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Feliz [49]
3 years ago
13

Trina's science club is building bluebird houses to put in the local park. For the first 1/6 dozen bird houses, Trina finds that

they used 3/4 foot of wooden trim. Trina needs to buy enough wooden trim for one dozen bird houses . Write a unit rate that describes how much wooden trim Trina needs.
Mathematics
1 answer:
meriva3 years ago
3 0

<u>Answer:</u>

4.5 foot is needed for a unit dozen bird house

<u>Explanation:</u>

Given Tina saw ¾ foot of wooden trim is being used for  first 1/6 dozen bird houses

Therefore, we can write as,4

1/6 dozen = ¾ foot

For 1 dozen bird houses, we have to use the unitary method

1/6 dozen = ¾ foot  

Therefore, 1 dozen or unit dozen = \frac{\frac{3}{4}}{\frac{1}{6}}

1 dozen = ¾ * 6 = \frac{9}{2} foot = 4.5 foot  

Therefore, 4.5 foot is needed for a unit dozen bird house

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Match each subtraction problem on the left with the equivalent simplified expression on the right.
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Step-by-step explanation:

A:

  • 5x-(3x+1)
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B:

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C:

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4 0
2 years ago
You earn $9 per hour babysitting. This is $2 more than what you earned per hour last year. What did you earn last year?
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4 0
3 years ago
Read 2 more answers
A certain firm has plants A, B, and C producing respectively 35%, 15%, and 50% of the total output. The probabilities of a non-d
Sliva [168]

Answer:

There is a 44.12% probability that the defective product came from C.

Step-by-step explanation:

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

P = \frac{P(B).P(A/B)}{P(A)}

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

-In your problem, we have:

P(A) is the probability of the customer receiving a defective product. For this probability, we have:

P(A) = P_{1} + P_{2} + P_{3}

In which P_{1} is the probability that the defective product was chosen from plant A(we have to consider the probability of plant A being chosen). So:

P_{1} = 0.35*0.25 = 0.0875

P_{2} is the probability that the defective product was chosen from plant B(we have to consider the probability of plant B being chosen). So:

P_{2} = 0.15*0.05 = 0.0075

P_{3} is the probability that the defective product was chosen from plant B(we have to consider the probability of plant B being chosen). So:

P_{3} = 0.50*0.15 = 0.075

So

P(A) = 0.0875 + 0.0075 + 0.075 = 0.17

P(B) is the probability the product chosen being C, that is 50% = 0.5.

P(A/B) is the probability of the product being defective, knowing that the plant chosen was C. So P(A/B) = 0.15.

So, the probability that the defective piece came from C is:

P = \frac{0.5*0.15}{0.17} = 0.4412

There is a 44.12% probability that the defective product came from C.

3 0
3 years ago
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