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Anastasy [175]
3 years ago
7

How do you find the product of -9over5 times 5over 3? please help.

Mathematics
1 answer:
Inessa [10]3 years ago
3 0
Here look at this for my answer...

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Can someone help me, I have no clue
enyata [817]

yes. which questions do you need me to answer?


6 0
3 years ago
SCREENSHOT OF MY QUESTION:
drek231 [11]

Answer in fraction form = 9/8

Answer in decimal form = 1.125

note: the improper fraction 9/8 converts to the mixed number 1 & 1/8

To get this answer, we divide both sides by 8 to isolate n. The expression 8n really means "8 times n". We divide to undo the multiplication.

3 0
3 years ago
Ok guys ik that this isn't a part of academic but i wanna say a story ... since i had 15 subs ive always wanted 100 .. like so b
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I'll sub my guy. Don't give up, keep climbing.

5 0
3 years ago
Lara randomly surveyed 15 students
Rina8888 [55]

Answer:

hii

Step-by-step explanation:

yes it represents school population

Hope this helps you

Thank you

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6 0
3 years ago
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.

No\ girl = \{BBB\}

P(No\ girl) = 0.125

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

7 0
3 years ago
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