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nekit [7.7K]
3 years ago
9

Sixty percent of the students at a certain school wear neither a ring nor a necklace. Twenty percent wear a ring and 30 percent

wear a necklace. If one of the students is chosen randomly, what is the probability that this student is wearing (a) a ring or a necklace?

Mathematics
1 answer:
Reptile [31]3 years ago
7 0

Answer:

0.4

Step-by-step explanation:

Given

60 % wear neither ring nor a necklace

20 % wear a ring

30 % wear necklace

This question can be Solved by using Venn diagram

If one person is choosen randomly  among the given student the probability that this student is wearing a ring or necklace is

P\left ( wear \ ring\ or \ necklace )+P\left ( neither\ ring\ or\ necklace)=1

P\left ( wear \ ring\ or\ necklace )=1-0.6=0.4

The sum of probabilty is equal to 1 because it completes the set

Therefore the required probabilty is 0.4

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oee [108]

Answer:

Option D (APEX)

Step-by-step explanation:

L= Student has lice

NL= Student has no lice

PT=Test Shows positive

NT=Test Shows negative

P(\frac{NT}{NL}) =0.7708 (Probability that a student has no lice and the test shows negative)

P(\frac{NT}{L}) = 0.0108 ( Probability that a student has lice and the test shows negative)

  1. P\frac{NT}{NL} = \frac{P\frac{NT}{NL} }{P\frac{NT}{NL} +P\frac{NT}{L} }
  2. \frac{0.7708}{0.7708+00108}
  3. \frac{0.7708}{0.7816}
  4. 0.98618219 (Simplify)

<u> Terms Of Percentages</u>

0.98618219 × 100

=98.6

Therefore your answer is 98.6% (Option D)

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Brut [27]

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Step-by-step explanation:

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Number 1 is A
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Answer:

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Step-by-step explanation:

Let the number of jars is x.

<u>80 liters distributed, each jar has:</u>

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<u>Redistribution with 4 less jars, each jar now has:</u>

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<u>Each jar has now twice the amount:</u>

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