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valentina_108 [34]

3 years ago

10

Rewrite this expression using the distributive property then simplify. 2/3(6x+12)

2 answers:

Nostrana [21]3 years ago

8 0

So distributive property of multiplication says
a(b+c)=(ab)+(ac)

so 2/3(6x+12)=((2/3)6x)+((2/3)12)=12/3x+24/3=4x+8=4(x+2)

Bezzdna [24]3 years ago

8 0

2/3(6x+12)

=12/3x+24/3

=4x+8

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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

SSSSS [86.1K]

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

3 0

3 years ago

In a carnival game, a person wagers $2 on the roll of two dice. if the total of the two dice is 2, 3, 4, 5, or 6 then the pe

Olin [163]

Answer: a loss of 4 cents

<u>Step-by-step explanation:</u>

The probability of rolling a sum of 2, 3, 4, 5, or 6 is $\dfrac{15}{36}$ which earns $2.00

The probability of rolling a sum of 28, 9, 10, 11, or 12 is $\dfrac{15}{36}$ which loses $2.00

The probability of rolling a sum of 7 is $\dfrac{6}{36}$ which loses $0.25

$\bigg(\dfrac{15}{36}\times \$2.00\bigg)+\bigg(\dfrac{15}{36}\times -\$2.00\bigg)+\bigg(\dfrac{6}{36}\times -\$0.25\bigg)=\boxed{-\$0.04}$

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The pair of points is on the graph of an inverse variation. Find the missing value.

rosijanka [135]

Answer:

The answer is 2

Step-by-step explanation:

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Juan and Gloria are comparing the two numbers 354.6 and 354.06. Juan says that 354.06 is equal to 10 x 354.6. Gloria says he is

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Answer: I agree with Gloria because 10 times 354.6 is 3546

Step-by-step explanation:

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3 years ago

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Answer:

Yea!

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