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zavuch27 [327]
4 years ago
11

Refer to the following distribution of commissions:Monthly Commissions Class Frequencies$600 up to $800 3800 up to 1,000 71,000

up to 1,200 111,200 up to 1,400 121,400 up to 1,600 401,600 up to 1,800 241,800 up to 2,000 92,000 up to 2,200 4For the preceding distribution, what is the midpoint of the class with the greatest frequency?
Engineering
1 answer:
erica [24]4 years ago
7 0

Answer:

21.81 %

Explanation:

given,

$600 up to $800        3

800 up to 1,000           7

$1,000 up to $1,200    11

$1,200 up to $1,400    12

$1,400 up to $1,600    40

$1,600 up to $1,800    24

$1,800 up to $2,000     9

$2,000 up to $2,200    4

                                                   

Total = 3 + 7 + 11 + 12 + 40 + 24 + 9 + 4 = 110

the frequency of $1,600 to $1800

 = \dfrac{24}{110}

 =21.81 %

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Answer:

15.99 ft/s

Explanation:

From Newton's equation of motion, we have

v = u + at

v = Final speed

u = initial speed

a = acceleration

t = time

now

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v = 17.6 ft/s

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thus,

17.6 = 13.2 + a(3)

or

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or

a = 1.467 m/s²

Thus,

For Points A and B

v = speed at B i.e v'

u = 13.2 ft/s

a = 1.467 ft/s²

t = 1.90 s

therefore,

v' = 13.2 + (1.467 × 1.90 )

v' = 13.2  + 2.7867

v' = 15.9867 ≈ 15.99 ft/s

3 0
3 years ago
Assume the transistor is biased in the saturation region at VGS 4 V. (a) Calculate the ideal cutoff frequency. (b) Assume that t
insens350 [35]

Answer:

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Answer: A) 5.17 GHz

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Answer:

The correct answer is letter "C": Both.

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Thus, using a sharp chisel could pry a seal out of a hole and a regular socket can often be used to force smaller metal-backed seals into place. Thus, technicians "A" and "B" are correct.

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