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Snezhnost
22 days ago
11

The binary value of the ASCII letter "c" is 0110 0011. Using the handy chart that we learned in the lesson, convert this number

to its decimal value. You'll need to use some math for this question 128,64,32,16,8,4,2
Mathematics
1 answer:
Zina [9.1K]22 days ago
5 0

Answer:

The decimal representation is 99

Step-by-step explanation:

To change 01100011 into a decimal number, we apply the zero index method, starting from the last digit

0 = 7

1 = 6

1 = 5

0 = 4

0 = 3

0 = 2

1 = 1

1 = 0

Next, we compute by multiplying the binary digits by 2 raised to the power of their zero-index

(0*2^7) + (1*2^6) + (1*2^5) + (0*2^4) + (0*2^3) + (0*2^2) + (1*2^1) + (1*2^0)

  =(0*128) + (1*64) + (1*32) + (0*16) + (0*8) + (0*4) + (1*2) + (1*1)

      =0 + 64 + 32 + 0 + 0 + 0 + 2 + 1

         = 64 + 32 +3

             =99

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12 hours ago
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Given :

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1 month ago
Suppose the time interval between two consecutive defective light bulbs from a production line has a uniform distribution over a
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Answer:

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d) There is an 11.11% likelihood that the time interval between two successive defective light bulbs will be under 10 minutes.

Step-by-step explanation:

A uniform probability situation occurs when all outcomes have an equal chance of happening.

In this context, we identify a lower and upper limit for the distribution known as 'a' and 'b', respectively.

The likelihood of finding a value X that is less than x is determined by this formula.

P(X < x) = \frac{x - a}{b-a}

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P(c \leq X \leq d) = \frac{d-c}{b-a}

The average of the uniform distribution is:

M = \frac{a+b}{2}

The standard deviation of the uniform distribution can be calculated as:

S = \sqrt{\frac{(b-a)^{2}}{12}}

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There is a 27.78% chance that the time interval between two successive defective light bulbs falls between 10 and 35 minutes.

c) What is the standard deviation of the time interval?

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The standard deviation of the time interval is 25.98 minutes.

d) What is the probability that the time interval between two consecutive defective light bulbs will be less than 10 minutes?

P(X < 10) = \frac{10 - 0}{90 - 0} = 0.1111

There is an 11.11% likelihood that the time duration between two consecutive defective light bulbs will be shorter than 10 minutes.

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