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olga_2 [115]
2 years ago
13

Help please help me

Mathematics
1 answer:
asambeis [7]2 years ago
8 0

Answer:

Step-by-step explanation:

Range = maximum value - minimum value

= 25 - 6 = 19

Option D is the correct answer

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The cost of 6 movie tickets is $57.00. What is the cost of 8 movie tickets?
Rudiy27

Answer:

Step-by-step explanation:

57 divided by 6 = 7.5

9.5  x  8 = $76

7 0
2 years ago
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
2 years ago
Simplify (Are all division)<br><br>2+4i<br>3i<br><br>3+2i<br>4+i<br><br>2i^11
Alla [95]

We\ know:\ i=\sqrt{-1}\to i^2=-1

\dfrac{2+4i}{3i}=\dfrac{2+4i}{3i}\cdot\dfrac{3i}{3i}=\dfrac{6i+12i^2}{9i^2}=\dfrac{6i+12(-1)}{9(-1)}\\\\=\dfrac{6i-12}{-9}=\dfrac{2i-4}{-3}=\dfrac{4}{3}-\dfrac{2}{3}i

\dfrac{3+2i}{4+i}=\dfrac{3+2i}{4+i}\cdot\dfrac{4-i}{4-i}=\dfrac{(3+2i)(4-i)}{4^2-i^2}\\\\=\dfrac{(3)(4)+(3)(-i)+(2i)(4)+(2i)(-i)}{16-(-1)}=\dfrac{12-3i+8i-2i^2}{16+1}\\\\=\dfrac{12+5i-2(-1)}{17}=\dfrac{12+5i+2}{17}=\dfrac{14+5i}{17}=\dfrac{14}{17}+\dfrac{5}{17}i

2i^{11}=2i^{10+1}=2i^{10}i^1=2i^{2\cdot5}i=2i(i^2)^5=2i(-1)^5=2i(-1)=-2i


Used:\ (a+b)(a-b)=a^2-b^2

6 0
3 years ago
Domain and Range for this equation? F(x)= 3x^7+0x^6-0x^5-2x^4-3x^3-2x^2-x^1+x^0
sp2606 [1]
The domain is equal all real numbers
D=\mathbb{R}

The degrees of the polynomial is odd (3x^7, 7 is odd), therefore the range is equal all real numbers

R=\mathbb{R}
8 0
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