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hoa [83]
3 years ago
9

I need this ASAP!!!!

Mathematics
1 answer:
lubasha [3.4K]3 years ago
8 0

Answer:

23/2 is already in the simplest form. It can be written as 11.5 in decimal form

Step-by-step explanation:

23 2 =    22 2 +    1 2 = 11 +    1 2 =  11  1 2

This fraction cannot be simplified any further because there is no common divisor for 1 and 2 other than 1

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H + 1/3 when h= 1 2/3
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2. Because if 1/3 + 2/3 is 1 then 1+1=2
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The letters of the alphabet are written on cards and places in a brown paper bag. What is the probability or drawing a vowel, re
ehidna [41]

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0.1%

Step-by-step explanation:

:D

3 0
3 years ago
The fine schedule for overdue books at the county library is modeled by the values in the table.
Tasya [4]

Answer:

The answer would be B!

Step-by-step explanation:

In the chart if you divide the number of days over due to the cost of the fine, you get 5/2 every time. therefore, if y is the cost in cents and x is the days overdue, you would multiply x5/2 or y=5/2x, so your answer is B (y=5/2x).  

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3 0
3 years ago
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A group of 32 players forms 4 volleyball teams if there are 96 players how many teams can be formed
kotegsom [21]

Answer:

12 teams

Step-by-step explanation:

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<h2><u><em>hope this helped : )</em></u></h2>
6 0
3 years ago
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Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%. Find
Papessa [141]

Answer:

1) \text{P(at least one boy and one girl)}=\frac{3}{4}

2) \text{P(at least one boy and one girl)}=\frac{3}{8}

3) \text{P(at least two girls)}=\frac{1}{2}

Step-by-step explanation:

Given : Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%.

To  Find : The probability of each event.  

1) P(at least one boy and one girl)

2) P(two boys and one girl)

3) P(at least two girls)        

Solution :

Let's represent a boy with B and a girl with G

Mr. and Mrs. Romero are expecting triplets.

The possibility of having triplet is

BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG

Total outcome = 8

\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}

1) P(at least one boy and one girl)

Favorable outcome =  BBG, BGB, BGG, GBB, GBG, GGB=6

\text{P(at least one boy and one girl)}=\frac{6}{8}

\text{P(at least one boy and one girl)}=\frac{3}{4}

2) P(at least one boy and one girl)

Favorable outcome =  BBG, BGB, GBB=3

\text{P(at least one boy and one girl)}=\frac{3}{8}

3) P(at least two girls)

Favorable outcome = BGG, GBG, GGB, GGG=4

\text{P(at least two girls)}=\frac{4}{8}

\text{P(at least two girls)}=\frac{1}{2}

4 0
3 years ago
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