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Anit [1.1K]
3 years ago
12

Plz help !!!!!!!!!!!

Mathematics
2 answers:
NNADVOKAT [17]3 years ago
4 0

Answer:

the first one

Step-by-step explanation:

the 36 and the 5 stay in the same places

the x with the negative 4 exponent moves to the bottom and combines with x squared so that makes x to the 6th power on the bottom.

The y  and x move to the top to combine with what is there.

vodomira [7]3 years ago
3 0

Answer:   \bold{a)\quad \dfrac{36y^5z^2}{5x^6}}

<u>Step-by-step explanation:</u>

\dfrac{36x^{-4}y^2z^0}{5x^2y^{-3}z^{-2}}\\\\\\=\dfrac{36x^{-4+4}y^{2+3}z^{0+2}}{5x^{2+4}y^{-3+3}z^{-2+2}}\\\\\\=\dfrac{36x^{0}y^5z^2}{5x^6y^{0}z^{0}}\\\\\\=\dfrac{36y^5z^2}{5x^6}

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Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children. Let BBG ind
Gemiola [76]

Answer:

(a)

S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}

(b)

i.

1\ girl = \{GBB, BBG, BGB\}

P(1\ girl) = 0.375

ii.

Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}

P(Atleast\ 2 \ girls) = 0.5

iii.

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Step-by-step explanation:

Given

Children = 3

B = Boys

G = Girls

Solving (a): List all possible elements using set-roster notation.

The possible elements are:

S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}

And the number of elements are:

n(S) = 8

Solving (bi) Exactly 1 girl

From the list of possible elements, we have:

1\ girl = \{GBB, BBG, BGB\}

And the number of the list is;

n(1\ girl) = 3

The probability is calculated as;

P(1\ girl) = \frac{n(1\ girl)}{n(S)}

P(1\ girl) = \frac{3}{8}

P(1\ girl) = 0.375

Solving (bi) At least 2 are girls

From the list of possible elements, we have:

Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}

And the number of the list is;

n(Atleast\ 2 \ girls) = 4

The probability is calculated as;

P(Atleast\ 2 \ girls) = \frac{n(Atleast\ 2 \ girls)}{n(S)}

P(Atleast\ 2 \ girls) = \frac{4}{8}

P(Atleast\ 2 \ girls) = 0.5

Solving (biii) No girl

From the list of possible elements, we have:

No\ girl = \{BBB\}

And the number of the list is;

n(No\ girl) = 1

The probability is calculated as;

P(No\ girl) = \frac{n(No\ girl)}{n(S)}

P(No\ girl) = \frac{1}{8}

P(No\ girl) = 0.125

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