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2 years ago

6

What is 2 to the 9th power

1 answer:

andrey2020 [161]2 years ago

5 0

Answer:

it is 512

Step-by-step explanation:

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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

2 years ago

Given that cos (x) = 1/3, find sin (90 - x)

ddd [48]

Answer:

$\sin(90^{\circ} - x)=\frac{1}{3}$

Step-by-step explanation:

Given: $\cos (x)=\frac{1}{3}$

We have to find the value of $\sin(90^{\circ} - x)$

Since Given $\cos (x)=\frac{1}{3}$

Using trigonometric identity,

$\sin(90^{\circ} - \theta)=\cos\theta$

Thus, for $\sin(90^{\circ} - x)$ comparing , we have,

$\theta=x$

We get,

$\sin(90^{\circ} - x)=\cos x=\frac{1}{3}$

Thus, $\sin(90^{\circ} - x)=\frac{1}{3}$

3 0

3 years ago

Read 2 more answers

What is 10 times by 10

Tpy6a [65]

100 because you make ten groups of ten and it makes 100 Hope that helped Have a good day! :)

7 0

3 years ago

Read 2 more answers

(easy) If ΔEFG ~ ΔLMN with a ratio of 3:1, which of the following is true?

alexandr402 [8]

Answer:

segment EG over segment LN equals segment FG over segment MN

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent

In this problem

The corresponding sides are

EF and LM

EG and LN

FG and MN

The corresponding angles are

∠E≅∠L

∠F≅∠M

∠G≅∠N

therefore

EF/LM=EG/LN=FG/MN=3/1

7 0

3 years ago

Read 2 more answers

A series of trade discounts

Inessa05 [86]

they offer a lot of discounts prety much

8 0

3 years ago