Answer:
is there a answer choice?
Answer:
Step-by-step explanation:
the sum of the interior angles of a hexagon is 720 degrees. All sides are the same length (congruent) and all interior angles are the same size (congruent). To find the measure of the interior angles, we know that the sum of all the angles is 720 degrees.
Now,
102+146+158+120+124+x=720
or,650+x=720
or,x=70
Therefore The value of X is 70 degree.
Answer:
x = 1.7
Step-by-step explanation:
Answer:
The angles are adjacent and x=100
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!!!
Answer:
There is a 32.22% probability that at least 3 flights arrive late.
Step-by-step explanation:
For each flight, there are only two possible outcomes. Either it arrives on time, or it arrives late. This means that we can solve this problem using binomial probability concepts.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinatios of x objects from a set of n elements, given by the following formula.
And is the probability of X happening.
In this problem, we have that:
There are 10 flights, so .
A success in this case is a flight being late. 80% of its flights arriving on time, so 100%-80% = 20% arrive late. This means that .
(a) Find the probability that at least 3 flights arrive late.
Either less than 3 flights arrive late, or at least 3 arrive late. The sum of these probabilities is decimal 1. This means that:
In which
So
Finally
There is a 32.22% probability that at least 3 flights arrive late.
The problem is solved by calculating the probability of the complementary event (0,1,2 flights arriving late) using the binomial distribution, then subtracting this from 1 to find the probability of at least 3 flights arriving late.
This problem is typically solved by using a binomial probability formula, which is used when there are exactly two mutually exclusive outcomes of a trial, often referred to as 'success' and 'failure'.
Here, our 'success' is a flight arriving late. The probability of success, denoted as p, is thus 20% or 0.2 (since 80% arrive on time, then 100%-80% = 20% arrive late). The number of trials, denoted as n, is 10 (the number of randomly selected flights).
We want to find the probability that at least 3 flights arrive late, in other words, 3,4,...,10 flights arrive late. The problem can be solved easier by considering the complementary event: 0,1,2 flights arrive late. Then subtract the sum of these probabilities from 1.
The binomial probability of exactly k successes in n trials is given by:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
Where C(n, k) is the binomial coefficient, meaning choosing k successes from n trials.
We calculate like so:
P(X=0) = C(10, 0) * (0.2)^0 * (0.8)^10
P(X=1) = C(10, 1) * (0.2)^1 * (0.8)^9
P(X=2) = C(10, 2) * (0.2)^2 * (0.8)^8
Sum these up and subtract from 1 to get the probability that at least 3 flights arrive late. This gives the solution to the question.
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