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3 years ago

What value of x is in the solution set of 9(2x+1)<9x-18

1 answer:

6 0

18x + 9 < 9x - 18

9x < -27

<span>x < -3

</span><span>If I managed to help you, please make sure to mark my answer as the "Brainliest" answer. Thanks! :)</span>

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For what values of $x$ is it true that $x^2 - 5x - 4 \le 10$? Express your answer in interval notation.

dybincka [34]

Answer:

x ∈ [-2, 7]

Step-by-step explanation:

The given equation ...

x^2 -5x -4 ≤ 10

can be rewritten as ...

x^2 -5x -14 ≤ 0

and factored as ...

(x +2)(x -7) ≤ 0

Clearly, the "or equal to" condition will be met when x=-2 and x=7. For values of x between these numbers, one factor is negative and the other is positive. Hence the product will be negative. So, numbers in that interval are the solution set.

x ∈ [-2, 7]

5 0

3 years ago

The blue dot is at what value on the number line?

Ksenya-84 [330]

Answer:

17.

Step-by-step explanation:

one dash is equal to 2. so just go by twos

4 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

Which expression represents the series 1+5+25+125+625

tankabanditka [31]

$a_n=5^n$
where n starts at 0.

$a_n=5^{(n-1)}$
where n starts at 1.

7 0

3 years ago

What is -4√12+√75 in the simplest radical form?

nekit [7.7K]

Answer: -3 $\sqrt{3}$

Step-by-step explanation:

-4 $\sqrt{12}$ + $\sqrt{75}$

$\sqrt{12}$ can be written as $\sqrt{4}$ x $\sqrt{3}$

which is the same as 2 $\sqrt{3}$

Also , $\sqrt{75}$ can be written as $\sqrt{25}$ x $\sqrt{3}$ , which is the same as 5 $\sqrt{3}$

Substituting into the main question , we have

-4 (2 $\sqrt{3}$ ) + 5 $\sqrt{3}$

= -8 $\sqrt{3}$ + 5 $\sqrt{3}$

= -3 $\sqrt{3}$

3 0

3 years ago