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3 years ago

PLEASE HELP ME, I know these are easy but I forgot to do this assignment and now i don't remember how to do any of it.

Mathematics

2 answers:

astra-53 [7]3 years ago

7 0

Okay so you know the first picture give me the name of the assignment and what’s under the name I’ll try to find the answers if you still need it just let me know

Hatshy [7]3 years ago

5 0

Answer:

Your answers will be down below;

1. 6

2. 7

3. 9.6

4. 13.2

5. 1.5

6. 2.1

7. 11

8. 3

9. 12

10. 6

11. 6

12. 9

Step-by-step explanation:

I had this assignment before and I will NOT add a step-by-step explanation because I'm currently lazy at this time -.-

But I am 100% sure that these are the answer, like I said I HAD this assignment before. Therefore the answers above are your answers to your assignment. =.=

Mark me brainliest if you can, if not then it's alright. Have a good day! :)

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Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a student is proficient in reading but not mathematics and $A \cap B$ is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

$B = b + (A \cap B)$

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

$(A \cup B) + C = 1$

In which

$(A \cup B) = a + b + (A \cap B)$

65% were found to be proficient in both reading and mathematics.

This means that $A \cap B = 0.65$

78% were found to be proficient in mathematics

This means that $B = 0.78$

$B = b + (A \cap B)$

$0.78 = b + 0.65$

$b = 0.13$

85% of the students were found to be proficient in reading

This means that $A = 0.85$

$A = a + (A \cap B)$

$0.85 = a + 0.65$

$a = 0.20$

Proficient in at least one:

$(A \cup B) = a + b + (A \cap B) = 0.20 + 0.13 + 0.65 = 0.98$

What is the probability that the student is proficient in neither reading nor mathematics?

$(A \cup B) + C = 1$

$C = 1 - (A \cup B) = 1 - 0.98 = 0.02$

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3 years ago

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Answer:

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3 years ago

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Answer:

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3 years ago

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Ganezh [65]

Answer:

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Substitute

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