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Tanya [424]
3 years ago
6

A dump truck brought 1⁄3 of a ton of rock on the first trip, 1⁄2 of a ton on the second trip, but on the third trip, had to take

back 4⁄5 of a ton. What was the total weight of the rock left at the construction site?
Mathematics
2 answers:
Allisa [31]3 years ago
7 0

Answer:

1/30 of a ton

Step-by-step explanation:

Write this as an expression:

1/3 of a ton and 1/2 of a ton were taken to the site.

4/5 of a ton were remove from the site.

Amount of a ton at the site  =   \frac{1}{3}+ \frac{1}{2}- \frac{4}{5}

Solve the equation by finding <u>common denominators</u> (when the bottom numbers are the same).

(\frac{1}{3}+ \frac{1}{2})- \frac{4}{5}       Focus on the first part. The least common multiple (LCM) of 3 and 2 is "6", which will become the denominator.

For <u>1/3 to become ?/6</u>, multiply top and bottom by 2.

For <u>1/2 to become ?/6</u>, multiply top and bottom by 3.

(\frac{2}{6}+ \frac{3}{6})- \frac{4}{5}

Now we can add the numerators together over the same denominator.

\frac{5}{6}- \frac{4}{5}

Do the same thing as before and change the denominators. The LCM of 6 and 5 is "30".

For <u>5/6 to become ?/30</u>, multiply top and bottom by 5.

For <u>4/5 to become ?/30</u>, multiply top and bottom by 6.

\frac{25}{30}- \frac{24}{30}

Subtract the numerators:

\frac{1}{30}                  Answer in tons

Therefore there was 1/30 of a ton of rock left at the construction site.

bija089 [108]3 years ago
5 0
Larry received 1/3 pound of candy from his grandmother and 2/5 pound of candy from his best friend. Larry's sister ate 1/2 pound of Larry's candy. How many pounds of candy does Larry have left?
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Suppose there are 4 defective batteries in a drawer with 10 batteries in it. A sample of 3 is taken at random without replacemen
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Answer:

a.) 0.5

b.) 0.66

c.) 0.83

Step-by-step explanation:

As given,

Total Number of Batteries in the drawer = 10

Total Number of defective Batteries in the drawer = 4

⇒Total Number of non - defective Batteries in the drawer = 10 - 4 = 6

Now,

As, a sample of 3 is taken at random without replacement.

a.)

Getting exactly one defective battery means -

1 - from defective battery

2 - from non-defective battery

So,

Getting exactly 1 defective battery = ⁴C₁ × ⁶C₂ =  \frac{4!}{1! (4 - 1 )!} × \frac{6!}{2! (6 - 2 )!}

                                                                            = \frac{4!}{(3)!} × \frac{6!}{2! (4)!}

                                                                            = \frac{4.3!}{(3)!} × \frac{6.5.4!}{2! (4)!}

                                                                            = 4 × \frac{6.5}{2.1! }

                                                                            = 4 × 15 = 60

Total Number of possibility = ¹⁰C₃ = \frac{10!}{3! (10-3)!}

                                                        = \frac{10!}{3! (7)!}

                                                        = \frac{10.9.8.7!}{3! (7)!}

                                                        = \frac{10.9.8}{3.2.1!}

                                                        = 120

So, probability = \frac{60}{120} = \frac{1}{2} = 0.5

b.)

at most one defective battery :

⇒either the defective battery is 1 or 0

If the defective battery is 1 , then 2 non defective

Possibility  = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 0 , then 3 non defective

Possibility   = ⁴C₀ × ⁶C₃

                   =  \frac{4!}{0! (4 - 0)!} × \frac{6!}{3! (6 - 3)!}

                   = \frac{4!}{(4)!} × \frac{6!}{3! (3)!}

                   = 1 × \frac{6.5.4.3!}{3.2.1! (3)!}

                   = 1× \frac{6.5.4}{3.2.1! }

                   = 1 × 20 = 20

getting at most 1 defective battery = 60 + 20 = 80

Probability = \frac{80}{120} = \frac{8}{12} = 0.66

c.)

at least one defective battery :

⇒either the defective battery is 1 or 2 or 3

If the defective battery is 1 , then 2 non defective

Possibility  = ⁴C₁ × ⁶C₂ = 60

If the defective battery is 2 , then 1 non defective

Possibility   = ⁴C₂ × ⁶C₁

                   =  \frac{4!}{2! (4 - 2)!} × \frac{6!}{1! (6 - 1)!}

                   = \frac{4!}{2! (2)!} × \frac{6!}{1! (5)!}

                   = \frac{4.3.2!}{2! (2)!} × \frac{6.5!}{1! (5)!}

                   = \frac{4.3}{2.1!} × \frac{6}{1}

                   = 6 × 6 = 36

If the defective battery is 3 , then 0 non defective

Possibility   = ⁴C₃ × ⁶C₀

                   =  \frac{4!}{3! (4 - 3)!} × \frac{6!}{0! (6 - 0)!}

                   = \frac{4!}{3! (1)!} × \frac{6!}{(6)!}

                   = \frac{4.3!}{3!} × 1

                   = 4×1 = 4

getting at most 1 defective battery = 60 + 36 + 4 = 100

Probability = \frac{100}{120} = \frac{10}{12} = 0.83

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