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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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Need help asap, thanks!!!ヾ(⌐■_■)ノ♪
Lelu [443]

Step-by-step explanation:

the probability of making one shot is 75% or 0.75 or 3/4.

the probability to make 2 shots is then

3/4 × 3/4 = 9/16 = 0.5625

and so on.

the probability to miss 1 shot is then 25% or 0.25 or 1/4.

the probability to make at least 4 out of 5 shots is the sum of the probability to make all 5 shots plus the probability to make 4 shots.

or the probability to miss 0 plus the probability to miss 1 shot.

anyway, we can go with the positive approach. it seems to be the same complexity as the negative approach.

the probability to make all 5 shots is

(3/4)⁵ = 3⁵/4⁵ = 243 / 1024 = 0.237304688...

the probability to e.g make the first 4 shots and miss the 5th is

(3/4)⁴×1/4 = 3⁴/4⁵ = 81 / 1024 = 0.079101563...

how many possibilities do we have to make 4 out of 5 shots ?

there is no repetition, and the sequence inside the "picked" 4 does not matter, so we need combinations :

C(5,4) = 5! / (4! × (5-4)!) = 5

so, the probability to make exactly 4 out 5 shots is

81/1024 × 5 = 405/1024 = 0.395507813...

and the total probability to make at least 4 shots is

243/1024 + 405/1024 = 648/1024 = 81/128 = 0.6328125

to control,

this plus making exactly 3 plus exactly 2 plus exactly 1 plus none must be the probability 1 (as there is no other possible outcome).

making 3 :

(3/4)³×(1/4)² = 3³/4⁵ = 27/1024

C(5,3) = 5×2 = 10

270/1024

making 2 :

3²/4⁵ = 9/1024

C(5,2) = 5×2 = 10

90/1024

making 1 :

3/1024

C(5,1) = 5

15/1024

making 0 :

1/1024

in sum

(270+90+15+1)/1024 = 376/1024

plus the 648/1024 = 1024/1024 = 1

correct.

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2 years ago
How would you solve 3x^2-12=0?
Gennadij [26K]

                                                                                 3x^2 - 12  =  0

Personally, I would divide each side by 3                x^2 - 4  =  0

Then I would add 4 to each side                              x^2        =  4

Then I would take the square root of each side:     x  =  +4
                                                                                     x  =  -4 .

Having found both solutions, I would relax with a cup of tea.

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4 years ago
120 tickets were sold in one hour and 50% were childrens tickets how many childrens tickets were sold?
Elanso [62]

Answer:

60

Step-by-step explanation:

because half of children is 50. 120÷50=60

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Answer:

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Step-by-step explanation:

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