ou write a short story, but you want to make sure your work is protected before you post it online. What should you do to help protect your copyright? (Site 1)

Answers

Answer 1

Answer: You can add the Copyright symbol. And write a notice below your work saying that

Related Questions

1260=(n-2)x180find the value of ncan you please give me a explaination
Find the missing segment in the image below
A study to determine the sensitivity and specificity of a new test for macular degeneration is conducted on 2430 people. Macular degeneration occurs at a rate of 16.72%. Your sample has the same prevalence of macular degeneration. You find that 377 people with macular degeneration tested positive with the new test. You also have a total of 561 positive test results in your study. CALCULATE THE SPECIFICITY of this test.Question 2 options:A)83.29%B)98.45%C)92.86%D)67.20%E)90.91%
Rita earns $18.00 per hr. If she gets a 6% raise, what will be her new hourly wage?
The graph of a function f is illustrated below. What is the graph of the inverse function of f?

What are the cordinates of the image of the point (11,1)after a translation 10 units down followed by a reflection over the y-axis

Answers

Answer:

(-11,-9)

Step-by-step explanation:

translate 10 units down: y value: 1-10=-9

reflect over y-axis: x-value becomes opposite: 11 becomes -11

HELP!!!!!In a group of 15 children, 5 said they are an only child, 3 said they have 1 sibling only, 4 said they have exactly 2 siblings, and 3 said they have exactly 3 siblings.

What is the average number of siblings for this group? Enter your answer as a decimal, rounded to the nearest tenth,
_______

Answers

The average number of siblings for the group is 1.3.

We must compute the total number of siblings and divide it by the total number of children in the group to determine the average number of siblings for the group.

There are a total of: siblings in the bunch.

3 + 8 + 9 = 20 (1 sibling times 3 children) + (2 siblings times 4 children) + (3 siblings times 3 children)

The total number of kids in the group is: 5 + 3 + 4 + 3 = 15

As a result, the group as a whole has an average of:

20/15 = 1.33 (rounded to the nearest tenth) (rounded to the nearest tenth)

So, the group as a whole has 1.3 siblings on average.

To know more about Average visit:

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Evaluate 8 − 12 ÷ 2 + 3. 1 −1 4 5

Answers

According to the order of operations, multiplication and division are performed before addition and subtraction.

... 8 - 12 ÷ 2 + 3

... = 8 - 6 + 3 . . . . . . 12÷2 = 6

... = 2 + 3 . . . . . . . . . 8-6 = 2

... = 5

Answer:-13

Step-by-step explanation:

Determine from the balanced chemical reaction ifthe derived mole ratio is correct or incorrect.

Answers

It's incorrect. There are 6 moles of NaBr produced for every 3 moles of Na₂S among the reactants (as well as 2 moles of FeBr₃), so the mole ratio of NaBr to Na₂S should be 2 : 1, not 1 : 1.

The white "Spirit" black bear (or Kermode) Ursus americanus kermodei, differs from the ordinary black bear by a single amino acid change in the melanocortin 1 receptor gene (MC1R). In this population, the gene has two forms (or alleles): the "white" allele b and the "black" allele B. The trait is recessive: white bears have two copies of the white allele of this gene (bb), whereas a bear is black if it has one or two copies of the black allele (Bb or BB). Both color morphs and all three genotypes are found together in the bear population of the northwest coast of British Columbia. If possessing the white allele has no effect on growth, survival, reproductive success, or mating patterns of individual bears, then the frequency of individuals with 0, 1, or 2 copies of the white allele (b) in the population will follow a binomial distribution. To investigate, Hedrick and Ritland (2011) sampled and genotype 87 bears from the northwest coast:42 were BB
24 were Bb
21 were bb
Assume that this is a random sample.
A formal hypothesis test was carried out to compare the observed and expected frequencies of genotypes. The procedure obtained P = 0.0001.
1. "The frequency distribution of genotypes has a binomial distribution in the population" is the________ hypothesis, whereas "The frequency distribution of genotypes does not have a binomial distribution" is the _________ hypothesis.
2. The degrees of freedom for the test statistic are __________.
Say whether the each of the following statements is true or false solely on the basis of these results:
3. The difference between the observed and expected frequencies is statistically significant.________
4. The test statistic exceeds the critical value corresponding to α = 0.05. ___________
5. The test statistic exceeds the critical value corresponding to α = 0.01. ____________

Answers

Answer:

1) i ) Null hypothesis, ii) Alternate hypothesis

2) Degree of Freedom = 14.9011

3) True

4) True

5) True

Step-by-step explanation:

1) "The frequency distribution of genotypes has a binomial distribution in the population" is the NULL hypothesis, whereas "The frequency distribution of genotypes does not have a binomial distribution" is the ALTERNATE hypothesis.

2) Degrees of freedom: see attached calculation for

3) True, the difference between the observed and expected frequencies is statistically significant since p-value < ∝alpha(0.05)

4) True

when, 14.9011 > 3.8415 where ∝ = 0.05

when, p-value < alpha,TS > critical value

here p-value < 0.0001 < 0.05

5) True

when, 14.9011 > 6.635 where ∝ = 0.01

when, p-value < alpha,TS > critical value

here p-value < 0.0001 < 0.01

Final answer:

The null hypothesis states that the genotype frequency distribution follows a binomial distribution, with the alternative hypothesis stating the contrary. The degrees of freedom for the test statistic are two. The P value of 0.0001 is statistically significant, thus the test statistic exceeds the critical values corresponding to both α = 0.05 and α = 0.01.

Explanation:

1. The frequency distribution of genotypes has a binomial distribution in the population is the null hypothesis, whereas The frequency distribution of genotypes does not have a binomial distribution is the alternative hypothesis.

2. For this chi-square test, since there are 3 possible outcomes (BB, Bb, bb) the degrees of freedom are 3-1 = 2.

3. The difference between the observed and expected frequencies is statistically significant, that is true. A P value of 0.0001 is highly significant, clearly less than 0.05, and denotes a significant difference.

4. The test statistic exceeds the critical value corresponding to α = 0.05. This is true, as the statistically significant low P value indicates the test statistic is in the critical region.

5. The test statistic exceeds the critical value corresponding to α = 0.01. This is also true.