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3 years ago

Plz help. N+ 4/3=1/3 what's a fact family just the N don't give a fraction

Mathematics

2 answers:

Ymorist [56]3 years ago

4 0

I believe the answer is n=-1

lesantik [10]3 years ago

3 0

I believe the answer is N = -1

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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

Write the ratio 48:30:42 in its simplest form

Zielflug [23.3K]

Answer- 8:5:7

Explanation:
Find the greatest common factor for all of them. Which is 6.
Then you just divide each number by 6
48/6= 8
30/6= 5
42/6= 7
Making it 8:5:7

7 0

3 years ago

If five times a number plus two is 17 what is the number? A. 3 B. 2 C. 7 D. 5

VLD [36.1K]

5*3+2=17 it's A, 3 in this equation.

3 0

3 years ago

Read 2 more answers

12x-5y=20 y=x+4<br> 3x+y=3 x=-y+3

Vladimir79 [104]

Standard form (which I assume is what they want you to turn these into) is y=mx+b.

12x-5y=20 is y=2.4x-4.
Subtract 12x from both sides, -5y=-12x+20.
Multiply both sides by -1, 5y=12x-20.
Divide both sides by 5, y=2.4x-4.

y=x+4 is already in standard form- I'm not quite sure what to do.

3x+y=3 is y=-3x+3.
Subtract 3x from both sides, y=-3x+3.

x=-y+3 is y=-x+3
Ad y to both sides, y+x=3.
Subtract x from both sides, y=-x+3.

4 0

3 years ago

Can somebody answer my question, I asked one like 1h and 30min ago and no one has answered!

Fiesta28 [93]

What is it? Can you comment the question?

7 0

3 years ago