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mote1985 [20]
2 years ago
12

In expanded notation, the hexadecimal 74AF16 is (7*4096) + (4*256) + (A*16) + (F*1). When converting from hexadecimal to decimal

, what value is assigned to F?
Mathematics
1 answer:
Feliz [49]2 years ago
6 0

Answer:

F is assigned the value of 15

74AF_{16} = 29871_{10}

Step-by-step explanation:

Hexadecimal number system is base 16 and it contain the following numbers:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F

A has a value of 10

B has a value of 11

C has a value of 12

D has a value of 13

E has a value of 14

F has a value of 15

By completing the expanded notation:

(7*4096) + (4*256) + (A * 16) + (F *1)\\= (7*4096) + (4*256) + (10 * 16) + (15 *1)\\= 28672 + 1024 + 160 + 15\\= 29871

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2 years ago
An automated egg carton loader has a 1% probability of cracking an egg, and a customer will complain if more than one egg per do
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Answer:

a) Binomial distribution B(n=12,p=0.01)

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d) P=0.366

Step-by-step explanation:

a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.

b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.

P(k=0)=\binom{12}{0}p^0(1-p)^{12}=1*1*0.99^{12}=1*0.886=0.886\\\\P(k=1)=\binom{12}{1}p^1(1-p)^{11}=12*0.01*0.99^{11}=12*0.01*0.895=0.107

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c) In this case, the distribution is B(1200,0.01)

P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924

d) In this case, the distribution is B(100,0.01)

We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.

P(k=0)=\binom{100}{0}p^0(1-p)^{100}=0.99^{100}=0.366

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