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stepladder [879]
3 years ago
13

Wavelength and Frequency

Mathematics
1 answer:
sp2606 [1]3 years ago
7 0

Answer:

B. 3 to 2

Step-by-step explanation:

I calculated it logically

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Find the volume of the figure round your answer to the nearest tenth if necessary
egoroff_w [7]

Answer:

56.5

I think this is right

7 0
3 years ago
A box contains 20 light box of which five or defective it for lightbulbs or pick from the box randomly what's the probability th
Snowcat [4.5K]

Answer:

1

Step-by-step explanation:

Given:-

- The box has n = 20 light-bulbs

- The number of defective bulbs, d = 5

Find:-

what's the probability that at most two of them are defective

Solution:-

- We will pick 2 bulbs randomly from the box. We need to find the probability that at-most 2 bulbs are defective.

- We will define random variable X : The number of defective bulbs picked.

Such that,               P ( X ≤ 2 ) is required!

- We are to make a choice " selection " of no defective light bulb is picked from the 2 bulbs pulled out of the box.

- The number of ways we choose 2 bulbs such that none of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 2:

        X = 0 ,       Number of choices = 15 C r = 15C2 = 105 ways

- The probability of selecting 2 non-defective bulbs:

      P ( X = 0 ) = number of choices with no defective / Total choices

                       = 105 / 20C2 = 105 / 190

                       = 0.5526

- The number of ways we choose 2 bulbs such that one of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 1 and out of defective n = 5 choose r = 1 defective bulb:

        X = 1 ,       Number of choices = 15 C 1 * 5 C 1 = 15*5 = 75 ways

- The probability of selecting 1 defective bulbs:

      P ( X = 1 ) = number of choices with 1 defective / Total choices

                       = 75 / 20C2 = 75 / 190

                       = 0.3947

- The number of ways we choose 2 bulbs such that both of them are defective, out of 5 available defective bulbs choose r = 2 defective.

        X = 2 ,       Number of choices = 5 C 2 = 10 ways

- The probability of selecting 2 defective bulbs:

      P ( X = 2 ) = number of choices with 2 defective / Total choices

                       = 10 / 20C2 = 10 / 190

                       = 0.05263

- Hence,

    P ( X ≤ 2 ) = P ( X =0 ) + P ( X = 1 ) + P (X =2)

                     = 0.5526 + 0.3947 + 0.05263

                     = 1

7 0
3 years ago
A point P has coordinates (3,2) what are the new coordinates after reflecting point P over the y-axis
lara [203]
The x-coordinate changes from +3 to -3.  The y-coordinate stays the same.

8 0
3 years ago
Read 2 more answers
How many numbers can you make between 0-99 using exactly four 4s?
MAVERICK [17]

Answer:

Number Four Fours Term Solved By

0 4 - 4 + 4 - 4 johnv

1 (4/4) / (4/4) johnv

2 4/4 + 4/4 pilot

3 (4+4+4) / 4 pilot

4 [ (4/4) ^4 ] * 4 catbells

5 [ 4*4 +4 ] / 4 pilot

6 [ (4+4) /4 ] +4 scruffy

7 4+4 - (4/4) Flora Flatbelly

8 4*4 - 4 - 4 catbells

9 4/4 +4 +4 scruffy

10 4 + √4 + √4 + √4 P Rubie

11 44 / (√4 + √4) christina

12 4 + 4 + √4 + √4 johnv

13 4!/(√4) + 4/4 tackline

14 4+4+4 + √4 D12Bit

15 (44/4)+4 D12Bit

16 4*4*4 / 4 D12Bit

17 4*4 + 4/4 D12Bit

18 4*4 + 4 / √4 D12Bit

19 4! - 4 - 4/4 ?

Solutions to the Second Challenge: 20 to 99

Number Four Fours Term Solved By

20 4 * 4 + √4 + √4

21 4! - 4 + 4/4

22 4! - (4! / 4 - 4)

23 4! - 4^(4-4)

24 4 * ( 4 + 4/√4 )

25 4! + 4^(4-4)

26 4! + (4! / 4 + 4)

27 4! + 4 - 4/4

28 4! + 4 * (4/4)

29 4! + 4 + 4/4

30 (4 + 4/4)! / 4

31 (4! + 4)/4 +4! MicheleR

32 4*4 + 4*4

33 (4 - .4)/.4 + 4! IceMantis

34 4*4 * √4 + √

35 4! + 44/4

36 4*4 * √4 + 4

37 4! + (4! + √4) / √4 christina

38 44 - 4 - √4

[39] 4! + 4! - 4/.4~

40 44 - √4 - √

41 (4*4 + .4) / .4

42 44 - 4/√4

43 44 - 4/4

44 44 + 4 - 4

45 44 + 4/4

46 44 + 4/√4

47 4! + 4! - 4/4

48 44 + √4 + √4

49 4! + 4! + 4/4

50 44 + 4! / 4

51 (4! - 4 + .4) / .4 Mr.B

52 44 + 4 + 4

[53] 44 + 4 / .4~

54 4! + 4! + (4! / 4)

55 [4! - ( 4 / √4 ) ] IceMantis

56 44 + 4! / √4

[57] 4! + 4! + 4/.4~

58 4! + 4! + 4/.4 IceMantis

59 4!/.4 - 4/4 FireMantis

60 44 + 4 * 4

61 4!/.4 + 4/4 FireMantis

62 (4 + 4) ^ √4 - √4

63 (4! + √4) / .4 - √4 FireMantis

64 4 * 4 * √4 * √4

65 (4^4 +4) / 4 Jeremy

66 (4 + 4) ^ [√4] + √4

67 [4! + √4] / .4 + √4 FireMantis

68 (4 + 4) ^ √4 + 4

69 (4!+4-.4)/.4 Partick

70 44 + 4! + √4 FireMantis

71 (4 + 4! + .4) / .4 IceMantis

72 [ (4! / 4) ^ √4 ] * √4

[73] [4! * √4 + √.4~)] / √.4~) capOCC

74 4! + 4! + 4! + √4 FireMantis

75 (4! + √4 + 4) / .4 FireMantis

76 4!*4 - 4! + 4 Jeremy

[77] (4/.4~) ^ √4 - 4 FireMantis

78 √4 / .4~) * [4! + √4] ArBeeJay Ruakura

79 (4/.4~) ^ √4 - √4

80 4! * 4 - 4*4 -

[81] (4/.4~) * (4/.4~) -

82 (4! / .4) + 4! - √4 Partick

[83] (4/.4~) ^ √4 + √4 -

84 4! * 4 - 4! / √4 -

[85] (4/.4~) ^ √4 + 4 -

86 4!*4 - 4/.4 IceMantis

87 4! * 4 - 4/.4~ -

88 4! * 4 - 4 - 4 -

89 4! + [4! + √4]/.4 RyVikOCC

90 4! * 4 - 4! / 4 -

91 4 * 4! - (√4 / .4) IceMantis

92 4! * 4 - √4*4) -

[93] 4 * 4! - [ √4 / √(.4~) ] IceMantis

94 4! * 4 - 4/√4 -

95 4! * 4 - 4/4 -

96 4! * 4 + 4 - 4 -

97 4! * 4 + 4/4 -

98 4! * 4 + 4/√4 -

[99] (4!*√4-4) / .4~ IceMantis

Step-by-step explanation:

7 0
3 years ago
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A die is rolled. What is the probability of rolling the following?
Rus_ich [418]

Answer:

2/3 chance (67%)

Step-by-step explanation:

Multiples of 5:

(5)

Multiples of 2:

(2)

(4)

(6)

This is a 4/6 chance or 2/3

6 0
3 years ago
Read 2 more answers
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