A traveler has 8 pieces of luggage. How many ways can he select 4 pieces of luggage for a trip

Answers

Answer 1
Answer: He can select it in 32 ways
Answer 2
Answer:

Answer:36

Step-by-step explanation:


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Throughout the US presidential election of 2012, polls gave regular updates on the sample proportion supporting each candidate and the margin of error for the estimates. This attempt to predict the outcome of an election is a common use of polls. In each case below, the proportion of voters who intend to vote for each candidate is given as well as a margin of error for the estimates. Indicate whether we can be relatively confident that candidate A would win if the election were held at the time of the poll. (Assume the candidate who gets more than of the vote wins.)

Answers

Answer:

1.) We cannot say for certain which candidate will win. But A has a statistical edge.

2.) We can say certainly that candidate A will win the election; albeit with a not so big margin.

3.) Candidate A will win this election based on the results of the final poll's before the election.

4.) We cannot say for certain which candidate will win. But A has a statistical edge.

The reasons are explained below.

Step-by-step explanation:

Confidence interval expresses a range of values in the distribution where the true proportion or mean can be found with some level of confidence.

Confidence Interval = (Sample Mean or Proportion) ± (Margin of error)

1. Candidate A: 54% & Candidate B:46% with Margin of error: + 5%

The confidence interval for candidate A

(54%) ± (5%) = (49%, 59%)

The confidence interval for candidate B

(46%) ± (5%) = (41%, 51%)

Since values greater than 50% occur in both intervals, we cannot say for certain that either of the two candidates will outrightly win the election. It just slightly favours candidate A who has A bigger range of confidence interval over 50% for the true sample proportion to exist in.

2. Candidate A: 52% & Candidate B:48% with Margin of error: + 1%

The confidence interval for candidate A

(52%) ± (1%) = (51%, 53%)

The confidence interval for candidate B

(48%) ± (1%) = (47%, 49%)

Here, it is outrightly evident that candidate A will win the elections based on the result of the final polls. The overall range of the confidence interval that contains the true sample proportion of voters that support candidate A is totally contained in a region that is above 50%. So, candidate A wins this one, easily; albeit with a close margin though.

3. Candidate A: 53% & Candidate B:47% with Margin of error: + 2%

The confidence interval for candidate A

(53%) ± (2%) = (51%, 55%)

The confidence interval for candidate B

(47%) ± (2%) = (45%, 49%)

Here too, it is outrightly evident that candidate A will win the elections based on the result of the final polls. The overall range of the confidence interval that contains the true sample proportion of voters that support candidate A is totally contained in a region that is above 50%. Hence, statistics predicts that candidate A wins this one.

4. Candidate A: 58% & Candidate B:42% with Margin of error: + 10%

The confidence interval for candidate A

(58%) ± (10%) = (48%, 68%)

The confidence interval for candidate B

(42%) ± (10%) = (32%, 52%)

Since values greater than 50% occur in both intervals, we cannot say for certain that either of the two candidates will outrightly win the election. It just slightly favours candidate A who has A bigger range of confidence interval over 50% for the true sample proportion to exist in.

Hope this Helps!!!

Find the probability for choosing a letter at random from the word: Probability P(not P) A.) 1/11 B.) 2/11 C.) 5/11 D.) 10/11​

Answers

Answer:

D.) 10/11

Step-by-step explanation:

Here the given word Probability has 11 letters in it. And we have to calculate the Probability of not selecting a letter P from the above word.

So the formula for calculating any probability is Total Favorable outcomes / Total number of outcomes.

Here total number of outcomes are 11 as word Probability has 11 letters.

So the probability of selecting letter P from word Probability =  (1)/(11)

Now the P(not P) = 1 - P(selecting letter P)

                            = 1 - (1)/(11) = (10)/(11)

A sample survey contacted an SRS of 2854 registered voters shortly before the 2012 presidential election and asked respondents whom they planned to vote for. Election results show that 51% of registered voters voted for Barack Obama. We will see later that in this situation the proportion of the sample who planned to vote for Barack Obama (call this proportion V) has approximately the Normal distribution with mean μ 0.52 and standard deviation σ = 0.009. (a) If the respondents answer truthfully, what is P (0.5くV < 0.54)? This is the probability that the sample proportion v estimates the population proportion 0.52 within plus or minus 0.02.
P (0.5<= V <=0.54) (±0.0001)=

(b) In fact, 50% of the respondents said they planned to vote for Barack Obama V = 0.5. If respondents answer truthfully, What is P(V <=0.5)?
P (V <=0.5) (±0.0001) =

Answers

Answer:

a) 97.37%

b) 1.31%

Step-by-step explanation:

a)  

Here we want to calculate the area under the Normal curve with mean 0.52 and standard deviation 0.009 between 0.5 and 0.54

This can be easily done with a spreadsheet and we get

P (0.5くV < 0.54) = 0.9737 or 97.37%

(See picture 1)

b)

Here we want the area under the Normal curve with mean 0.52 and standard deviation 0.009 to the left of 0.5.

P(V ≤ 0.5) = 0.0131 or 1.31%

(See picture 2)

Find the slope of the line that passes through the points (2,12)and(-2,0)

Answers

Answer:

Step-by-step explanation:

(0 - 12)/(-2 - 2)= -12/-4= 3

y - 0 = 3(x + 2)

y = 3x + 6

Please solve this equation​

Answers

Answer:

x = 115° , y = 140° , z = 40°

Step-by-step explanation:

40° , x and 25° lie on a straight line and sum to 180° , that is

x + 40° + 25° = 180°

x + 65° = 180° ( subtract 65° from both sides )

x = 115°

z and 40° are vertically opposite angles and are congruent , then

z = 40°

y and z lie on a straight line and sum to 180° , that is

y + 40° = 180° ( subtract 40° from both sides )

y = 140°

Problem 9-10 The elongation of a steel bar under a particular tensile load may be assumed to be normally distributed, with a mean of .06 in. and standard deviation of .008 in. A sample of n=100 bars is subjected to the test. Find the probability that the sample mean elongation is between .0585 in. and .0605 in.

Answers

Answer:

The answer is 0.7036.

Step-by-step explanation:

Check the attached file for the computations.

The probability that the mean life of a random sample mean elongation is between .0585 in. and .0605 in. is 70.57%

What is z score?

Z score is used to determine by how many standard deviations the raw score is above or below the mean.

It is given by:

z = (raw score - mean) / (standard deviation÷√sample)

Mean = 0.06, standard deviation = 0.008, sample = 100.

For x = 0.0585:

z = (0.0585 - 0.06)/ (0.008 ÷√100) = -1.88

For x = 7.2:

z = (0.0605 - 0.06)/ (0.008 ÷√100) = 0.623

P(-1.88 < z < 0.63) = P(z < 0.63) - P(z < -1.88) = 0.7357 - 0.03 = 0.7057

The probability that the mean life of a random sample mean elongation is between .0585 in. and .0605 in. is 70.57%

Find out more on z score at: brainly.com/question/25638875