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Evgesh-ka [11]
3 years ago
15

How to do this ,I don’t understand

Mathematics
2 answers:
kondor19780726 [428]3 years ago
6 0

Answer:

idk how to do this

Step-by-step explanation:

Sorry tho

Aneli [31]3 years ago
6 0
I’m thinking maybe you have to solve dtep by step
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HURRY!<br> What are all the possible rational solutions for x^4-3x^2+2x5=0
xxTIMURxx [149]

Answer: x = 1

x =(-3-√-15)/4=(-3-i√ 15 )/4= -0.7500-0.9682i

x =(-3+√-15)/4=(-3+i√ 15 )/4= -0.7500+0.9682i

x2 = 0

Step-by-step explanation:

4 0
2 years ago
HELP ASSSAP WITH THIS QUESTION
bulgar [2K]
The answer is linear and increasing. look at it.... it goes up. 

8 0
3 years ago
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
Jack is a professional weightlifter. his personal lifting record was 125.8kg. at a weightlifting competition, he broke his own r
Softa [21]

Answer:

126.34kg

Step-by-step explanation:

125.8+0.54=126.34kg

Convert 540 grams to kilograms` 0.54 kilograms, and add to the previous record

5 0
3 years ago
Solve for x and y please help!!
Pie

Answer:

the answer is

y= 26.57

x= 35.09

5 0
3 years ago
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