plzzzzz help me I can fail algebra. on Sunday, Sean ran 5/7 of the distance he ran on Thursday (9/10). fill in the number of miles he ran on Sunday

Answers

Answer 1
Answer:

Answer:

9/14

Step-by-step explanation:

You multiply the two fractions. So 5/7*9/10, you just multiply across. This would give you 45/70. Then, after simplifying, this fraction is 9/14. So, Sean ran 9/14 of a mile on Sunday. Hope this helps, Sorry if im wrong!!


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What is the ninth term in the binomial expansion of (x – 2y)13?329,472x5y8
–329,472x5y8
–41,184x8y5
41,184x8y5

Answers

Answer:

It's A

Step-by-step explanation:

On Edg

Complete the number sentence to solve. Three students share 5 feet of ribbon equally. How many feet of ribbon does each student get? 5 + 3 = Each student's share is 1 feet of ribbon.

Answers

Answer:

The answer is that each student gets 1.666667 feet of ribbon

Step-by-step explanation:

Help me please!!!!!!!!!!!

Answers

b. it is equivalent to 2/100

Answer:

B. It is equivalent to 20/100

C. It is equivalent to the fraction represented by this picture

Step-by-step explanation:

20/100 simplified is 2/10

The graph also contains the right amount of boxes to represent 2/10 or 20/100

Hope this helps!

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.

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Question 2A square room of side length x is made 5 m longer (its width stays the same). What is the new area
of the room?​

Answers

Answer:

To find the width, multiply the length that you have been given by 2, and subtract the result from the perimeter. You now have the total length for the remaining 2 sides. This number divided by 2 is the width.

Step-by-step explanation:

The area of any quadrilateral can be determined by multiplying the length of its base by its height. Since we know the shape here is square, we know that all sides are of equal length. From this we can work backwards by taking the square root of the area to find the length of one side. hope this helps you :)

What is the greatest common factor of 4xy2 and 20x2y4?4xy
4xy2
24xy
24xy2

Answers

The greatest common factor of 4xy² and 20x2y^4 is: 4xy².

What is the Greatest Common Factor?

Greatest common factor of a set of terms is the factor that all terms in the set share together.

Given:

4xy² and 20x2y^4

4xy² can divide both terms, meaning both terms share a factor.

Therefore, the greatest common factor of 4xy² and 20x2y^4 is: 4xy².

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