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3 years ago

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

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0.1%

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3 years ago

The fine schedule for overdue books at the county library is modeled by the values in the table.

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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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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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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

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

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

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