P-value is less than the level of significance α = 0.05, we reject the null hypothesis and conclude that the mean time to assemble at the end of a 3-week training period is less for the new training procedure.
a. Performing the test using the classical approach α=0.05
Hypothesis test:
Null hypothesis H₀: The mean time for the new procedure = the mean time for the standard procedure
Alternative hypothesis H₁: The mean time for the new procedure < the mean time for the standard procedure (one-tailed test)
Level of significance α = 0.05
Since population standard deviations are unknown, we will use the t-test to conduct the hypothesis test.
Below is the calculation of the t-test.(For detailed calculation of t-test, please refer to the image attached)
Using a t-distribution table with df = 14 (n1 + n2 - 2), the critical value for a one-tailed test at α = 0.05 is -1.761.
The test statistic (-2.029) is less than the critical value (-1.761), so we reject the null hypothesis and conclude that the mean time to assemble at the end of a 3-week training period is less for the new training procedure.
b.Performing the above test assuming population variances are equal
Hypothesis test
Null hypothesis H₀: The mean time for the new procedure = the mean time for the standard procedure
Alternative hypothesis H₁: The mean time for the new procedure < the mean time for the standard procedure (one-tailed test)Level of significance α = 0.05
Since population variances are assumed to be equal, we will use the pooled t-test to conduct the hypothesis test.
Below is the calculation of the pooled t-test.(For detailed calculation of pooled t-test, please refer to the image attached)
Using a t-distribution table with df = 16 (n₁ + n₂ - 2), the critical value for a one-tailed test at α = 0.05 is -1.746.
The test statistic (-2.029) is less than the critical value (-1.746), so we reject the null hypothesis and conclude that the mean time to assemble at the end of a 3-week training period is less for the new training procedure.
c. Performing the above test (#2) using the p-value approach.
Hypothesis test:
Null hypothesis H₀: The mean time for the new procedure = the mean time for the standard procedure
Alternative hypothesis H₁: The mean time for the new procedure < the mean time for the standard procedure (one-tailed test)
Level of significance α = 0.05
Since population variances are assumed to be equal, we will use the pooled t-test to conduct the hypothesis test.
The test statistic is -2.029, which corresponds to a p-value of 0.0328.
Since the p-value is less than the level of significance α = 0.05, we reject the null hypothesis and conclude that the mean time to assemble at the end of a 3-week training period is less for the new training procedure.
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GIVING BRAINLIEST PLEASE DUE TODAY
A. For babysitting, Nicole charges a flat fee of $3, plus $5 per hour.
Write an equation for the cost, C, after h hours of babysitting.
B. How much money will she make if she babysits for 5 hours?
C. If Nicole earned $48.00, how many hours did she babysit?
In the figure below, m<3 = 136. Find m <1, m<2, and m<4 please explain
Answer:
angle 3=angle (vertical opposite angle)
angle 3 +angle 2=180°(by linear pair)
136°+angle 2=180°
angle 2=180°-136°
angle 2=44°
angle 2=angle 4(by vertical opposite angle)
A pressure vessel has a design pressure of 50 bar. However the safety case for the chemical plant on which it is to be used requires that the pressure vessels have a 95% probability of surviving a pressure of 70 bar. Computer codes have generated an estimate of only 0.80 for the probability that any such pressure vessel, picked at random, will survive at 70 bar. However, they have also calculated that of the 20% of the pressure vessels that will not survive a pressure of 70 bar, 40% will fail under a pressure of 58 bar or less, while 80% will fail under a pressure of 65 bar or less. It is decided that an over-pressure test needs to be used to give reassurance on the behaviour of this particular pressure vessel. This test may be carried out at either 58 bar or 65 bar. The lower pressure test is considerably less difficult and cheaper to administer. (i) Suppose that you are brought in as a consultant. By calculating the probability of the pressure vessel being able to support the 70 bar maximum pressure if the over-pressure test is passed, advise on which over-pressure test should be administered.
It is advised that lower pressure test should be administered.
Probability of survival of pressure vessel the pressure vessel is tested under the lower pressure test at 58 bar, the probability of survival is given by the sum of the probability of survival if the vessel is one of the 60% that will survive 70 bar and the probability of survival if the vessel is one of the 40% that will fail at 70 bar but will survive 58 bar or less. If P1 represents the probability of survival of the vessel if it is one of the 60% that will survive 70 bar, and P2 represents the probability of survival if it is one of the 40% that will fail at 70 bar but will survive 58 bar or less, then the probability of survival of the vessel, if it is tested under the lower pressure test at 58 bar, is given by:
P = 0.60 x 1 + 0.40 x (1 - 0.60) = 0.76
If the pressure vessel is tested under the higher pressure test at 65 bar, the probability of survival is given by the sum of the probability of survival if the vessel is one of the 60% that will survive 70 bar, and the probability of survival if the vessel is one of the 40% that will fail at 70 bar but will survive 65 bar or less. If P3 represents the probability of survival of the vessel if it is one of the 40% that will fail at 70 bar but will survive 65 bar or less, then the probability of survival of the vessel, if it is tested under the higher pressure test at 65 bar, is given by:
P = 0.60 x 1 + 0.40 x (1 - 0.80 x P3)
The condition that the probability of survival of the vessel, if it is tested under the higher pressure test at 65 bar, is at least 0.95 is therefore:
0.60 + 0.40 x (1 - 0.80 x P3) ≥ 0.95
This simplifies to: P3 ≤ 0.625
Using the above values for P1, P2, and P3, it is clear that the probability of the vessel surviving if tested at the lower pressure of 58 bar is greater than 0.95. Therefore, the lower pressure test should be carried out.
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choose the correct simplification of (4x3 − 3x − 7) (3x3 5x 3).
a. 7x3 − 2x − 4
b. x3 − 8x − 10
c. 7x3 2x − 4
d. x3 8x 10
The answer is not provided in the given options.
To simplify the expression (4x^3 - 3x - 7)(3x^3 + 5x + 3), we can use the distributive property of multiplication.
Multiplying each term in the first expression by each term in the second expression, we get:
(4x^3)(3x^3) + (4x^3)(5x) + (4x^3)(3) + (-3x)(3x^3) + (-3x)(5x) + (-3x)(3) + (-7)(3x^3) + (-7)(5x) + (-7)(3)
Simplifying each term, we have:
12x^6 + 20x^4 + 12x^3 - 9x^4 - 15x^2 - 9x - 21x^3 - 35x - 21
Combining like terms, we get:
12x^6 + (20x^4 - 9x^4) + (12x^3 - 21x^3) + (-15x^2) + (-9x - 35x) + (-21)
Simplifying further, we have:
12x^6 + 11x^4 - 9x^3 - 15x^2 - 44x - 21
Therefore, the correct simplification of (4x^3 - 3x - 7)(3x^3 + 5x + 3) is:
12x^6 + 11x^4 - 9x^3 - 15x^2 - 44x - 21.
Therefore, the answer is not provided in the given options.
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A SHS student conducted a survey to test the claim that "less than half of all the adults are annoyed by the violence on television" . Suppose that from a poll of 2,400 surveyed adults, 1,152 indicated their annoyance with television violence. Test this claim using 0.10 level of significance.
The null hypothesis (H0) assumes that the proportion of adults annoyed by television violence is equal to or greater than 0.5, while the alternative hypothesis (Ha) assumes that the proportion is less than 0.5.
In this case, the sample proportion is calculated as the number of adults indicating annoyance divided by the total sample size: 1,152/2,400 = 0.48.
Next, we can calculate the test statistic, which follows a standard normal distribution under the null hypothesis. The test statistic formula is z = (p - P) / sqrt(P(1-P)/n), where p is the sample proportion, P is the hypothesized proportion under the null hypothesis (0.5 in this case), and n is the sample size.
Using the given values, we can calculate the test statistic:
z = (0.48 - 0.5) / sqrt(0.5(1-0.5)/2400) ≈ -1.67.
Finally, we compare the test statistic to the critical value. At a significance level of 0.10, the critical value for a one-tailed test is approximately -1.28. Since the test statistic (-1.67) is smaller than the critical value, we reject the null hypothesis. This suggests that there is evidence to support the claim that less than half of all adults are annoyed by violence on television.
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C⊃D
~(A∨B)∨C
~B∨D
Show that each of the following arguments is valid by
constructing a proof
The given arguments are proved using logical inference rules.
To show that each of the following arguments is valid, we need to construct a proof using logical inference rules. Here is a proof for the given arguments:
Argument 1:
1. C ⊃ D (Premise)
2. ~(A ∨ B) ∨ C (Premise)
3. ~B ∨ D (Premise)
4. ~(A ∨ B) (Assumption)
5. ~A ∧ ~B (De Morgan's Law, 4)
6. ~B (Simplification, 5)
7. D (Disjunctive Syllogism, 3, 6)
8. ~(A ∨ B) ∨ D (Disjunction Introduction, 7)
9. C (Disjunction Elimination, 2, 8)
10. ~(A ∨ B) ∨ C (Disjunction Introduction, 9)
Therefore, the argument is valid.
Argument 2:
1. C ⊃ D (Premise)
2. ~(A ∨ B) ∨ C (Premise)
3. ~B ∨ D (Premise)
4. ~A ∨ ~B (Assumption)
5. ~(A ∨ B) (De Morgan's Law, 4)
6. C (Disjunction Elimination, 2, 5)
7. C ⊃ D (Premise)
8. D (Modus Ponens, 6, 7)
9. ~B ∨ D (Disjunction Introduction, 8)
10. ~(A ∨ B) ∨ D (Disjunction Introduction, 9)
Therefore, the argument is valid.
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PLEASE HELP I’m bad at math
A.Find the greatest common factor GCF of 42 and 12
B.Use the GCF to factor 42 + 12
Please be quick if you can
Answer:
6
Step-by-step explanation:
The GCF of 42 and 12 is 6
After six rolls of a standard die, the experimental probability of rolling a 3 is 26. What do you expect will happen to the experimental probability if the die is rolled 90 more times? Explain.
Answer:
The experimental probability should get closer to the theoretical probability of 1/6 with more trials.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
Experimental probability:
The number of desired outcomes is taken from the results of an experiment.
Theoretical probability:
Found before the experiment happens.
For a large number of trials, the experimental probability will be closer to the theoretical probability.
In this question:
A standard die has 6 sides, one which is 3. So the theoretical probability of rolling a 3 is 1/6.
After six rolls of a standard die, the experimental probability of rolling a 3 is 2/6.
The experimental probability, after six rolls, is 2/6 = 1/3.
What do you expect will happen to the experimental probability if the die is rolled 90 more times?
As the number of trials increase, the experimental probability is expected to get closer to the theoretical probability, which in this case is 1/6.
Suppose that a random variable X satisfies E[X] = 0, E[X2] = 1, E[X3] = 0, E[X4] = 3 and let Y = a + bx+cX? Find the correlation coefficient p(X,Y).
Given the random variable X with specific expected values and the equation Y = a + bx + cX, we are asked to find the correlation coefficient p(X,Y).
The correlation coefficient between two random variables X and Y is given by the formula:
p(X,Y) = Cov(X,Y) / sqrt(Var(X) * Var(Y))
To calculate the correlation coefficient, we need to find the covariance (Cov(X,Y)) and the variances (Var(X) and Var(Y)).
Given the expected values, we can calculate the required values as follows:
Cov(X,Y) = E[XY] - E[X]E[Y]
Var(X) = E[[tex]X^2[/tex]] - [tex](E[X])^2[/tex]
Var(Y) = E[tex][(Y - E[Y])^2][/tex]
Using the provided expected values, we can substitute them into the formulas:
Cov(X,Y) = E[XY] - E[X]E[Y] = E[(a + bx + cX)X] - (0)(E[a + bx + cX]) = E[aX + b[tex]X^2[/tex] + c[tex]X^2[/tex]] = a(E[X]) + b(E[[tex]X^2[/tex]]) + c(E[[tex]X^3[/tex]])
Var(X) = E[[tex]X^2[/tex]] - [tex](E[X])^2[/tex] = 1 - [tex](0)^2[/tex] = 1
Var(Y) = E[(Y - [tex]E[Y])^2[/tex]] = E[(a + bx + cX - [tex](E[a + bx + cX]))^2[/tex]] = E[[tex](a + bx + cX)^2[/tex]]
Using the provided values for E[[tex]X^3[/tex]] and E[[tex]X^4[/tex]], we can simplify the expressions further and calculate the values.
Once we have the values of Cov(X,Y), Var(X), and Var(Y), we can substitute them into the correlation coefficient formula to find p(X,Y).
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What is the surface area of this right rectangular prism with dimensions of 8 inches by 4 inches by 14 inches?
a. 310
b. 400
c. 525
d. 650
Answer:
400
Step-by-step explanation:
The area of sides and add then up
divide 32x3 48x2 − 40x by 8x. 4x2 − 6x 5 4x2 6x − 5 4x3 − 6x2 5 4x3 6x2 − 5
The division of 32x^3 - 48x^2 - 40x by 8x results in the quotient 4x^2 - 6x - 5 on solving the given equation.
To divide 32x^3 - 48x^2 - 40x by 8x, we divide each term of the dividend by the divisor, 8x.
Dividing 32x^3 by 8x gives us 4x^2, as x^3/x = x^2 and 32/8 = 4.
Dividing -48x^2 by 8x gives us -6x, as -48x^2/8x = -6x.
Dividing -40x by 8x gives us -5, as -40x/8x = -5.
Combining these results, the quotient is 4x^2 - 6x - 5.
The quotient represents the result of dividing the dividend by the divisor, resulting in a polynomial expression without any remainder. Therefore, when dividing 32x^3 - 48x^2 - 40x by 8x, the quotient is 4x^2 - 6x - 5.
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If arc QT = (27x + 3) arc RT = (9x – 5) and RST = (102 – 2) find arc RT.
Answer:
Step-by-step explanation:
x = 6
arc RT = 49°
What is tangent?"It is a line that intersects the circle exactly at one point."
What is secant?"It is a line that intersects circle at two points."
For given example,
arc QT = (27x + 3)°
arc RT = (9x – 5)°
∠RST = (10x – 2)°
From figure we can observe that line ST is tangent and line SQ is secant.
∠RST is the angle subtended by tangent ST and secant SQ
We know, the angle subtended by the tangent and the secant is half the difference of the measures of the intercepted arcs.
⇒ ∠RST = (QT - RT)/2
⇒ 10x - 2 = [(27x + 3) - (9x - 5)] /2
⇒ 2(10x - 2) = 27x + 3 - 9x + 5
⇒ 20x - 4 = 18x + 8
⇒ 20x - 18x = 8 + 4
⇒ x = 6
So, arc RT would be,
⇒ 9x - 5 = 9(6) - 5
⇒ 9x - 5 = 49°
Therefore, arc RT = 49°
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PLEASE HELP ME ITS AN EMERGENCY!
Answer:
Number 1 is correct
4.5 x 12 = 54
Number 3 is wrong,
Formula of a triangle:
BH x 1/2(basically dividing by 2)
8 x 15 = 120,
120 DIVIDED BY 2 = 60 is your area, not 120.
Your plug in would be for the triangle:
8 x 15 x 1/2
Number 5 is wrong.
11 + 4 = 15
15 x 6 (you forgot to multiply by the height!) = 90
90 divided by 2 ( x 1/2) = 45 is your area, NOT 90.
Your formula for a trapezoid is:
(b1 + b2) x h x 1/2 Don't forget your height next time!
Plug in: (4 + 11) x 6 x 1/2
What is the best approximation for √29? 5 5.2 5.9 6
Answer:
Hello, Brainly users, hi hows your day going. Great. Yeah thanks for asking. Anyway, the answer is 5.2.
Step-by-step explanation:
Will provide step-by-step explanation as to how I figured it out if I can get brainliest *HINT HINT*. Have a good day. And keep good vibes amid the pandemic
:D
Answer:
5.2
Plz mark me as brainliest.
Evaluate the following expression for P = -3 and S = 2
Answer: -9
2 to the power of 0 = 1 then all you have to do is plug in the numbers and simplify
Answer:
-9
Step-by-step explanation:
Well you just subsitute it all and solve from there
s^0= 2^0
p^-2= -3^-2
Anything squared to the power of 0 is 1
so its already 1/smth
the second part is just 3^-2 first which is 1/9 then the negative sign which is -1/9
The heights of the female adults in a country can be represented by a random variable that follows the normal distribution N(170,30) Answer these questions:
1. To enter the tallest 20% of the female adults, a man must be at least [....] cm tall.
2. To enter the tallest 1% of the female adults, a man must be at least [....] cm tall.
Given the heights of the female adults in a country is represented by a random variable that follows the normal distribution N(170,30)1. To enter the tallest 20% of the female adults, a man must be at least 184.87 cm tall.
Solution:It is given that, the heights of the female adults in a country can be represented by a random variable that follows the normal distribution N(170,30)Let X be the height of female adults, then X ~ N(170, 30)
Let P be the probability of the tallest 20% female adults.To find the value of x we need to use the standard normal distribution formula which is given byz = (x - μ) / σWhere,z = standard score or z-scorex = the raw scoreμ = the meanσ = the standard deviation
Now, the probability of the tallest 20% female adults is P = 0.20 or 20%We know that the total area under the normal curve is 1 which means P(X < μ) = 0.5So, P( X > μ) = 1 - P(X < μ) = 1 - 0.5 = 0.5Therefore, 0.5 = P(Z < z) at z = 0.84 from standard normal distribution table,0.84 = (x - μ) / σOn substituting the values,0.84 = (x - 170) / 30x - 170 = 0.84 x 30x - 170 = 25.2x = 195.2So, to enter the tallest 20% of the female adults, a man must be at least 184.87 cm tall.2.
To enter the tallest 1% of the female adults, a man must be at least 201.17 cm tall.
Solution: It is given that, the heights of the female adults in a country can be represented by a random variable that follows the normal distribution N(170,30)Let X be the height of female adults, then X ~ N(170, 30)
Let P be the probability of the tallest 1% female adults.
Now, the probability of the tallest 1% female adults is P = 0.01 or 1%We know that the total area under the normal curve is 1 which means P(X < μ) = 0.5So, P( X > μ) = 1 - P(X < μ) = 1 - 0.5 = 0.5Therefore, 0.5 = P(Z < z) at z = 2.33 from standard normal distribution table,2.33 = (x - μ) / σOn substituting the values,2.33 = (x - 170) / 30x - 170 = 2.33 x 30x - 170 = 69.9x = 239.9 cm
So, to enter the tallest 1% of the female adults, a man must be at least 201.17 cm tall.
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To enter the tallest 1% of the female adults, a man must be at least 104.1 cm tall.
The heights of female adults in a country can be represented by a random variable that follows the normal distribution N(170,30).
The questions and their solutions are:
1. To enter the tallest 20% of female adults, a man must be at least [ ] cm tall.
To solve this, we can use the standard normal distribution table.
Let Z be the standard normal distribution.
To find the corresponding Z-score to the 20th percentile, we use the standard normal distribution table.
P(Z < z) = 0.20, where P(Z < z) is the area under the standard normal distribution curve to the left of z.
z = -0.84 (rounded to 2 decimal places).
Using the formula z = (X - µ) / σ, we can solve for X, the height of the woman:
[tex]z = (X - µ) / σX = σz + µX = 30(-0.84) + 170X = 147.8[/tex] (rounded to the nearest tenth of a cm)
Therefore, to enter the tallest 20% of the female adults, a man must be at least 147.8 cm tall.
2. To enter the tallest 1% of female adults, a man must be at least [ ] cm tall.
P(Z < z) = 0.01
z = -2.33 (rounded to 2 decimal places).
Using the formula z = (X - µ) / σ, we can solve for X, the height of the woman:
[tex]z = (X - µ) / σX = σz + µX = 30(-2.33) + 170X = 104.1[/tex] (rounded to the nearest tenth of a cm)
Therefore, to enter the tallest 1% of the female adults, a man must be at least 104.1 cm tall.
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you randomly select 100 drivers ages 16 to 19 from example 4. what is the probability that the mean distance traveled each day is between 19.4 and 22.5 miles?
Given that we randomly select 100 drivers ages 16 to 19 from example 4. We are to determine the probability that the mean distance traveled each day is between 19.4 and 22.5 miles. The probability that the mean distance traveled each day is between 19.4 and 22.5 miles is approximately 1.00.
Probability distribution is a function which represents the probabilities of all possible values of a random variable.
When the probability distribution of a random variable is unknown, we can use the Central Limit Theorem (CLT) to estimate the mean of the population.
Let X be the mean distance traveled each day by the 100 drivers ages 16 to 19.
Then, the distribution of X is approximately normal with the mean μ = 20.4 miles and the standard deviation σ = 3.8 miles.
Therefore, we can calculate the z-score as follows; z = (X - μ) / (σ / √n), where X = 19.4 and n = 100.
z₁ = (19.4 - 20.4) / (3.8 / √100)
z₁ = -2.63 and
z₂ = (22.5 - 20.4) / (3.8 / √100)
z₂ = 5.53
Hence, the probability that the mean distance traveled each day is between 19.4 and 22.5 miles is;
P(19.4 < X < 22.5) = P(z₁ < z < z₂).
Using the z-table, the probability is found to be; P(-2.63 < z < 5.53) ≈ 1.00.
Therefore, the probability that the mean distance traveled each day is between 19.4 and 22.5 miles is approximately 1.00.
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two points on a parabola are (-3,5) and (11,5) what is the equation of the axis of symmetry
Answer:
I don't know how to do it the subject
What is the area of a circle with a radius of 20 inches?
Group of answer choices
1256 square inches
314 square inches
31.4 square inches
125.6 square inches
Answer:
1256 square inches
Step-by-step explanation:
Area of a circle:
A = πr²
Given:
r = 20
Work:
A = πr²
A=(3.14)20²
A = 3.14(400)
A = 1256
Answer:
1256 square inches.
Step-by-step explanation:
20 squared * π = 1256 square inches
The graph of the function is shown below
Which of the following functions best represents the graph ?
A) y= 0.5(2.5)^x
B) y= 3.5x^2 + 0.5
C) y= 0.5(6)^x
D) y= 0.5x+2.5
Answer:
B) y=3.5x^2 +0.5
Step-by-step explanation:
the (0,0.5) tells you what the y-intercept is :)
hope this helps :)
Sophia went to see a play at the theater downtown. 8:30 PM. The first act was 55 minutes long. Intermission lasted for 20 minutes, and the second act was an hour long. What time was it when the play finished?
Five-sixths of the students at a nearby college live in dormitories. If 6000 students at the college live in dormitories, how many students are there in the college?
The number of students that are in college is 7200 if Five-sixths of the students at a nearby college live in dormitories. If 6000 students at the college live in dormitories.
What is a fraction?Fraction number consists of two parts, one is the top of the fraction number which is called the numerator and the second is the bottom of the fraction number which is called the denominator.
It is given that:
Five-sixths of the students at a nearby college live in dormitories. If 6000 students at the college live in dormitories
Let x be the number of students that are in college.
Then from the question:
The value of x can be found as follows:
x = (6000/5)×6
x = (1200)×6
x = 7200
Thus, the number of students that are in college is 7200 if Five-sixths of the students at a nearby college live in dormitories. If 6000 students at the college live in dormitories.
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HELP ME ASAP!!!!!!!!
See picture below.
Answer:
Step-by-step explanation:
x+3 - x = 3 = width
2(x+3) = 2x + 6 = length
2x + 6 +2x +6 +3 +3 = 4x + 18 = Perimeter of T
Assume that a sample is used to estimate a population mean u. Find the margin of error M.E. that corresponds to a sample of size 23 with a mean of 37.6 and a standard deviation of 16.1 at a confidence level of 95%.
Report ME accurate to one decimal place because the sample statistics are presented with this accuracy. M.E. ______
Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.
Based on the illustration above, the value of margin of error M.E is 6.961
Margin of error (M.E) is calculated as the product of critical value (CV) and standard error (SE) of sample mean.
The formula for standard error of sample mean is:
SE = σ/√n
where σ is the population standard deviation and n is the sample size. The formula for margin of error is:
M.E. = CV x SE
where CV is the critical value.
The critical value for a 95% confidence level with 22 degrees of freedom (sample size 23 - 1) is 2.074 (rounded to 3 decimal places).
The sample mean is 37.6 and the population standard deviation is 16.1.
Sample size, n = 23.
Using the formula,
SE = σ/√n
SE = 16.1/√23
SE = 3.365 (rounded to 3 decimal places)
Now, using the calculated value of SE and CV,
ME = CV x SE
ME = 2.074 × 3.365
ME = 6.961 (rounded to 1 decimal place)
Therefore, the margin of error (M.E.) is 6.961 (rounded to 1 decimal place).
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solve the given differential equation by undetermined coefficients. 1 4 y'' y' y = x2 − 4x
The given second-order linear differential equation, 1y'' + 4y' + y = x^2 - 4x, can be solved using the method of undetermined coefficients. The particular solution is obtained by assuming a form for the solution and determining the coefficients based on the right-hand side of the equation.
To solve the given differential equation by undetermined coefficients, we first consider the homogeneous equation, which is obtained by setting the right-hand side equal to zero: 1y'' + 4y' + y = 0. The characteristic equation associated with this homogeneous equation is [tex]r^2[/tex]+ 4r + 1 = 0, where r represents the roots of the equation. Solving this quadratic equation, we find two complex conjugate roots: r = -2 ± i.
Since the right-hand side of the original equation is a polynomial of degree 2, we assume a particular solution of the form y_p = A[tex]x^{2}[/tex] + Bx + C. Substituting this assumed form into the original equation, we differentiate it twice to obtain the expressions for y''_p and y'_p, and substitute them back into the original equation. This allows us to equate the coefficients of like powers of x on both sides of the equation.
By comparing coefficients, we find that A = 1 and B = -2. However, the term C is a constant and does not contribute to the differential equation. Hence, the particular solution is y_p = [tex]x^{2}[/tex] - 2x.
Finally, the general solution of the differential equation is given by the sum of the homogeneous solution and the particular solution: y = y_h + y_p. Since the homogeneous solution contains complex roots, it can be expressed as y_h =[tex]e^{-2x}[/tex](C_1cos(x) + C_2sin(x)), where C_1 and C_2 are arbitrary constants. Thus, the complete solution is y = [tex]e^{-2x}[/tex]C_1cos(x) + C_2sin(x)) + [tex]x^2[/tex] - 2x.
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One year of classes at the University of Texas at Austin costs $10,700.
Georgio has received a grant that will pay $700 and a scholarship for
$5,500. He wants to get a job to pay 40% of the remainder of the costs
and borrow the rest of the money. How much does he need to earn on
his job, and how much will he need to borrow?
O
Georgio has to earn $1,600 and borrow $2,700.
Georgio has to earn $1,800 and borrow $2,700.
Georgio has to earn $1,800 and borrow $2,300.
Georgio has to earn $1,600 and borrow $2,300.
Answer:
Georgio has to earn $1,800 and borrow $2,700.
What is the value of a?
A. -18
B. -14
C. 14
D. 18
The AIC strikes a balance between:
The AIC, or the Akaike Information Criterion, strikes a balance between model complexity and goodness of fit.
In statistical modeling, it is crucial to find a balance between the complexity of a model and its ability to accurately capture the underlying patterns in the data. On one hand, a complex model with numerous parameters may be able to fit the data very closely, resulting in a low error or residual.
However, such a model runs the risk of overfitting, meaning it may become too specific to the training data and perform poorly when applied to new, unseen data.
On the other hand, a simpler model with fewer parameters may not capture all the nuances of the data and may have a higher error or residual. This is known as underfitting, as the model fails to capture the underlying complexity of the data.
The AIC addresses this trade-off by considering both the goodness of fit and the complexity of the model. It penalizes models with a higher number of parameters, encouraging a balance between model complexity and goodness of fit.
The AIC takes into account the residual sum of squares (RSS) or the likelihood of the model, and adjusts it based on the number of parameters used. The goal is to select the model with the lowest AIC value, indicating a good compromise between complexity and fit.
By striking this balance, the AIC provides a reliable criterion for model selection, allowing researchers and statisticians to choose the most appropriate model for their data while avoiding both overfitting and underfitting.
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simplify this (√7+√3)²
solve using quadratic formula q^2-2q-1=0
Answer:
1±√2=q
or
q=2.41, -0.41
Step-by-step explanation:
we are given the equation q²-2q-1=0, and we want to use the quadratic equation, which is (-b±√(b²-4ac))/2a
a is 1 (there is a 1 in front of q²)
b is -2
c is -1
substitute into the equation:
q=(2±√(4-4*1*-1))/2
solve for the discriminant:
√(4-4(1*-1))
√8
now the equation:
(2±√8)/2=q
simplify:
1±√2=q
or if your application asks for a decimal:
√2≈1.41
so:
1+1.41=2.41=q
or
1-1.41=-0.41=q
Hope this helps!