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Lerok [7]
2 years ago
7

tle=" \frac{1}{ \sec(1 - \sin(x) ) } dx" alt=" \frac{1}{ \sec(1 - \sin(x) ) } dx" align="absmiddle" class="latex-formula">
Any help with this? ​

Mathematics
1 answer:
Gnoma [55]2 years ago
4 0

\displaystyle\dfrac{1}{\sec x(1-\sin x)}=\dfrac{\cos x}{1-\sin x}\\\\\\\int \dfrac{dx}{\sec x(1-\sin x)}=\int\dfrac{\cos x\, dx}{1-\sin x}\\\\\\u=1-\sin x\\du=-\cos x\, dx\\\\\int \dfrac{\cos x\, dx}{1-\sin x}=\int \dfrac{-du }{u}=-\int \dfrac{du}{u}=-\ln |u|+C=-\ln|1-\sin x|+C

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Graph the line<br> y=-4x+2<br> ‼️will make brainliest‼️
Gekata [30.6K]

Answer:It would be -4 across and go two up

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
30 points: Answer the questions in the picture provided
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5 0
3 years ago
The product of 3 and a number x is 23 . What is the value of x?
Scrat [10]
I believed there's a typo in your question
<span>The product of 3 and a number x is 2/3
if so
</span>3x = 2/3
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answer 
x = 2/9


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
Question 5
Darina [25.2K]

Answer:

the answer is 12.24 .

Step-by-step explanation:

that answer is right is because when you add of those you get 530 , then you divide by how many numbers there are which is 7 , which equals to 75.7 . when you get that number ( 75.7) you subtract that number to 65,90,85,70,70,95, and 55 . when you get the total of all of those you add those . which is 10.7, 14.3 , 9.3 , 5.7 , 5.7 , 19.3 , and 20.7 . from adding those you get 85.7. you divide by 7 and get 12.24 . hope this helps :) .

8 0
2 years ago
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