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3 years ago

At first, Kai earned a score of 80 on a quiz. He discovered that 5 points had been taken

Mathematics

2 answers:

DedPeter [7]3 years ago

8 0

Answer:

The two expressiones that could be uses to find Kai's new quiz score are:

1. 80 + 5

2. 80 - ( -5 )

Step-by-step explanation:

The initial score in the Kai quiz was 80, but when the professor noticed his mistake, he raised the 5 points that he had subtracted by mistake.

80 + 5; It shows that the professor corrected the negative points that Kai had in his Quiz.

80 - (-5); By subtracting a negative amount, what is done is to transform the negative number into a positive number and add it. 80 - (-5) = 80 +5 = 85

mario62 [17]3 years ago

5 0

Initial score was 80. There was an accidental 5 point deduction. This means anything that gives 85 is the correct answer.

80 + 5 = 85

80 - (-5) = 85

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Step-by-step explanation:

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3 years ago

Suppose box A contains 4 red and 5 blue poker chips and box B contains 6 red and 3 blue poker chips. Then a poker chip is chosen

sergejj [24]

Answer:

0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Coin chosen from box B is red.

Event B: Blue poker chip transferred.

Probability of choosing a red coin:

7/10 of 4/9(red coin from box A)

6/10 of 5/9(blue coin from box A). So

$P(A) = \frac{7}{10}*\frac{4}{9} + \frac{6}{10}*\frac{5}{9} = \frac{28 + 30}{90} = 0.6444$

Blue chip transferred, red coin chosen:

6/10 of 5/9. So

$P(A \cap B) = \frac{6}{10}*\frac{5}{9} = \frac{30}{90} = 0.3333$

What is the probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3333}{0.6444} = 0.5172$

0.5172 = 51.72% probability a blue poker chip was transferred from box A to box B, given that the coin just chosen from box B is red

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