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Dovator [93]
3 years ago
8

1) When the plumber comes to my house to fix the leaky faucet, he charges a flat rate of

Mathematics
2 answers:
lyudmila [28]3 years ago
8 0
EXPLANATION:
Money charging per hour = $22
Additional charge= $50
Final bill= $116
Final bill removing the additional charge= $(116 - 50) = $66
Time the plumber was in the house= (66/22)hr = 3hr.

ANSWER: The plumber was there for 3 hours.

Hope it helps u!
Travka [436]3 years ago
4 0

Answer:

3 hours

Step-by-step explanation:

To find the number of hours the plumber was at the house, we need to solve for x.

22x + 50 = 116

22x = 116 - 50

22x = 66

22x/22 = 66/22

x = 3

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2 years ago
A farmer needs to ship 71 pumpkins to a grocery store if each crates can hold 19 pumpkins how many crates Will the farmer need ?
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Answer:

The farmer will need 7 crates.

Step-by-step explanation:

<em>It is given that there are 71 pumpkins in total.</em>

<em>And each crate holds 19 pumpkins.</em>

Let the number of crates required be<em> "n".</em>

The total number of pumpkins can be calculated by multiplying number of pumpkins in each crate with total number of crates.

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3 0
4 years ago
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}

7 0
3 years ago
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