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DIA [1.3K]
3 years ago
8

Assume that we have seeded a program with 5 defects before testing. After the test, 20 defects were detected, of which, 2 are fr

om the seeded defects and 18 are real, non-seeded defects. We know there are 3 more seeded defects remaining in the software. Assuming a linear relationship, what is the estimated real, non-seeded defects that may still be remaining in the program
Computers and Technology
1 answer:
melamori03 [73]3 years ago
5 0

Answer:

see explaination

Explanation:

Given that we have;

Defects before testing = Defects planted = 5

Defects after testing = 20

Seeded defects found = 2

Real, non seeded defects = 18

Hence, Total number of defects = (defects planted / seeded defects found) * real, non seeded defects = (5/2)*18 = 45

Therefore, Estimated number of real defects still present = estimated total number of defects - number of real, non seeded defects found = 45-18 = 27

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What could be one possible reason where the recipient is not guaranteed that the data being streamed will not get interrupted?
anyanavicka [17]

Answer

The senders PC is using UDP protocol

Explanation

UDP is the User Datagram Protocol which is used as an alternative communication protocol to the TCP which is used primarily for establishing low latency and loss of tolerating connections between applications on the internet UDP is normally used by the programs running on different computers on a network. Its purposes is to send short messages which are datagrams. It is not much reliable because of its occasional loss of packet. Due to this packet loss  the recipient is not guaranteed that the data being streamed will not get interrupted. This is because If a router on the Internet starts getting overloaded, or a packet gets corrupted due to interference or anything, the packet will be dropped unlike the TCP (Transmission control  protocol)which resend the packets and keeps re sending. The UDP does not resend the packets which are dropped. Once they are dropped that all.


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3 years ago
Express the following binary numbers in hexadecimal. (a) %100011100101 (b) %1011001111 (show work)
Lapatulllka [165]

Answer:

(100011100101)_{2} = (8E5)_{16} = %8E5

(1011001111) = (2CF)_{16} = %2CF

Explanation:

Binary and hexadecimal values have the following pair equivalences.

(0000)_{2} = (0)_{16}

(0001)_{2} = (1)_{16}

(0010)_{2} = (2)_{16}

(0011)_{2} = (3)_{16}

(0100)_{2} = (4)_{16}

(0101)_{2} = (5)_{16}

(0110)_{2} = (6)_{16}

(0111)_{2} = (7)_{16}

(1000)_{2} = (8)_{16}

(1001)_{2} = (9)_{16}

(1010)_{2} = (A)_{16}

(1011)_{2} = (B)_{16}

(1100)_{2} = (C)_{16}

(1101)_{2} = (D)_{16}

(1110)_{2} = (E)_{16}

(1111)_{2} = (F)_{16}

We convert from binary to hexadecimal selecting groups of 4 binary from the binary code, from the least significant bits(at the right) to the most significant bits(at the left). The conversion is an hexadecimal "string" from the last group you converted to the first. So:

(a) %100011100101

(0101)_{2} = (5)_{16}

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(1000)_{2} = (8)_{16}

So

(100011100101)_{2} = (8E5)_{16}

(b) %1011001111

(1111)_{2} = F_{16}

(1100)_{2} = C_{16}

(10)_{2} = (0010)_{2} = 2_{16}

(1011001111) = (2CF)_{16}

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