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Find a diginacci sequence with no equal term to 11: quick answer please

Answers

Answer 1

Answer:

Answer:

uhh hope this helps

Step-by-step explanation:

A Diginacci sequence is created as follows.

• The ﬁrst two terms are any positive whole numbers.

• Each of the remaining terms is the sum of the digits of the previous

two terms.

For example, starting with 5 and 8 the Diginacci sequence is

5, 8, 13, 12, 7, 10,. . .

The calculations for this example are

5 + 8 = 13, 8 + 1 + 3 = 12, 1+ 3 +1+ 2 = 7, 1 + 2 + 7 = 10.

a) List the ﬁrst 26 terms of the Diginacci sequence above.

b) Find, with explanation, two starting terms for a Diginacci sequence

so that its 2021st term is 11.

c) Find, with explanation, a Diginacci sequence that has no term equal

to 11.

d) Find, with explanation, a sequence with two diﬀerent starting terms

which contains ﬁve consecutive terms that are even and not all identical

Both 2 and 3 is correct.

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Suppose that X is a random variable with probability function x 0 1 2 3 4 5 P(x) 0.00032 0.0064 0.0512 0.2048 0.4096 0.32768 and that the random variable Y = 9X2 − 36X + 4. (a) What is the range of Y ? (b) Calculate P{Y = 4}. (c) Calculate P{Y = 0}.

Answers

(a) For each value of X we have:

$X=0\quad\Rightarrow\quad Y=4\n\nX=1\quad\Rightarrow\quad Y=-23\n\nX=2\quad\Rightarrow\quad Y=-32\n\nX=3\quad\Rightarrow\quad Y=-23\n\nX=4\quad\Rightarrow\quad Y=4\n\nX=5\quad\Rightarrow\quad Y=49$

so the range of Y = {-32, -23, 4, 49}

(b)

$P\{Y=4\}=P\{X=0\}+P\{X=4\}=0.00032+0.4096=\boxed{0.40992}$

(c)

Y cannot equal 0, so

$P\{Y=0\}=0$

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%)

Hope this Helps!!!

Y varies directly with x Part 119. If y = 4 when x = -2, find x when y = 6
20. If y = 6 when x = 2, find x when y = 12
21. If y = 7 when x = 2, find x when y =3

Answers

Answer:

19. -3

20. 4

21. 6/7

Step-by-step explanation:

Since the variables are proportional, you can multiply the given x value by the new/old ratio of y-values.

19. x = 6/4(-2) = -3

20. x = (12/6)(2) = 4

21. x = (3/7)(2) = 6/7

What is the image of (0, -8) after a reflection over the line y = -x?
Submit Answer

Answers

The required reflection of the point (0, -8) on the line y = -x is (8, 0).

To determine the image of (0, -8) after a reflection over the line y = -x.

What is the equation?

the equation is the relationship between variables and represented as y = ax + b is an example of a polynomial equation.

Here,
For the image of the point of the line y = -x
The reflection of point (0, - 8) is given as by swaping the value and change the sign, So the image of the point is (8, 0).

Thus, the required reflection of the point (0, -8) on the line y = -x is (8, 0).

Learn more about equation here:

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Answer:

(8,0)

Step-by-step explanation:

could be negative 8 but it said positive 8

positive works

A limited edition poster increases in value each year. After 1 year, the poster is worth $20.70. After 2 years, it is worth $23.81. Which equation can be used to find the value,y, after x years? (Round money values to the nearest penny.)

Answers

y= 3.11x+ 17.59

I got this equation by doing 23.81-20.70 to find m, which is 3.11.
Then to find b, or y-intercept, I subtracted 3.11 from 20.70 to get the origin all price and got 17.59

A manufacturing company regularly conducts quality control checks at specified periods on the products it manufactures. Historically, the failure rate for LED light bulbs that the company manufactures is 5%. Suppose a random sample of 10 LED light bulbs is selected. What is the probability that a) None of the LED light bulbs are defective? b) Exactly one of the LED light bulbs is defective? c) Two or fewer of the LED light bulbs are defective? d) Three or more of the LED light bulbs are not defective?

Answers

Answer:

a) There is a 59.87% probability that none of the LED light bulbs are defective.

b) There is a 31.51% probability that exactly one of the light bulbs is defective.

c) There is a 98.84% probability that two or fewer of the LED light bulbs are defective.

d) There is a 100% probability that three or more of the LED light bulbs are not defective.

Step-by-step explanation:

For each light bulb, there are only two possible outcomes. Either it fails, or it does not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)$

In which $C_(n,x)$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_(n,x) = (n!)/(x!(n-x)!)$

And p is the probability of X happening.

In this problem we have that:

$n = 10, p = 0.05$

a) None of the LED light bulbs are defective?

This is P(X = 0).

$P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)$

$P(X = 0) = C_(10,0)*(0.05)^(0)*(0.95)^(10) = 0.5987$

There is a 59.87% probability that none of the LED light bulbs are defective.

b) Exactly one of the LED light bulbs is defective?

This is P(X = 1).

$P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)$

$P(X = 1) = C_(10,1)*(0.05)^(1)*(0.95)^(9) = 0.3151$

There is a 31.51% probability that exactly one of the light bulbs is defective.

c) Two or fewer of the LED light bulbs are defective?

This is

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = 2) = C_(10,2)*(0.05)^(2)*(0.95)^(8) = 0.0746$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5987 + 0.3151 + 0.0746 0.9884$

There is a 98.84% probability that two or fewer of the LED light bulbs are defective.

d) Three or more of the LED light bulbs are not defective?

Now we use p = 0.95.

Either two or fewer are not defective, or three or more are not defective. The sum of these probabilities is decimal 1.

$P(X \leq 2) + P(X \geq 3) = 1$

$P(X \geq 3) = 1 - P(X \leq 2)$

In which

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = 0) = C_(10,0)*(0.95)^(0)*(0.05)^(10)\cong 0$

$P(X = 1) = C_(10,1)*(0.95)^(1)*(0.05)^(9) \cong 0$

$P(X = 2) = C_(10,1)*(0.95)^(2)*(0.05)^(8) \cong 0$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0$

$P(X \geq 3) = 1 - P(X \leq 2) = 1$

There is a 100% probability that three or more of the LED light bulbs are not defective.

Final answer:

The question relates to binomial distribution in probability theory. The probabilities calculated include those of none, one, two or less, and three or more LED bulbs being defective out of a random sample of 10.

Explanation:

This question relates to the binomial probability distribution. A binomial distribution is applicable because there are exactly two outcomes in each trial (either the LED bulb is defective or it's not) and the probability of a success remains consistent.

a) In this scenario, 'none of the bulbs being defective' means 10 successes. The formula for probability in a binomial distribution is p(x) = C(n, x) * [p^x] * [(1-p)^(n-x)]. Plugging in the values, we find p(10) = C(10, 10) * [0.95^10] * [0.05^0] = 0.5987 or 59.87%.

b) 'Exactly one of the bulbs being defective' implies 9 successes and 1 failure. Following the same formula, we get p(9) = C(10, 9) * [0.95^9] * [0.05^1] = 0.3151 or 31.51%.

c) 'Two or less bulbs being defective' means 8, 9 or 10 successes. We add the probabilities calculated in (a) and (b) with that of 8 successes to get this probability. Therefore, p(8 or 9 or 10) = p(8) + p(9) + p(10) = 0.95.

d) 'Three or more bulbs are not defective' means anywhere from 3 to 10 successes. As the failure rate is low, it's easier to calculate the case for 0, 1 and 2 successes and subtract it from 1 to find this probability. This gives us p(>=3) = 1 - p(2) - p(1) - p(0) = 0.98.

Learn more about Binomial Probability here:

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