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leonid [27]
3 years ago
6

6. As part of their employee benefits, all workers at Light and Power Electric Company

Mathematics
1 answer:
Semenov [28]3 years ago
8 0

Answer:

  $1769.06

Step-by-step explanation:

The sum of Mia's three highest salaries is ...

  $92,000 +94,800 +96,250 = $283,050

so the average is ...

  $283,050/3 = $94,350

Her pension is then ...

  0.01875 × $94,350 = $1769.06

_____

<em>Additional comment</em>

This is equivalent to the total of those salaries, divided by 160.

  T/3 × 0.01875 = T × (0.01875/3) = T × 0.00625 = T/160

Sometimes, there are simpler ways to calculate things like this.

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In a certain Algebra 2 class of 30 students, 19 of them play basketball and 12 of them play baseball. There are 8 students who p
Alenkinab [10]

Answer:

Probability that a student chosen randomly from the class plays basketball or baseball is  \frac{23}{30} or 0.76

Step-by-step explanation:

Given:

Total number of students in the class = 30

Number of students who plays basket ball = 19

Number of students who plays base ball = 12

Number of students who plays base both the games = 8

To find:

Probability that a student chosen randomly from the class plays basketball or baseball=?

Solution:

P(A \cup B)=P(A)+P(B)-P(A \cap B)---------------(1)

where

P(A) = Probability of choosing  a student playing basket ball

P(B) =  Probability of choosing  a student playing base ball

P(A \cap B) =  Probability of choosing  a student playing both the games

<u>Finding  P(A)</u>

P(A) = \frac{\text { Number of students playing basket ball }}{\text{Total number of students}}

P(A) = \frac{19}{30}--------------------------(2)

<u>Finding  P(B)</u>

P(B) = \frac{\text { Number of students playing baseball }}{\text{Total number of students}}

P(B) = \frac{12}{30}---------------------------(3)

<u>Finding P(A \cap B)</u>

P(A) = \frac{\text { Number of students playing both games }}{\text{Total number of students}}

P(A) = \frac{8}{30}-----------------------------(4)

Now substituting (2), (3) , (4) in (1), we get

P(A \cup B)= \frac{19}{30} + \frac{12}{30} -\frac{8}{30}

P(A \cup B)= \frac{31}{30} -\frac{8}{30}

P(A \cup B)= \frac{23}{30}

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