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3 years ago

7

Is it possible to prove these triangles congruent by the HL Theorem?

2 answers:

aleksandr82 [10.1K]3 years ago

4 0

since by the provided tickmarks WE ≅ ER, and QE ≅ ET, then both triangles are congruent by LL theorem, Leg Leg.

anyhow, since those legs are congruent, their hypotenuses must also be congruents, and we could use the HL theorem as well.

Damm [24]3 years ago

4 0

Answer: I am preety sure it is yes im taking the quiz now but they are both right angels so

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3 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

3 years ago