In a random sample of 8 people, the mean commute time to work was 33.5 minutes and the standard deviation was 7.3 minutes. A 95% confidence interval using the t-distribution was calculated to be (27.4,39.6). After researching commute times to work, it was found that the population standard deviation is 9.6 minutes. Find the margin of error and construct a 95% confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare the results. .... The margin of error of u is: (Round to two decimal places as needed.) A 95% confidence interval using the standard normal distribution is (A D (Round to one decimal place as needed.) Compare the results. Choose the correct answer below. O A. The confidence interval found using the standard normal distribution is wider than the confidence interval found using the student's t-distribution. O B. The confidence interval found using the standard normal distribution has smaller lower and upper confidence interval limits. O C. The confidence interval found using the standard normal distribution is the same as the confidence interval found using the student's t-distribution. O D. The confidence interval found using the standard normal distribution is narrower than the confidence interval found using the student's t-distribution.

Q: (-14) + (–7)

A:

Q: V(3278)3 + 6230 – 12e90i

A:

Q: #1 only

A: Amarjeet 2mr find x.

Q: (1-2=) dr

A:

Q: a3-16a

A: a³−16a ⇒a(a²−16) ⇒a(a² −4²) ⇒a(a+4)(a−4)

Q: 10 14 9.) Find O 18

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Q: 1.25a +3> 0.5a-6 and 2.5a 129-1.5a

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Q: A-1 does not exist.

A:

Q: こ (0) of

A:

Q: 9 – 4x > –3

A: The given inequality is 9-4x>-3

Q: х3 — 2х dx х — 1 -3 10. 2

A: Given: I=∫23x3-2xx-1dx For evaluating given integral, we substitute x-1=t…

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A: The expression is given as 1+–3m+–9m+4+5+–10m. Simplifying the given expression. 1+–3m+–9m+4+5+–10m…

Q: 18) 2x y

A:

Q: (4 – a2)

A: To integrate.

Q: Evaluate. (3- 5)² + 4(1 – 5)

A: Given, (3-5)2 + 4(1-5)

Q: 0 1 2 4 0 2 4 0 0 3 9 12

A:

Q: 2(3+d-1)

A: Solve the given expression  given expression is 2(3+d-1)

Q: 5)235

A: Answer : 47

Q: Find the LU factorization of A = A = 1 1 -3 0 9 3 -17 5 2

A:

Q: 2.)(-6)-(5(1-3)-6)-4+9)

A: The given equation is as  (-6)-(5((-3)-6)-(-4)+9)

Q: 9. (x² – 2)3

A:

Q: 1 + 2n < n° for all integers n² 2

A:

Q: 1) 6dx 3/3

A: ∫33361+x2dx=6∫33311+x2dx=6tan-1x333=6tan-13-tan-133=6tan-13-tan-113=6π3-π6=2π-π=π

Q: Find A3 if A = {1, 2, a, b}

A: Given,      A=1, 2, a, b

Q: Evalvate S (4sin co) -8cos co) )do

A: Given that: ∫0π(4sinθ-8cosθ)dθ

Q: (1– 3csc(

A:

Q: S√4 4+² +3 dt

A:

Q: 75 + フチ + k「3

A:

Q: –4(–6m – 7).

A:

Q: 3. (x³ + 4x² –x – 1) + (x + 3)

A: Here we have to divide two polynomials as shown.

Q: Z4×Z6 has an element of order 24. true or false ?

A:

Q: 9. x+8

A: For two intersecting chords , the  multiples of the lengths of the line segments on each chord are…

Q: 2. (1 + 3/2x − 1)² = 3 + √3/2x − 1

A:

Q: f 3 +4 F 27/-1)dt +

A:

Q: 4 -3 4m n 16) 22 4m n

A:

Contingency Table

A contingency table can be defined as the visual representation of the relationship between two or more categorical variables that can be evaluated and registered. It is a categorical version of the scatterplot, which is used to investigate the linear relationship between two variables. A contingency table is indeed a type of frequency distribution table that displays two variables at the same time.

Binomial Distribution

Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.