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MariettaO [177]
2 years ago
10

What are the step for this {-5x-5}{10}=11 ? Help Me Please!!

Mathematics
2 answers:
Leviafan [203]2 years ago
6 0

Answer:

Step-by-step Let's solve your equation step-by-step.

(−5x−5)(10)=11

Step 1: Simplify both sides of the equation.

(−5x−5)(10)=11

(−5x)(10)+(−5)(10)=11(Distribute)

−50x+−50=11

−50x−50=11

Step 2: Add 50 to both sides.

−50x−50+50=11+50

−50x=61

Step 3: Divide both sides by -50.

−50x\ −50 = 61 /−50

x= −61 /50

Answer:

x= −61 /50

Alekssandra [29.7K]2 years ago
3 0

Answer:

-61/50

Step-by-step explanation:

Let us start by <u>expanding</u> the expression on the left side:

(-5x)*10 - 5(10)=11\\-50x-50=11

We want to find x, so we should<u> isolate</u> x on the left side:

-50x=11+50\\-50x=61\\x=-\frac{61}{50}

<em>I hope this helps! Please let me know if you have any further questions :)</em>

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15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.
Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = 30C5 = \frac{30 !}{(30-5)!5!} \\

30C5 = 142506 ways

Number of ways of selecting 5 good bulbs  from 26 bulbs = 26C5 = \frac{26 !}{(26-5)!5!} \\

26C5 = 65780 ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = \frac{26C4  * 4C1}{30C5}

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) =  \frac{26C3  * 4C2}{30C5}

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) +  Pr(2 defective) +  Pr(3 defective) +  Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) =  \frac{26C2  * 4C3}{30C5}

Pr(3 defective) = 0.009

Pr(4 defective) =  \frac{26C1  * 4C4}{30C5}

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

3 0
3 years ago
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