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baherus [9]
3 years ago
9

Happy halloween every one sorry moderators today i want to pass my spirit today

Mathematics
2 answers:
Kamila [148]3 years ago
7 0

Answer:

Happy Halloween!!!

Step-by-step explanation:

Hope I made your day better buddy.

Brainliest??

oksano4ka [1.4K]3 years ago
7 0

Answer:

Happy Halloween Dude have a Good one

Step-by-step explanation:

-Dracula

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4 years ago
What does 1/9-(-2/5-8/9) equal?
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1/9 - (-2/5 - 8/9) = 1/9 + 2/5 + 8/9 = 1 + 2/5 = 7/5 or 1 2/5
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A whole pie is cut into 8 equal slices. Three pieces are served. What fraction of the pie is left?
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1 pie is cut into eighths or 1/8 a slice. 3 are taken out.
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4 0
3 years ago
Abby, Deborah, Mei-Ling, Sam, and Roberto work in a firm's public relations office. Their employer must choose two of them to at
sattari [20]

Answer:

a) Ω = { {Abby, Deborath}, {Abby, Mei-Ling}, {Abby, Sam}, {Abby,Roberto}, {Deborah, Mei-Ling}, {Deborah, Sam}, {Deborah, Roberto}, {Mei-Ling, Sam}, {Mei-Ling, Roberto}, {Sam, Roberto} }

b) 0.1

c) 0.4

d) 0.3

Step-by-step explanation:

a) The sample space must contain every possible combination of two names. Since we dont care about the <em>order</em> of the chosen names, we can describe every element of the sample space as a <em>subset</em> of 2 elements of the set {Abby, Deborah, Mei-Ling, Sam, Roberto}. That subset will represent the names of the chosen persons. With this in mind, we conclude that the sample space is

Ω = { {Abby, Deborath}, {Abby, Mei-Ling}, {Abby, Sam}, {Abby,Roberto}, {Deborah, Mei-Ling}, {Deborah, Sam}, {Deborah, Roberto}, {Mei-Ling, Sam}, {Mei-Ling, Roberto}, {Sam, Roberto} }

b) The cardinality of the sample space Ω is 10. Since all choices are <em>equally likely</em>, any choice will have probability \frac{1}{10} = 0.1 , because the 10 of them combined must add up 1.

c) We need to find all possible choices that includes Mei-Ling, those will be our <em>favourable cases. </em>The amount of favourable cases must be divided to the total amount of cases (the cardinality of Ω) in order to obtain the probability of Mei-Ling being chosen. Mei-Ling is included on 4 choices (one for each of her partners), this means that she has a probability of \frac{4}{10} = 0.4  to being chosen.

d) We have <em>3 favourable cases </em>,the choices {Abby, Deborah}, {Abby, Mei-Ling} and {Deborah, Mei-Ling}, which neither of them contain a man. By dividing that number to the total number of cases, we obtain a probability of \frac{3}{10} = 0.3  that neither of the two men are chosen

7 0
4 years ago
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