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lilavasa [31]
3 years ago
5

Y = (x + 3)^2 -4 and identify the axis of symmetry

Mathematics
1 answer:
Vikki [24]3 years ago
7 0

Answer:

Step-by-step explanation:

the line for equation : x = - 3

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What is 1/3x+4=6?<br> Also show steps plz if you can...
Sever21 [200]

Answer:

x = 6

Step-by-step explanation:

6 0
3 years ago
Is this corrrecct??? answe quick!?​
jok3333 [9.3K]

Answer: x = 27

Step-by-step explanation:

2x+3x+45=180º because all of the angles of a triangle add up to 180

5x=180-45

5x=135

x=135/5

x=27

5 0
3 years ago
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The ___________ of a lens or mirror is a rotational symmetry axis of the surfaces.
densk [106]

Answer:

Optical axis

Step-by-step explanation:

Optical axis is the rotational symmetry axis of the surfaces.

A line with a certain degree of rotational symmetry is called as the optical axis in an optical system.

It is the straight line that passes through the geometric center of the lens and joins two curvature centers of its surfaces.

It is also called as the principal axis.

5 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
Find solution of the linear equation y = 2/3x + 1/3
ankoles [38]

Answer:

3/3=1

Step-by-step explanation:


6 0
3 years ago
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