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uysha [10]
4 years ago
5

Tiana can clear a football field of debris in 3 hours. Jacob can clear the same field in 2 hours. If they decide to clear the fi

eld together, how long will it take them?
Mathematics
2 answers:
Inessa [10]4 years ago
7 0

For this case, the first thing we must do is define a variable.

We have then:

t: time it will take Tiana and Jacob to finish the work together.

We have then that the equation that models the problem is given by:

\frac{1}{3} + \frac{1}{2} = \frac{1}{t}

From here, we clear the value of t.

For this, we follow the following steps:

1) Multiply both sides of the equation by 6t:

\frac{6t}{3} + \frac{6t}{2} = \frac{6t}{t}

2) Simplify both sides of the equation:

2t + 3t = 6

3) Add similar terms:

5t = 6

4) Clear the value of t.

t = \frac{6}{5}

Answer:

If they decide to clear the field together they will take them about 6/5 hours

Lapatulllka [165]4 years ago
4 0
Let us determine each of their rates of work:

Since Tiana can clear the field in 3 hours, meaning she can do 1/3 of the work per hour.

In the same way, Jacob can do 1/2 of the work per hour. 

Now, what would be the rate of work if they were working together? Let's look at it like this:

Hours to complete the job:
Tiana = 3
Jacob = 2
Together = t

Work done per hour:
Tiana = 1/3
Jacob = 1/2
Together = 1/t

If you add their labor together:

1/3 + 1/2 = 1/t
5/6 = 1/t
t = 6/2 = 1.2

Together, they can clear the field in 1.2 hours.
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Step-by-step explanation:

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combine like terms

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Determine if the following system of equations has no solutions, infinitely many
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3 years ago
Suppose your boss wants you to obtain a sample to estimate a population mean. Based on previous analyses, you estimate that 49 i
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Answer:

the minimum sample size n = 11.03

Step-by-step explanation:

Given that:

approximate value of the population standard deviation \sigma = 49

level of significance ∝ = 0.01

population mean = 38

the minimum sample size n = ?

The minimum sample size required can be determined by calculating the margin of error which can be re[resented by the equation ;

Margin of error = Z_{ \alpha /2}} \times \dfrac{\sigma}{\sqrt{n}}

38 = \dfrac{2.576 \times 49}{\sqrt{n}}

\sqrt{n} = \dfrac{2.576 \times 49}{38}

\sqrt{n} = \dfrac{126.224}{38}

\sqrt{n} = 3.321684211

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Thus; the minimum sample size n = 11.03

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3 years ago
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