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siniylev [52]
3 years ago
7

Cassandra had a pair of mice. The female gave birth to eight pups, for male and four female. In the next term, the five female m

ice each gave birth to eight pups, again four male and four female. If, in the next term, each female does the same and no mice die, how many mice, male and female, will Cassandra now have?
Mathematics
1 answer:
geniusboy [140]3 years ago
4 0
The first pair
Male - 1
Female - 1

First generation by 1 female
Male - 4
Female - 4

Second generation by 5 females
Male - 4 x 5 = 20
Female - 4 × 5 = 20

Third generation by 25 females
Male - 4 × 25 = 100
Female - 4 × 25 = 100

Total males = 1 + 4 + 20 + 100 = 125
Total females = 1 + 4 + 20 + 100 = 125
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Marginal distribution: This distribution gives the probability for each possible value of the Random variable ignoring other random variables. Basically, the values of other variables is not considered in the marginal distribution, they can be any value possible. For example, if you have two variables X and Y, the probability of X being equal to a value, lets say, 4, contemplates every possible scenario where X is equal to 4, independently of the value Y has taken. If you want the probability of a dice being a multiple of 3, you are interested that the dice is either 3 or 6, but you dont care if the dice is even or odd.

Conditional distribution: This distribution contrasts from the previous one in the sense that we are restricting the universe of events to specific condition for other variable, making a modification of our marginal results. If we know that throwing a dice will give us a result higher than 2, then to in order to calculate the probability of the dice being a multiple of 3 using that condition, we have two favourable cases (3 and 6) from 4 total possible results (3,4,5 and 6) discarding the impossible values (1 and 2) from this universe since they dont match the condition given (note that the restrictions given can also reduce the total of favourable cases).

The joint distribution calculates the probabilities for two different events (related to two different random variables) occuring simultaneously. If we want to calculate the joint probability of a dice being multiple of 3 and greater than 2 at the same time, our possible cases in this case are 3 and 6 from 6 possible results. We are not discarding 1 or 2 as possible results because we are not assuming, that the dice is greater than 2, that is another condition that we should met in the combination of events.

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Simplify each expression:<br> ( 5a^2b^3)(4ab^5)
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Answer:

(22•5a3b4)

Step-by-step explanation:

STEP

1

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Equation at the end of step 1

4ab • (5a2 • b3)

STEP

2

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Multiplying exponential expressions :

2.1 a1 multiplied by a2 = a(1 + 2) = a3

Multiplying exponential expressions :

2.2 b1 multiplied by b3 = b(1 + 3) = b4

Final result :

(22•5a3b4)

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