Answer:
he binomial distribution, the probability is the same for every trial. For the hypergeometric distribution, each trial changes the probability for each subsequent trial because there is no replacement.
Explanation:
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The Cornell System for notetaking
a was developed by Dr. Walter Pauk at Cornell University.
b involves dividing your paper into three sections: Notes in the right-hand ccolumn; labels or possible questions for recall on the left-hand column; and a summary at the bottom.
c is easily done on a computer.
d provides a handy way to review and test yourself using your notes.
e All of the above.
The Cornell System for notetaking was developed by Dr. Walter Pauk at Cornell University.
What is Cornell method in note taking?It involves dividing your paper into three sections: Notes in the right-hand column; labels or possible questions for recall on the left-hand column; and a summary at the bottomThe Cornell Note Taking method is known to be a type of short cut that hinders the use of long sentences.
Conclusively, Here, one have to use short notes that are often write down by a person using the right-hand column by applying the use of recognizable abbreviations and symbols.
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What the best way to prepare for essay portion ?
How Do You Begin an Essay?
Decide on your essay type and topic. Sometimes, you will already have an assigned essay type or topic, so that will save you one step. ...
Brainstorm and research the topic you’ve chosen. Once you’ve chosen your topic, brainstorm all the different supporting ideas that you can talk about for the topic.
Develop your thesis statement. After you have brainstormed and researched, write down your thesis statement. ...
Write your outline. Once you have your thesis statement, you can start to prepare your outline. Many people skip the outline process, thinking it’s a waste of time.
Start writing. Using your outline, you can now begin writing your essay. Some writers choose to write their paragraphs in order, beginning with the hook.
Which example best describes the term carrying capacity?
O A. The amount of food and water available for prairie dogs in a given
area
O B. The maximum number of young prairie dogs that a population can
care for over time
O C. The maximum prairie dog population size that an area can sustain
over time
D. The amount of space available for prairie dogs in a given area
Answer:
C. The maximum prairie dog population size that an area can sustain over time
Explanation:
Just got it right on my quiz :)
A landscaping company currently has an inventory of 8 fruit trees, 13 pine trees, and 14 maple trees. It plans to give 4 trees away at next
Saturday's lawn and garden show in the city park. The 4 winners can select which type of tree they want. Assume they select randomly.
a. What is the probability that all 4 winners will select the same type of tree?
b. What is the probability that 3 pine trees will be selected and the other tree will be a maple?
c. What is the probability that no fruit trees and 2 of each of the others will be selected?
a. The probability is
(Round to four decimal places as needed.)
The distribution of the trees in the landscape company is an illustration of probability
The probability that all 4 winners will select the same type of tree is 0.0341The probability that 3 pine trees will be selected and the other tree will be a maple is 0.0710The probability that no fruit trees and 2 of each of the others will be selected is 0.0904a. The probability that all 4 winners will select the same type of tree?The distribution of the trees are:
Fruit trees = 8Pine trees = 13Maple trees = 14Total = 8 + 13 + 14
Total = 35
The probability is then calculated as:
P(Same type) = P(All fruit trees) + P(All pine trees) + P(All maple trees)
So, we have:
P(Same type) = (8/35 * 7/34 * 6/33 * 5/32) + (13/35 * 12/34 * 11/33 * 10/32) + (14/35 * 13/34 * 12/33 * 11/32)
Evaluate
P(Same type) = 0.0341
Hence, the probability that all 4 winners will select the same type of tree is 0.0341
b. The probability that 3 pine trees will be selected and the other tree will be a maple?The probability is then calculated as:
P(3 pine and one maple) = 4 * P(Three maple and One Pine)
So, we have:
P(3 pine and one maple) = 4 * (13/35 * 12/34 * 11/33 * 13/32)
Evaluate
P(3 pine and one maple) = 0.0710
Hence, the probability that 3 pine trees will be selected and the other tree will be a maple is 0.0710
c. The probability that no fruit trees and 2 of each of the others will be selected?The probability is then calculated as:
P(2 pine and 2 maple) = 4 * P(Two pine and Two maple)
So, we have:
P(2 pine and 2 maple) = 4 * (13/35 * 12/34 * 14/33 * 13/32)
Evaluate
P(2 pine and 2 maple) = 0.0904
Hence, the probability that no fruit trees and 2 of each of the others will be selected is 0.0904
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suppose a firms operating under cournots duoploy model sell a product to the market total market share is determined as 10,000 units what is the amount in units of the product that each firm can supply to the market
Answer: 10,000
Explanation:
A comparison between species: Biologists wish to compare the prevalence of a genetic mutation across
two species of Neema Toads. In a sample of 9 Neema Toads of Species I, 0.6% had the mutation while in a
sample of 12 Neema Toads of Species II, 1% showed the mutation. Is there significant evidence to suggest a
difference in the prevalence of the mutation across the two species? Conduct a hypothesis test at a
significance level of a = 0.01.
The result from the hypotheis test below shows that there is no significant evidence existing to show difference in the mutation.
How to solve for the z statisticsThe hypothesis for this test has been stated below:
H0: pI - pII = 0
Ha: pI - pII ≠ 0
n1 = 9
p1 = 0.6%
n2 = 12
p2 = 1%
Using the z statistics formula
[tex]\frac{0.006-0.01}{\sqrt{0.000663+0.000825} } \\\\\frac{-0.004}{\sqrt{0.001488} }[/tex]
z = -0.103
p value = z(-0.103) = 0.4589
The pvalue is greater that 0.01
The decision is to fail to reject the null hypothesis.
There is no significant evidence to show that there is a difference in the prevalence of mutation.
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