(FIC) = 0.32 0.19 P(C) = a.
The probability that the student earned a "C" grade who was a Fresh man is 0.32/a. The probability that the student was a Fresh man who earned a "C" grade in STA 2023 is 1.28.
The probability that the student earned a "C" grade who was a Fresh man and the probability that the student was a Fresh man who earned a "C" grade in STA 2023 are to be determined based on the given information.
Let us consider the events: C : Student was a Fresh man F : Student earned a "C" grade P(F) = 0.25 (Probability that a student earned a "C" grade)P(FIC) = 0.32 (Probability that a student who was a Freshman earned a "C" grade)P(C) = a (Probability that a student earned a "C" grade)
We need to determine the following probabilities .P(F|C)P(C|F)We know the following from the conditional probability formula. P(FIC) = P(F and C) = P(F|C) P(C)Substitute the given probabilities. P(F|C)P(C) = P(F and C) = P(FIC) = 0.32P(C) = aP(F|C) = 0.32/a ------ (1)P(FIC) = P(F and C) = P(C|F) P(F)Substitute the given probabilities. P(C|F)P(F) = P(F and C) = P(FIC) = 0.32P(C|F) = 0.32/0.25 = 1.28Using Bayes' theorem, P(F|C) = [P(C|F)P(F)]/P(C)
Substitute the values of P(F|C), P(C|F), P(F), and P(C) in the above equation. P(F|C) = [1.28 × 0.25]/a = 0.32/aThe probability that the student earned a "C" grade who was a Fresh man is 0.32/a.
the probability that the student was a Fresh man who earned a "C" grade in STA 2023 is 1.28.
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Find the solution to the linear system of differential equations {x′y′==10x−6y9x−5y satisfying the initial conditions x(0)=6 and y(0)=8.
Solution to the given linear system of differential equations {x′y′==10x−6y9x−5y} is given by x = 6e^{3t} and y = 8e^{2t}.Let's solve the given system of differential equations {x′y′==10x−6y9x−5y} :Given system of differential equations is {x′y′==10x−6y9x−5y}
Differentiating both the sides of the equation w.r.t. "t", we get: x′y′ + xy′′ = 10x′ − 6y′ + 9xy′ − 5y′′ …(1)Putting the value of x′ from the first equation of the system into (1), we get: y′′ − 9y′ + 5y = 0 …(2)This is a linear homogeneous differential equation, whose auxiliary equation is given by: r^2 - 9r + 5 = 0(r - 5)(r - 1) = 0 => r = 5, 1Hence, the general solution to the differential equation (2) is given by: y = c1e^{5t} + c2e^{-t}Let's solve for the constants c1 and c2:Given initial conditions are: x(0) = 6 and y(0) = 8Putting t = 0 in the first equation of the system, we get: x′(0)y′(0) = 10x(0) - 6y(0)=> 6y′(0) = 40 => y′(0) = 20/3Putting t = 0 and y = 8 in the general solution of the differential equation (2), we get:8 = c1 + c2 …(3)Differentiating the general solution and then putting t = 0 and y′ = 20/3, we get:20/3 = 5c1 - c2 …
Solving equations (3) and (4), we get: c1 = 16/3 and c2 = 8/3Hence, the solution to the differential equation (2) is given by: y = (16/3)e^{5t} + (8/3)e^{-t}Putting this value of y in the first equation of the system, we get: x = (6/5)e^{3t}Putting both the values of x and y in the given system of differential equations {x′y′==10x−6y9x−5y}, we can verify that they satisfy the given system of differential equations.Hence, the required solution to the given linear system of differential equations {x′y′==10x−6y9x−5y} is given by x = 6e^{3t} and y = 8e^{2t}.
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