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AleksandrR [38]
3 years ago
15

A painter contractor determined the average amount of time needed to paint a house. when 2 painters are used it takes 20 hours t

o paint the exterior of a house. when 5 painters are used it takes 8 hours to paint the exterior of a house. let p represent the number of painters. let t represent the amount of time required to paint a house.
Mathematics
1 answer:
skad [1K]3 years ago
3 0
\text {2 painters take} \longrightarrow 20 \text { hours}

\text {1 painters will take} \longrightarrow 20 \times 2 = 40 \text { hours}

P = number of painters.

\text {P painters will take} \longrightarrow 40 \div p = \dfrac{40}{p}  \text { hours}

t = total of times needed,

\boxed{\boxed {\text {Answer: }t =  \dfrac{40}{p}}}

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15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.
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Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = 30C5 = \frac{30 !}{(30-5)!5!} \\

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Pr(no defective) has been calculated above = 0.46

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Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) =  \frac{26C3  * 4C2}{30C5}

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) +  Pr(2 defective) +  Pr(3 defective) +  Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) =  \frac{26C2  * 4C3}{30C5}

Pr(3 defective) = 0.009

Pr(4 defective) =  \frac{26C1  * 4C4}{30C5}

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

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3 years ago
How do you write 6,007,200 in words
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