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Georgia [21]
4 years ago
10

An air mass that originates in the Pacific ocean, west of Brazil, is most likely

Biology
1 answer:
kykrilka [37]4 years ago
7 0

An air mass that originates in the Pacific ocean, west of Brazil, is most likely WARM AND WET.

A. warm and wet

<u>Explanation:</u>

The pacific ocean has the tropical maritime air mass so it is warm and wet. they affect the United States originate over the Caribbean sea, Mexico, pacific and the western Brazil.

So the air masses in the pacific ocean, western Brazil are wet and warm.the air masses have the fairly uniform temperature and the moisture content in the horizontal direction.

Air masses are characterized by their temperature and the humidity properties. The air mass where the originate is based on the surface properties.

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Answer:

C

Explanation:

All forms of matter (even the acids and bases) are formed from the chemical elements listed on the periodic table.

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What are some physical features of kelp forests
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A neuroscientist cuts a brain in half, along the division between the hemispheres. this cut is referred to as a _____________ cu
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Answer;

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A neuroscientist cuts a brain in half, along the division between the hemispheres. this cut is referred to as a midsagittal cut.

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4 0
3 years ago
In humans there is a dominant allele (A) for the absence of moles; while the recessive allele (a) results in the presence of mol
Art [367]

Answer:

a. (p + q)^7 = p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7

b. i. 0.75

   ii. 0.000061

   iii. 0.012

   iv. 0.17

c. 0.67

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a. The expansion of the binomial (p + q)7 would be such that:

(p + q)^7 = p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7

b. Both couples are heterozygous:

             Aa    x    Aa

         AA   Aa   Aa   aa

Since A is dominant over a,

probability of having mole (aa) = 1/4

probability of not having moles = 3/4

<em>Therefore, the probability of the first child not having moles </em>= 3/4 or 0.75

ii. Let the probability of not having mole = p and the probability of having mole = q. From the binomial expansion:

(p + q)^7 = p^7q^0+7p^6q+21p^5q^2+35p^4q^3+35p^3q^4+21p^2q^5+7pq^6+p^0q^7

<em>Probability that all of the children will have moles </em>= p^0q^7

since p = 3/4 and q = 1/4

p^0q^7 = (3/4)^0(1/4)^7 = 0.000061

iii. <em>Probability that the first two children will have no moles and the last five will have moles</em> = 21p^2q5

                       = 21(3/4)^2(1/4)^5

                         = 0.012

iv. <em>Probability that 4 will have no moles and 3 will have moles out of the 7 children</em> = 35p^4q^3

               = 35(3/4)^4(1/4)^3

                      = 0.17

c. <em>Probability that the child born without moles is a carrier of the a-allele  = probability of heterozygou</em>s.

From the cross in (b), the genotypes of those born without moles are AA and 2Aa. Therefore, the probability of not having moles and be Aa is:

      = 2/3 or 0.67

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