HELP HELP HELP First to "help" rather than just answer with a random thanks for the points kinda response gets brainliest
2 answers:
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Answer:
Option (B)
Step-by-step explanation:
Length of PR = 4
RS = 4
QS = 4
For the length of PT,
PT² = RT² + PR² [Since, PT is the diagonal of rectangle PRT]
PT² = QS² + PR² [Since, RT ≅ QS]
PT² = 4² + 7²
PT² = 16 + 49
PT² = 65
Now for the length of PQ,
PQ² = QT² + PT²
PQ² = RS² + PT² [Since, QT ≅ RS]
PQ² = 4² + 65
PQ² = 16 + 65
PQ = √81
PQ = 9
Therefore, length of diagonal PQ is 9 units.
Option (B) will be the answer.
Answer:
(1) The probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.
(2) The probability at least 5 computers are infected is 0.949.
Step-by-step explanation:
The probability that a computer is defective is, <em>p</em> = 0.40.
(1)
Let <em>X</em> = number of computers to be tested before the 1st defect is found.
Then the random variable .
The probability function of a Geometric distribution for <em>k</em> failures before the 1st success is:
Compute the probability that the technician tests at least 5 computers before the 1st defective computer is found as follows:
P (X ≥ 5) = 1 - P (X < 5)
= 1 - [P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)]
Thus, the probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.
(2)
Let <em>Y</em> = number of computers infected.
The number of computers in the company is, <em>n</em> = 20.
Then the random variable .
The probability function of a binomial distribution is:
Compute the probability at least 5 computers are infected as follows:
P (Y ≥ 5) = 1 - P (Y < 5)
= 1 - [P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3) + P (Y = 4)]
Thus, the probability at least 5 computers are infected is 0.949.