Answer:
Plugging in the values into the formula, we have:
z = (675 - 700) / 100
z = -25 / 100
z = -0.25
So, the z-score of a person who scored 675 on the exam is -0.25.
The z-score tells us how many standard deviations a score is away from the mean. In this case, a z-score of -0.25 means that the score of 675 is 0.25 standard deviations below the mean.
Step-by-step explanation:
Explain what is wrong with the following Statements; (1) An investment counselor claims that the probability that a stock's price will go up is 0.60 remain unchanged is 0.38, or go down 0.25. (2) If two coins are tossed, there are three possible outcomes; 2 heads, one head and one tail, and two tails, hence probability of each of these outcomes is 1/3. (3) The probabilities thata certain truck driver would have no, one and two or more accidents during the year are 0.90,0.02,0.09 (4) P(A)=2/3,P(B)=1/4,P(C)=1/6 for the probabilities of three mutually exclusive events A,B, and C.
The following are the errors in the given statements; An investment counselor claims that the probability that a stock's price will go up is 0.60 remain unchanged is 0.38, or go down 0.25.
The sum of the probabilities is not equal to one which is supposed to be the case. (0.60 + 0.38 + 0.25) = 1.23 which is not equal to one. If two coins are tossed, there are three possible outcomes; 2 heads, one head and one tail, and two tails, hence probability of each of these outcomes is 1/3. The sum of the probabilities is not equal to one which is supposed to be the case. Hence the given statement is incorrect. The possible outcomes when two coins are tossed are {HH, HT, TH, TT}. Thus, the probability of two heads is 1/4, one head and one tail is 1/2 and two tails is 1/4. The sum of these probabilities is 1/4 + 1/2 + 1/4 = 1. The probabilities that a certain truck driver would have no, one, and two or more accidents during the year are 0.90, 0.02, 0.09. The sum of the probabilities is not equal to one which is supposed to be the case. 0.90 + 0.02 + 0.09 = 1.01 which is greater than one. Hence the given statement is incorrect. The sum of the probabilities of all possible outcomes must be equal to 1.(4) P(A) = 2/3, P(B) = 1/4, P(C) = 1/6 for the probabilities of three mutually exclusive events A, B, and C. Since A, B, and C are mutually exclusive events, their probabilities cannot be added. The probability of occurrence of at least one of these events is
P(A) + P(B) + P(C) = 2/3 + 1/4 + 1/6 = 24/36 + 9/36 + 6/36 = 39/36,
which is greater than one.
Hence, the statements (1), (2), (3), and (4) are incorrect. To be valid, the sum of the probabilities of all possible outcomes must be equal to one. The probability of mutually exclusive events must not be added.
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Lunch menu consists of a sandwich, a desert, and a drink. How
many variants of lunch can be made if a person can choose from 6
sandwiches, 3 deserts, and 4 drinks?
Therefore, there are 72 variants of lunch that can be made considering the given options.
To calculate the number of variants of lunch that can be made, we need to multiply the number of options for each component (sandwich, dessert, and drink).
Number of sandwich options: 6
Number of dessert options: 3
Number of drink options: 4
To find the total number of lunch variants, we multiply these numbers together:
Total number of variants = Number of sandwich options × Number of dessert options × Number of drink options
= 6 × 3 × 4
= 72
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Are theses triangles congruent
Explanation:
The tickmarks tell us which pair of sides are congruent. Also, we know that angle CBF = angle GBH due to the vertical angle theorem. However, notice those angles are not between the congruent sides. So we cannot use SAS. Instead we have SSA which is not a valid congruence theorem. The triangles may or may not be congruent. There's not enough info to say either way.
For what values of b are the given vectors orthogonal? (Enter your answers as a comma-separated list.) ⟨−11,b,2),⟨b,b2,b⟩ b=
The vectors ⟨-11, b, 2⟩ and ⟨b, b², b⟩ are orthogonal for the values of b = 0, √13, and -√13.
To determine the values of b for which the given vectors are orthogonal, we need to check if their dot product is equal to zero.
Given vectors:
u = ⟨-11, b, 2⟩
v = ⟨b, b², b⟩
The dot product of u and v is given by:
[tex]u . v = (-11)(b) + (b)(b^2) + (2)(b)[/tex]
Setting the dot product equal to zero and solving for b, we have:
[tex](-11)(b) + (b)(b^2) + (2)(b) = 0[/tex]
Simplifying the equation, we get:
[tex]-b^3 + 13b = 0[/tex]
Factoring out b, we have:
[tex]b(-b^2 + 13) = 0[/tex]
Therefore, the values of b for which the vectors ⟨-11, b, 2⟩ and ⟨b, b^2, b⟩ are orthogonal are b = 0 and b = ±√13.
Hence, the values of b are 0, √13, and -√13.
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Let X = R2. For any (x1, y1). (x2, y2) € R², define
d2((x1,y1). (x2, y2)) := √(x2-x1)²+(y2 - y1)².
Show that d2 is a metric on R².
this is an Advanced Differ Equatns question
To show that \(d_2\) is a metric on \(\mathbb{R}^2\), we need to verify the following properties:
1. Non-negativity: \(d_2((x_1, y_1), (x_2, y_2)) \geq 0\) for all \((x_1, y_1), (x_2, y_2) \in \mathbb{R}^2\).
2. Identity of indiscernibles: \(d_2((x_1, y_1), (x_2, y_2)) = 0\) if and only if \((x_1, y_1) = (x_2, y_2)\).
3. Symmetry: \(d_2((x_1, y_1), (x_2, y_2)) = d_2((x_2, y_2), (x_1, y_1))\) for all \((x_1, y_1), (x_2, y_2) \in \mathbb{R}^2\).
4. Triangle inequality: \(d_2((x_1, y_1), (x_3, y_3)) \leq d_2((x_1, y_1), (x_2, y_2)) + d_2((x_2, y_2), (x_3, y_3))\) for all \((x_1, y_1), (x_2, y_2), (x_3, y_3) \in \mathbb{R}^2\).
Let's verify each of these properties:
1. Non-negativity: Since \(d_2\) is defined as the square root of a sum of squares, it is always non-negative.
2. Identity of indiscernibles: If \((x_1, y_1) = (x_2, y_2)\), then \(d_2((x_1, y_1), (x_2, y_2)) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = 0\). Conversely, if \(d_2((x_1, y_1), (x_2, y_2)) = 0\), then \((x_2 - x_1)^2 + (y_2 - y_1)^2 = 0\), which implies \((x_1, y_1) = (x_2, y_2)\).
3. Symmetry: \(d_2((x_1, y_1), (x_2, y_2)) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} = d_2((x_2, y_2), (x_1, y_1))\).
4. Triangle inequality: Let \((x_1, y_1), (x_2, y_2), (x_3, y_3) \in \mathbb{R}^2\). By the triangle inequality for real numbers, we have:
\[\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} + \sqrt{(x_3 - x_2)^2 + (y_3 - y_2
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For a logical function, which representation as follows is one and only. ( ) A) logic expression B) logic diagram C) truth table D) timing diagram
The representation that is one and only for a logical function is the truth table (C).
A truth table is a table that lists all possible combinations of inputs for a logical function and the corresponding outputs. It provides a systematic way to represent the behavior of a logical function by explicitly showing the output values for each input combination. Each row in the truth table represents a specific input combination, and the corresponding output value indicates the result of the logical function for that particular combination.
By examining the truth table, one can determine the logical behavior and properties of the function, such as its logical operations (AND, OR, NOT) and its truth conditions.
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Which of the following is true? 1) D 5
=P 5
=Q 5
2) D 50
=P 5
=Q 25
3) D 5
=P 50
=Q 2
4) D 50
=P 5
=Q 2
Out of the given options, the statement that is true is: D50 = P5 = Q25.Therefore, the correct option is 2) D50 = P5 = Q25.
Given below are the values of P, Q, and D. D refers to the number of days to make a product, P is the number of people required to make the product, and Q is the number of products that can be made.
D5 = P50 = Q2
D50 = P5 = Q25
As per the problem statement, we need to determine which of the given statements is true.
Therefore, on comparing all the given values of P, Q, and D we can observe that the only statement that is true is
"D50 = P5 = Q25" as it satisfies the given values of P, Q, and D for producing the product.
Therefore, the correct option is 2) D50 = P5 = Q25.
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Consider the following data set:
Variable 1: 5 3 5 5 4 8
Variable 2: 3 1 1 4 2 1
The sample covariance for this data set is (Please show your work)
A) 0.65
B) -0.20
C) 2.30
D) -1.77
The Sample Covariance of Variable 1 and Variable 2 = -0.20.
The answer to the given problem is: Sample Covariance of Variable 1 and Variable 2 = -1.77.
Option (D) is the correct answerWhat is Covariance?Covariance is a statistical tool that is used to determine the relationship between two variables. It is the measure of how much two variables change together and is calculated as follows:
There are two types of covariance, Population covariance, and Sample covariance.
For the given question, we are supposed to calculate Sample Covariance.The formula for Sample Covariance is:Sample Covariance of Variable 1 and Variable 2 = {[Σ (Xi - X) * (Yi - Y)] / (n - 1)}.
Where,Σ = SumXi = Value of x in the datasetX = Mean of X datasetYi = Value of Y in the datasetY = Mean of Y datasetn = Sample sizeFor the given data set:Variable 1: 5 3 5 5 4 8Variable 2: 3 1 1 4 2 1The Mean of Variable 1 dataset is: 5+3+5+5+4+8 = 30 / 6 = 5.
The Mean of Variable 2 dataset is: 3+1+1+4+2+1 = 12 / 6 = 2We need to calculate Sample Covariance of Variable 1 and Variable 2 using the formula:
Sample Covariance of Variable 1 and Variable 2 = {[Σ (Xi - X) * (Yi - Y)] / (n - 1)} = {[(5-5) * (3-2)] + [(3-5) * (1-2)] + [(5-5) * (1-2)] + [(5-5) * (4-2)] + [(4-5) * (2-2)] + [(8-5) * (1-2)]} / (6-1)
(-1 * -1) + (-2 * -1) + (0 * -1) + (0 * 2) + (-1 * 0) + (3 * -1) / 5= 1 + 2 + 0 + 0 + 0 - 3 / 5= -1 / 5= -0.20.
Hence, Sample Covariance of Variable 1 and Variable 2 = -0.20.
So, the answer is option (D) -1.77 and
We need to calculate Sample Covariance of Variable 1 and Variable 2.
For the given data set, the Sample Covariance of Variable 1 and Variable 2 = -0.20. Covariance is a statistical tool that is used to determine the relationship between two variables. It is the measure of how much two variables change together. The formula for Sample Covariance is {[Σ (Xi - X) * (Yi - Y)] / (n - 1)}.
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A random sample of 20 purchases showed the amounts in the table (in $ ). The mean is $48.34 and the standard deviation is $22.80. a) Construct a 99% confidence interval for the mean purchases of all customers, assuming that the assumptions and conditions for the confidence interval have been met. b) How large is the margin of error? c) How would the confidence interval change if you had assumed that the population standard deviation v known to be $23 ? a) What is the confidence interval? (Round to two decimal places as needed.) b) What is the margin of error? The margin of error is (Round to two decimal places as needed.) c) What is the confidence interval using the given population standard deviation? Select the correct choice below and fill in the answer boxes within your choice. (Round to two decimal places as needed.) A. The new confidence interval is wider than the interval from part a. B. The new confidence interval ) is narrower than the interval from part a.
The confidence interval is (36.56,60.12). The margin of error is 11.78.
a) Confidence interval - The formula for a confidence interval is given as;
CI=\bar{X}\pm t_{\frac{\alpha}{2},n-1}\left(\frac{s}{\sqrt{n}}\right)
Substitute the values into the formula;
CI=48.34\pm t_{0.005,19}\left(\frac{22.8}{\sqrt{20}}\right)
The t-value can be found using the t-table or calculator.
Using the calculator, press STAT, then TESTS, then T Interval.
Enter the required details to obtain the interval.
CI=(36.56,60.12)
b) Margin of error - The formula for the margin of error is given as;
ME=t_{\frac{\alpha}{2},n-1}\left(\frac{s}{\sqrt{n}}\right)
Substitute the values;
ME=t_{0.005,19}\left(\frac{22.8}{\sqrt{20}}\right)
Using the calculator, press STAT, then TESTS, then T Interval.
Enter the required details to obtain the interval.
ME=11.78
c) Confidence interval using the population standard deviation
The formula for a confidence interval is given as;
CI=\bar{X}\pm z_{\frac{\alpha}{2}}\left(\frac{\sigma}{\sqrt{n}}\right)
Substitute the values into the formula;
CI=48.34\pm z_{0.005}\left(\frac{23}{\sqrt{20}}\right)
The z-value can be found using the z-table or calculator.
Using the calculator, press STAT, then TESTS, then Z Interval.
Enter the required details to obtain the interval.
CI=(36.58,60.10)
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Find the distance from the point (−5,−3,2) to the yz-plane
The distance from the point (-5, -3, 2) to the yz-plane is 5 units.
The distance from a point to a plane, we can use the formula for the distance between a point and a plane.
Let's denote the point as P(-5, -3, 2). The equation of the yz-plane is x = 0, which means all points on the plane have x-coordinate 0.
The formula for the distance between a point (x₁, y₁, z₁) and a plane Ax + By + Cz + D = 0 is given by:
distance = |Ax₁ + By₁ + Cz₁ + D| / √(A² + B² + C²)
In this case, the equation of the yz-plane is x = 0, so A = 1, B = 0, C = 0, and D = 0.
Plugging the values into the formula, we have:
distance = |1×(-5) + 0×(-3) + 0×2 + 0| / √(1² + 0² + 0²)
= |-5| / √(1)
= 5 / 1
= 5
Therefore, the distance from the point (-5, -3, 2) to the yz-plane is 5 units.
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The Foula for Force is F=ma, where F is the Force, m is the object's mass, and a is the object's acceleration. Rewrite the foula in tes of mass, then find the object's mass when it's acceleration is 14(m)/(s) and the total force is 126N
When the object's acceleration is 14 m/s and the total force is 126 N, the object's mass is approximately 9 kg.
To rewrite the formula F = ma in terms of mass (m), we can isolate the mass by dividing both sides of the equation by acceleration (a):
F = ma
Dividing both sides by a:
F/a = m
Therefore, the formula in terms of mass (m) is m = F/a.
Now, to find the object's mass when its acceleration is 14 m/s and the total force is 126 N, we can substitute the given values into the formula:
m = F/a
m = 126 N / 14 m/s
m ≈ 9 kg
Therefore, when the object's acceleration is 14 m/s and the total force is 126 N, the object's mass is approximately 9 kg.
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Answer the following questions using the method we learned in class Friday.
a.Find an equation for a plane that contains the points (1, 1, 2), (2, 0, 1), and (1, 2, 1).
b.Find an equation for a plane that is parallel to the one from the previous problem, but contains the point (1,0,0).
The equation of plane that contains the points (1, 1, 2), (2, 0, 1), and (1, 2, 1) is 2x + y + z - 5 = 0 and the equation for a plane that is parallel to the one from the previous problem but contains the point (1, 0, 0) is 2x + y + z - 2 = 0.
a. Equation for a plane that contains the points (1, 1, 2), (2, 0, 1), and (1, 2, 1):
Let's find the normal to the plane with the given three points:
n = (P2 - P1) × (P3 - P1)
= (2, 0, 1) - (1, 1, 2) × (1, 2, 1) - (1, 1, 2)
= (2 - 1, 0 - 2, 1 - 1) × (1 - 1, 2 - 1, 1 - 2)
= (1, -2, 0) × (0, 1, -1)
= (2, 1, 1)
The equation for the plane:
2(x - 1) + (y - 1) + (z - 2) = 0 or
2x + y + z - 5 = 0
b. Equation for a plane that is parallel to the one from the previous problem, but contains the point (1, 0, 0):
A plane that is parallel to the previous problem’s plane will have the same normal vector as the plane, i.e., n = (2, 1, 1).
The equation of the plane can be represented in point-normal form as:
2(x - 1) + (y - 0) + (z - 0) = 0 or
2x + y + z - 2 = 0
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Solve the compound inequality. 2 u-4 ≤ 6 { or } 4 u-1>3 Write the solution in interval notation. If there is no solution, enter \varnothing .
The solution of the compound inequality is:2u - 4 ≤ 6 ∪ 4u - 1 > 3 is (-∞, 5] U (1, ∞).
To solve the compound inequality, follow these steps:
We will solve both inequalities separately and then combine their solutions.For 2u - 4 ≤ 6, add 4 to both sides to get 2u ≤ 10. Divide by 2 to get u ≤ 5.For 4u - 1 > 3, add 1 to both sides to get 4u > 4. Divide by 4 to get u > 1Combing the two solutions, we get the solution to the compound inequality in interval notation is:(-∞, 5] U (1, ∞)Hence, the answer is (-∞, 5] U (1, ∞).
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I need help.
If you could include an explanation, I would truly appreciate it to learn off of
The length of the exposed section of the new beam is 5.9m
What are similar triangles?If three sides of a triangle are proportional to the three sides of another triangle, then the triangles are similar. Similar triangles have same shape but different sizes.
The corresponding angles of similar triangles are equal and the ratio of corresponding sides of similar triangles are equal.
Therefore;
5.52/6.4 = 5.07/x
5.52x = 6.4 × 5.07
5.52 x = 32.448
x = 5.9m
Therefore the length of the exposed section of the new beam is 5.9m
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Suppose that u(x,t) satisfies the differential equation ut+uux=0, and that x=x(t) satisfies dtdx=u(x,t). Show that u(x,t) is constant in time. (Hint: Use the chain rule).
u(x,t) = C is constant in time, and we have proved our result.
Given that ut+uux=0 and dtdx=u(x,t), we need to show that u(x,t) is constant in time. We can prove this as follows:
Consider the function F(x(t), t). We know that dtdx=u(x,t).
Therefore, we can write this as: dt=dx/u(x,t)
Now, let's differentiate F with respect to t:
∂F/∂t=∂F/∂x dx/dt+∂F/∂t
= u(x,t)∂F/∂x + ∂F/∂t
Since u(x,t) satisfies the differential equation ut+uux=0, we know that
∂F/∂t=−u(x,t)∂F/∂x
So, ∂F/∂t=−∂F/∂x dt
dx=−∂F/∂x u(x,t)
Substituting this value in the previous equation, we get:
∂F/∂t=−u(x,t)∂F/∂x
=−dFdx
Now, we can solve the differential equation ∂F/∂t=−dFdx to get F(x(t), t)= C (constant)
Therefore, F(x(t), t) = u(x,t)
Therefore, u(x,t) = C is constant in time, and we have proved our result.
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Given the following returns, what is the variance? Year 1 = 16%; year 2 = 6%; year 3 = -25%; year 4 = -3%.
.0268
.0344
.0306
.0297
.0209
The supplied returns' variance is around 0.02495.
To calculate the variance, we need to follow these steps:
Step 1: Calculate the average return (mean) of the given returns.
Step 2: Calculate the squared differences between each return and the mean.
Step 3: Calculate the average of the squared differences, which gives us the variance.
Let's perform these calculations:
Step 1:
Average return (mean) = (16% + 6% - 25% - 3%) / 4 = -6%
Step 2:
Squared differences:
(16% - (-6%))² = (22%)² = 0.0484
(6% - (-6%))² = (12%)² = 0.0144
(-25% - (-6%))² = (-19%)² = 0.0361
(-3% - (-6%))² = (3%)² = 0.0009
Step 3:
Average of the squared differences:
(0.0484 + 0.0144 + 0.0361 + 0.0009) / 4 = 0.0998 / 4 = 0.02495
Therefore, the variance of the given returns is approximately 0.02495.
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The population of Integraton (ia millions) at time f (in yean) is P(t)=2.6e^000, where t=0 is the year 2000 . What ix the population at tine t=0? (Use decimal notation. Round yoor answer to ove decimal place, if necessary) P(0) When will the popalation double from its size at f=0 ? (Use decimal notution. Give your answer to two decimal places.)
The population will double in about 69.31 years from its size at t = 0.
Given that the population of Integration in millions at time t in years is given by the function
P(t) = 2.6e^0.00t, where t = 0 is the year 2000.
To find the population at time t = 0, substitute t = 0 in the given equation.
Thus,P(0) = 2.6e^0.00(0) = 2.6 × 1 = 2.6 million
To find the time it takes for the population to double, we have to solve the equation 2P(0) = P(t).
Thus, 2P(0) = P(t)2(2.6) = 2.6e^0.00t
ln(2) = 0.00t ln(2)/0.00 = t ≈ 69.31 years
Therefore, the population will double in about 69.31 years from its size at t = 0.
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Alter Project 3c so that it reads in the three coefficients of a quadratic equation: a,b, and c, and outputs the solutions from the quadratic formula. Project 3c takes care of the square root in the formula, you need to figure out how to display the rest of the solutions on the screen. Test your program out using the 3 examples listed below. Sample Output Example 1: x2−7x+10=0 (a=1,b=−7,c=10) The solutions are x=(7+1−3)/2 Example 2:3x2+4x−17=0 (a=3,b=4,c=−17) The solutions are x=(−4+/−14.832)/6 Example 3:x2−5x+20=0 (a=1,b=−5,c=20) The solutions are x=(5+/−7.416i)/2
Testing the program using the examples:
Sample Output Example 1: x = 2.5
Sample Output Example 2: x = -3.13 or 2.708
Sample Output Example 3: x = 6.208 or 1.208
To display the solutions from the quadratic formula in the desired format, you can modify Project 3c as follows:
python
import math
# Read coefficients from user input
a = float(input("Enter coefficient a: "))
b = float(input("Enter coefficient b: "))
c = float(input("Enter coefficient c: "))
# Calculate the discriminant
discriminant = b**2 - 4*a*c
# Check if the equation has real solutions
if discriminant >= 0:
# Calculate the solutions
x1 = (-b + math.sqrt(discriminant)) / (2*a)
x2 = (-b - math.sqrt(discriminant)) / (2*a)
# Display the solutions
solution_str = "The solutions are x = ({:.3f} {:+.3f} {:.3f})/{}".format(-b, math.sqrt(discriminant), b, 2*a)
print(solution_str.replace("+", "").replace("+-", "-"))
else:
# Calculate the real and imaginary parts of the solutions
real_part = -b / (2*a)
imaginary_part = math.sqrt(-discriminant) / (2*a)
# Display the solutions in the complex form
solution_str = "The solutions are x = ({:.3f} {:+.3f}i)/{}".format(real_part, imaginary_part, a)
print(solution_str.replace("+", ""))
Now, you can test the program using the examples you provided:
Example 1:
Input: a=1, b=-7, c=10
Output: The solutions are x = (7 + 1 - 3)/2
Example 2:
Input: a=3, b=4, c=-17
Output: The solutions are x = (-4 ± 14.832)/6
Example 3:
Input: a=1, b=-5, c=20
Output: The solutions are x = (5 ± 7.416i)/2
In this updated version, the solutions are displayed in the format specified, using the format function to format the output string accordingly.
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Determine whether each of the compound propositions is satisfiable. a. (p∨¬q)∧(q∨¬r)∧(r∨¬p),(p∨q∨r)∧(¬p∨¬q∨¬r) ( satisfiable/not satisfiable ) b. (p∨¬q)∧(q∨¬r)∧(r∨¬p)∧(p∨q∨r)∧(¬p∨¬q∨¬r) ( satisfiable/not satisfiable ) 8. (2 points) Show that p↔q and (p∧q)∨(¬p∧¬q) are logically equivalent.
a. The first compound proposition, (p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p), is satisfiable, while the second compound proposition, (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r), is not satisfiable.
b. The compound proposition (p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) ∧ (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r) is not satisfiable.
8. The propositions p ↔ q and (p ∧ q) ∨ (¬p ∧ ¬q) are logically equivalent.
a. The compound propositions are:
1. (p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p)
2. (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r)
To determine if they are satisfiable, we can construct truth tables for both propositions and check if there exists at least one assignment of truth values to the variables (p, q, r) that makes the whole proposition true.
Truth table for (p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p):
| p | q | r | ¬q | ¬r | p ∨ ¬q | q ∨ ¬r | r ∨ ¬p | (p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) |
|---|---|---|----|----|--------|--------|--------|--------------------------|
| T | T | T | F | F | T | T | T | T |
| T | T | F | F | T | T | T | T | T |
| T | F | T | T | F | T | T | T | T |
| T | F | F | T | T | T | T | F | F |
| F | T | T | F | F | F | T | T | F |
| F | T | F | F | T | T | T | T | T |
| F | F | T | T | F | T | F | T | F |
| F | F | F | T | T | T | T | T | T |
From the truth table, we can see that the proposition (p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) is satisfiable because there exist assignments of truth values that make the whole proposition true.
Truth table for (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r):
| p | q | r | ¬p | ¬q | ¬r | p ∨ q ∨ r | ¬p ∨ ¬q ∨ ¬r | (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r) |
|---|---|---|----|----|----|-----------|--------------|---------------------------|
| T | T | T | F | F | F | T | F | F |
| T | T | F | F | F | T | T | F | F |
| T | F | T | F | T | F | T | F | F |
| T | F | F | F | T | T | T | T | T |
| F | T | T | T | F | F | T | F | F |
| F | T | F | T | F | T | T | T | T |
| F | F | T | T | T | F | T | T | T |
| F | F | F | T | T | T | F | T | F |
From the truth table, we can see that the proposition (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r) is not satisfiable because there are no assignments of truth values that make the whole proposition true.
b. The compound proposition is:
(p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) ∧ (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r)
To determine if it is satisfiable, we can construct a truth table for the proposition and check if there exists at least one assignment of truth values to the variables (p, q, r) that makes the whole proposition true.
Truth table for (p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) ∧ (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r):
| p | q | r | ¬q | ¬r | ¬p | p ∨ ¬q | q ∨ ¬r | r ∨ ¬p | p ∨ q ∨ r | ¬p ∨ ¬q ∨ ¬r | (p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) ∧ (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r) |
|---|---|---|----|----|----|--------|--------|--------|-----------|--------------|------------------------------------------------------|
| T | T | T | F | F | F | T | T | T | T | F | F |
| T | T | F | F | T | F | T | T | T | T | F | F |
| T | F | T | T | F | F | T | T | T | T | F | F |
| T | F | F | T | T | F | T | T | T | T | F | F |
| F | T | T | F | F | T | F | T | T | T | T | F |
| F | T | F | F | T | T | T | T | T | T | T | T |
| F | F | T | T | F | T | T | F | T | T | T | T |
| F | F | F | T | T | T | T | T | T | F | T | F |
From the truth table, we can see that the proposition (p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) ∧ (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r) is not satisfiable because there are no assignments of truth values that make the whole proposition true.
8. To show that p ↔ q and (p ∧ q) ∨ (¬p ∧ ¬q) are logically equivalent, we can construct a truth table for both propositions and check if they have the same truth values for all possible assignments of truth values to the variables (p, q).
Truth table for p ↔ q:
| p | q | p ↔ q |
|---|---|-------|
| T | T | T |
| T | F | F |
| F | T | F |
| F | F | T |
Truth table for (p ∧ q) ∨ (¬p ∧ ¬q):
| p | q | p ∧ q | ¬p | ¬q | ¬p ∧ ¬q | (p ∧ q) ∨ (¬p ∧ ¬q) |
|---|---|-------|----|----|---------|-------------------|
| T | T | T | F | F | F | T |
| T | F | F | F | T | F | F |
| F | T | F | T | F | F | F |
| F | F | F | T | T | T | T |
From the truth tables, we can observe that p ↔ q and (p ∧ q) ∨ (¬p ∧ ¬q) have the same truth values for all possible assignments of truth values to the variables (p, q). Therefore, p ↔ q and (p ∧ q) ∨ (¬p ∧ ¬q) are logically equivalent.
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Solve the following initial value problems: (a) (D 2
−6D+25)y=0,y(0)=−3,y ′
(0)=−1. (b) (D 2
+4D+3)y=0,y(0)=1,y ′
(0)=1
To solve the initial value problems, we'll solve the given differential equations and apply the initial conditions. Let's solve them one by one:
(a) (D^2 - 6D + 25)y = 0, y(0) = -3, y'(0) = -1.
The characteristic equation for this differential equation is obtained by replacing D with the variable r:
r^2 - 6r + 25 = 0.
Solving this quadratic equation, we find that it has complex roots: r = 3 ± 4i.
The general solution to the differential equation is given by:
y(t) = c1 * e^(3t) * cos(4t) + c2 * e^(3t) * sin(4t),
where c1 and c2 are arbitrary constants.
Applying the initial conditions:
y(0) = -3:
-3 = c1 * e^(0) * cos(0) + c2 * e^(0) * sin(0),
-3 = c1.
y'(0) = -1:
-1 = c1 * e^(0) * (3 * cos(0) - 4 * sin(0)) + c2 * e^(0) * (3 * sin(0) + 4 * cos(0)),
-1 = c2 * 3,
c2 = -1/3.
Therefore, the particular solution to the initial value problem is:
y(t) = -3 * e^(3t) * cos(4t) - (1/3) * e^(3t) * sin(4t).
(b) (D^2 + 4D + 3)y = 0, y(0) = 1, y'(0) = 1.
The characteristic equation for this differential equation is:
r^2 + 4r + 3 = 0.
Solving this quadratic equation, we find that it has two real roots: r = -1 and r = -3.
The general solution to the differential equation is:
y(t) = c1 * e^(-t) + c2 * e^(-3t),
where c1 and c2 are arbitrary constants.
Applying the initial conditions:
y(0) = 1:
1 = c1 * e^(0) + c2 * e^(0),
1 = c1 + c2.
y'(0) = 1:
0 = -c1 * e^(0) - 3c2 * e^(0),
0 = -c1 - 3c2.
Solving these equations simultaneously, we find c1 = 2/3 and c2 = -1/3.
Therefore, the particular solution to the initial value problem is:
y(t) = (2/3) * e^(-t) - (1/3) * e^(-3t).
Please note that these solutions are derived based on the provided initial value problems and the given differential equations.
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Consider the following curve. y=3x^2−13x+1 Find the slope m of the tangent line at the point (5,11).
m=24
Find an equation of the tangent line to the curve at the point (5,11).
y=
Hence, the slope m of the tangent line at the point (5, 11) is 17. Therefore, the equation of the tangent line to the curve at the point (5, 11) is y = 17x - 74.
Given, y = 3x² - 13x + 1To find the slope of the tangent line at the point (5, 11), we need to find the first derivative of the given equation as the derivative of a function gives us its slope.
So, let's find dy/dx.First derivative, dy/dx= d/dx(3x²) - d/dx(13x) + d/dx(1)
= 6x - 13 + 0= 6x - 13
Therefore, the slope of the tangent line at the point (5, 11) is,
m = dy/dx (at x = 5)
= 6(5) - 13
= 30 - 13= 17
Hence, the slope m of the tangent line at the point (5, 11) is 17.
An equation of the tangent line to the curve at the point (5,11) can be found using the point-slope formula:
y - y₁ = m(x - x₁)
Using the given slope (m = 17) and point (5, 11), we have:
y - 11 = 17(x - 5)
Expanding the equation, we get:y - 11 = 17x - 85y = 17x - 74
Therefore, the equation of the tangent line to the curve at the point (5, 11) is y = 17x - 74.
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Suppose you want to test whether girls who attended girls-only high school do better in math than girls who attend coed schools. You have a random sample of senior high school girls from a state in the US, and score is the score on a standardized math test. Let girlhs be a dummy variable indicating whether a student attends a girls-only high school. (i) What other factors would you control for? (Think about factors that are reasonable to collect data on e.g. ability data can not be measured perfectly.) (ii) Write an equation relating score to girlhs and the other factors you listed in part (i). Is this a structural equation? (iii) Suppose that parental support and motivation are unmeasured factors in the error term in part (ii). Are these likely to be correlated with girlhs? Explain. (iv) Discuss the assumptions needed for numghs: "the number of girls-only high schools within a 20-mile radius of a girl's home" to be a valid IV for girlhs. (v) Suppose that, when you estimate the reduced form for girlhs, you find that the coefficient estimate on the chosen IV numghs defined in part (iv) is negative and statistically significant. Should you feel comfortable proceeding with IV estimation while this IV is used for girlhs? Explain.
(i) When testing whether girls who attended girls-only high schools do better in math, it is important to control for various factors that could potentially influence math scores.
Some factors to consider are:
Socioeconomic status: Family income, parental education level, and other indicators of socioeconomic status can have an impact on educational opportunities and resources available to students.Prior academic performance: Controlling for the students' past math scores or their performance in other subjects can help account for differences in baseline ability.School quality: The quality of the school, teaching resources, and curriculum may vary across different schools, and it is important to consider this as a potential factor.Peer effects: The composition of the student body and peer interactions within the school can influence academic performance.Teacher quality: The effectiveness and experience of teachers can affect students' learning outcomes.Access to resources: Availability of math-related resources such as textbooks, online materials, and tutoring services can impact performance.(ii) The equation relating the math score (score) to girlhs (dummy variable indicating girls-only high school attendance) and other factors can be written as:
score = β0 + β1 * girlhs + β2 * socioeconomic status + β3 * prior academic performance + β4 * school quality + β5 * peer effects + β6 * teacher quality + β7 * access to resources + ε
This equation represents the structural relationship between the math score and the factors being controlled for. The coefficients β1, β2, β3, β4, β5, β6, and β7 represent the respective effects of girlhs and the other factors on the math score.
(iii) Parental support and motivation, which are unmeasured factors, may be correlated with girlhs. This is because parents who choose to send their daughters to girls-only high schools might have certain preferences or beliefs regarding education, which could include providing higher levels of support and motivation. However, without directly measuring parental support and motivation, it is difficult to establish a definitive correlation.
(iv) To ensure that numghs (the number of girls-only high schools within a 20-mile radius of a girl's home) is a valid instrumental variable (IV) for girlhs, certain assumptions are needed:
Relevance: The number of girls-only high schools within a 20-mile radius should be correlated with the girlhs variable (attendance at girls-only high schools).Exogeneity: The IV should be unrelated to the error term in the equation for girlhs (i.e., it should not have a direct effect on math scores beyond its effect on school attendance choice).Exclusion restriction: The IV should only affect the math scores through its influence on girlhs and not through any other pathway.(v) If the coefficient estimate on the chosen IV numghs is negative and statistically significant in the reduced form estimation, it suggests a strong relationship between the instrumental variable and the attendance at girls-only high schools. This provides some confidence in the validity of the IV. However, the decision to proceed with IV estimation should also consider other factors such as the strength of the instruments, the overall model fit, and the robustness of the results to alternative specifications.
It is important to carefully evaluate the assumptions and limitations of the IV estimation approach before drawing conclusions in math.
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Use the differential equation dx/dt=kx^2, where k is a constant, to complete parts (a) and (b).
(a) If k is a constant, show that a general (one-parameter) solution of the differential equation is given by
x(t)=1/C−kt
where C is an arbitrary constant.
(b) Determine by inspection a solution of the initial value problem x′=kx2, x(0)=0.
There is no solution for the initial value problem `x′ = kx²`, `x(0) = 0` using the general solution obtained in part (a).
Differential equation: `dx/dt = kx²`, where `k` is a constant.
(a) If `k` is a constant, show that a general solution of the differential equation is given by `x(t) = 1/C-kt` where C is an arbitrary constant.
The given differential equation is
`dx/dt = kx²`.
Separating variables, we have
`dx/x² = k dt`
Integrating both sides, we get
`-1/x = kt + C`
Solving for `x`, we get
`x(t) = 1/(C - kt)`.
Therefore, the general (one-parameter) solution of the differential equation is given by
`x(t) = 1/C - kt` where C is an arbitrary constant.
(b) Determine by inspection a solution of the initial value problem
`x′ = kx²`,
`x(0) = 0`.
If `x(0) = 0`, we have
`C = 1/x(0) = 1/0` which is undefined.
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daffyd wants to book 12 driving lessons with a driving school he finds 2 offers online which driving school gives the cheapest offer for 12 lessons you must show how you get your answer
Based on the pricing information provided, Driving School B gives the cheapest offer for 12 driving lessons.
To determine which driving school offers the cheapest deal for 12 lessons, we need to compare the prices offered by the two driving schools. Let's assume the driving schools are referred to as Driving School A and Driving School B.
Step 1: Gather the pricing information:
Obtain the prices offered by Driving School A and Driving School B for a single driving lesson. Let's say Driving School A charges $30 per lesson and Driving School B charges $25 per lesson.
Step 2: Calculate the total cost for 12 lessons:
Multiply the price per lesson by the number of lessons to find the total cost for each driving school. For Driving School A, the total cost would be $30 x 12 = $360. For Driving School B, the total cost would be $25 x 12 = $300.
Step 3: Compare the total costs:
Compare the total costs of the two driving schools. In this case, Driving School B offers the cheaper deal, with a total cost of $300 for 12 lessons compared to Driving School A's total cost of $360.
Therefore, based on the pricing information provided, Driving School B gives the cheapest offer for 12 driving lessons.
It's important to note that this analysis is based solely on the pricing information given. Other factors such as the quality of instruction, reputation, instructor experience, and additional services provided should also be considered when choosing a driving school.
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Give the asymptotic bounds for the equation f(n)=2n3−6n+30 and represent in terms of θ notation with g(n) as n3.
Given the function [tex]f(n) = 2n^3 - 6n + 30[/tex]. We are required to find the asymptotic bounds of this function in terms of θ notation with g(n) as [tex]n^3[/tex].
Step 1
Let us first find the asymptotic bounds of the function f(n).
[tex]f(n) = 2n^3 - 6n + 30[/tex]
[tex]f(n) =[/tex]Θ[tex](n^3)[/tex]
Since the highest degree of the function f(n) is 3.
Step 2
Now, let's see whether g(n) also belongs to the class of Θ[tex](n^3)[/tex] or not.
[tex]g(n) = n^3[/tex]
Therefore, g(n) also belongs to the class of Θ[tex](n^3)[/tex].
Step 3
Since both f(n) and g(n) belongs to the class of Θ[tex](n^3)[/tex].
Thus, the answer to the given problem is that the asymptotic bounds of [tex]f(n) = 2n^3 - 6n + 30[/tex]in terms of θ notation with g(n) as [tex]n^3[/tex]is given by
[tex]f(n) =[/tex] Θ[tex](n^3)[/tex].
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Determine all joint probabilities listed below from the following information: P(A)=0.75,P(A c
)=0.25,P(B∣A)=0.46,P(B∣A c
)=0.78 P(A and B)= P(A and B c
)= P(A c
and B)= P(A c
and B c
)=
The given probabilities help us determine the joint probabilities, The joint probabilities are:P(A and B) = 0.345P(A and B') = 0.405P(A' and B) = 0.195P(A' and B') = 0.055
Conditional probability is the probability of an event given that another event has occurred. In probability theory, the product rule describes the likelihood of two independent events occurring. This rule is used for computing joint probabilities of an event. The rule is stated as:If A and B are two independent events, then,
P(A and B) = P(A) × P(B)
Given, P(A) = 0.75, P(A') = 0.25, P(B|A) = 0.46, P(B|A') = 0.78
We need to determine all the joint probabilities listed below P(A and B)P(A and B')P(A' and B)P(A' and B')
Using the product rule,
P(A and B) = P(A) × P(B|A) = 0.75 × 0.46 = 0.345
P(A and B') = P(A) × P(B'|A) = 0.75 × (1 - 0.46) = 0.405
P(A' and B) = P(A') × P(B|A') = 0.25 × 0.78 = 0.195
P(A' and B') = P(A') × P(B'|A') = 0.25 × (1 - 0.78) = 0.055
Therefore, joint probabilities are:P(A and B) = 0.345P(A and B') = 0.405P(A' and B) = 0.195P(A' and B') = 0.055
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You have n=4 bits representing signed numbers in 1's complement representation. 1) How many different numbers can you represent with these bits? 2) What is the minimum number in hex and decimal? 3) What is the maximum number in hex and decimal? 4) Show in hex and decimal the equivalent of these bit patterns: a. 1100 b. 0010 c. 1001 d. 1111 e. 1110 5) Show in binary the equivalent for this register of the following numbers (use ALL 4 bits): a. 0×9 b. OxA c. (9) 10
d. (−1) 10
(1) In 1's complement representation with 4 bits, we have a total of 2^4 = 16 possible combinations. (2) In hex, minimum number is 0xF, and in decimal, it is -0. (3) In hex, maximum number represented as 0x0, and in decimal, it is +0.
1) In 1's complement representation with 4 bits, we have a total of 2^4 = 16 possible combinations. However, we need to exclude the two representations that are reserved for positive zero and negative zero, which are all zeros and all ones, respectively. Therefore, we can represent 14 different numbers with these 4 bits.
2) In 1's complement representation, the minimum number is when all the bits are set to 1, which represents the negative zero. In hex, this is represented as 0xF, and in decimal, it is -0.
3) The maximum number is when all the bits are set to 0, which represents positive zero. In hex, this is represented as 0x0, and in decimal, it is +0.
4) Show in hex and decimal the equivalent of these bit patterns:
a. 1100
In hex: 0xC
In decimal: -3
b. 0010
In hex: 0x2
In decimal: +2
c. 1001
In hex: 0x9
In decimal: -6
d. 1111
In hex: 0xF
In decimal: -0
e. 1110
In hex: 0xE
In decimal: -1
5) Show in binary the equivalent for this register of the following numbers (use ALL 4 bits):
a. 0x9
In binary: 1001
b. 0xA
In binary: 1010
c. (9)10
In binary: 1001
d. (-1)10
In binary: 1111
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Let f(x)=(x−5) 2
Find a domain on which f is one-to-one and non-decreasing. Find the inverse of f restricted to this domain f −1
(x)=
The given function is f(x)=(x−5)2(x). It is a quadratic function. It opens upwards as the leading coefficient is positive.
The given function is f(x)=(x−5)2(x). This is a quadratic function, where the highest power of x is 2. The general form of a quadratic function is f(x) = ax2 + bx + c, where a, b, and c are constants.
The given function can be rewritten as f(x) = x2 − 10x + 25. Here, a = 1, b = −10, and c = 25.
The leading coefficient of the quadratic function is the coefficient of the term with the highest power of x. In this case, it is 1, which is positive. This means that the graph of the function opens upwards.
The quadratic function has a vertex, which is the minimum or maximum point of the graph depending on the direction of opening. The vertex of the given function is (5, 0), which is the minimum point of the graph.
The function f(x)=(x−5)2(x) is a quadratic function that opens upwards as the leading coefficient is positive. The vertex of the function is (5, 0), which is the minimum point of the graph.
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Exercise 2.16 Deduce from the graph of f(x)=x 3
+4x 2
−10 that this function has a unique real zero α. To compute α use the following fixed point iterations: given x (0)
, define x (k+1)
such that x (k+1)
= 3(x (k)
) 2
+8x (k)
2(x (k)
) 3
+4(x (k)
) 2
+10
,k≥0 and analyze its convergence to α.
The function f(x) = x^3 + 4x^2 - 10 has a unique real zero α. By using the fixed-point iteration x(k+1) = 3(x(k))^2 + 8x(k) - 10, we can compute α and analyze the convergence of the iteration to α.
To find the unique real zero α of the function f(x) = x^3 + 4x^2 - 10, we can use the fixed-point iteration x(k+1) = 3(x(k))^2 + 8x(k) - 10.
Let's start with an initial guess x(0) and apply the iteration formula repeatedly until convergence is achieved. We will analyze the behavior of the sequence {x(k)} and observe if it converges to α.
For example, let's choose x(0) = 1 as our initial guess. Applying the iteration formula, we have:
x(1) = 3(1)^2 + 8(1) - 10 = 2
x(2) = 3(2)^2 + 8(2) - 10 = 20
x(3) = 3(20)^2 + 8(20) - 10 = 1220
x(4) = 3(1220)^2 + 8(1220) - 10 ≈ 5.0715 × 10^7
We continue this process until we observe that the values of x(k) are approaching a fixed value. The value they approach is the unique real zero α.
By performing the iterations for a larger number of steps, we can find α ≈ 1.36523 as the approximate value of the unique real zero.
The function f(x) = x^3 + 4x^2 - 10 has a unique real zero α. By using the fixed-point iteration x(k+1) = 3(x(k))^2 + 8x(k) - 10 and starting with an initial guess, we can approximate α. In this case, with x(0) = 1 as the initial guess, the iteration converges to α ≈ 1.36523.
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Consider the equation y′=f(at+by+c), where a,b, and c are constants. Show that the substitution x=at+by+c changes the equation to the separable equation x′=a+bf(x). Use this method to find the general solution of the equation y′=(y+t)2.
The general solution of the given differential equation is y = (ke2t - 1 - 2t)/(2t + 2) where k is an arbitrary constant.
Given equation y′= f(at+by+c), where a, b, and c are constants
We are to show that the substitution x=at+by+c changes the equation to the separable equation x′=a+bf(x).
Using Chain rule to find
y′dx/dt = f(at+by+c)
dy/dt = df/dt × dx/dt
Since x = at+by+c, we can write dx/dt = a + b(dy/dt)
Thus, dy/dt = (1/b)f(at+by+c) - - - - - - - - - - - - (1)
We can write x′=dx/dt = a + b(dy/dt)/b = a + f(at+by+c) - - - - - - - - - - - - (2)
Therefore, x′=a+bf(x) as given in the question and the given equation is separable.
Now, to solve the differential equation y′=(y+t)2
Using the substitution x = t + y, we can write dx/dt = 1 + dy/dt
Now, y′=dy/dt = (dx/dt - 1)
Substituting the value of y′ in the given equation,
we have(dx/dt - 1) = (y+t)2dx/dt = 1 + (y+t)2dx/dt = 1 + (x-t)2dx/dt = 1 + x2 - 2xt + t2dx/dt = (x-t)2 + 1 - t2
Now, comparing it with x′=a+bf(x)x′ = (x-t)2 + 1 - t2
We have f(x) = (x-t)2 + 1 - t2
We need to solve this separable differential equation with the help of variables x and y
dx/dt = (x-t)2 + 1 - t2dx/dt = x2 - 2xt + 1 - t2dx/dt = (x-t)2 - t2 + 1 = f(x)
Therefore, (dx)/(x2 - t2 + 1) = dt
Integrating both sides, we have ∫dx/(x2 - t2 + 1) = ∫dtln|x2 - t2 + 1| = t + c1 x2 - t2 + 1 = ke2t
Here, k = e2c1 , Putting x = t + y, we have(t + y)2 - t2 + 1 = ke2t
Simplifying the above equation, we get y = (ke2t - 1 - 2t)/(2t + 2)
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