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Mashutka [201]
3 years ago
6

Find the value of i77. A) 1 B) i C) −1 D) −i

Mathematics
2 answers:
Colt1911 [192]3 years ago
8 0

assuming you mean i^{77} where i=\sqrt{-1}

ok, notice a pattern in the exponent

i^1=i=\sqrt{-1}

i^2=-1

i^3=-\sqrt{-1}=-i

i^4=1

i^5=i=\sqrt{-1}

hum, so it goes 1,2,3,4, then repeats

ok, so every 4, the cycle repeats

how far  up is 77 from a multipule of 4?

4*19=76

76 is 1 away from 77

so i^{77}=(i^{76})(i^1)=

(i^4)^{19}(i)=(1)^{19}i=i

the answer is B

Marianna [84]3 years ago
6 0

Answer:

The correct answer is B

Step-by-step explanation: I remember from my algebra 2 class, you divide the powers by 4, and it should give you 194.5. When your decimal is .5 that means it is a positve "i".

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Lucy throws a fair six sided dice . so like the letter that matches the probability of the dice landing on a number between 1 an
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Answer:

Letter B matches the probability.

Step-by-step explanation:

Lucy throws a fair six sided dice.

Probability of the dice landing on a number between 1 and 6 = \frac{\text{Favorable outcome}}{\text{Number of events}}

= \frac{1}{6}

On a number line fraction \frac{1}{6} can be represented by,

An small section (A to B) will represent the fraction = \frac{1}{6}

Therefore, letter B will match the probability of a dice landing on a number between 1 and 6.

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In a certain assembly plant, three machines B1, B2, and B3, make 30%, 20%, and 50%, respectively. It is known from past experien
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Answer:

The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.

Step-by-step explanation:

Using Bayes' Theorem

P(A|B) = \frac{P(B|A)P(A)}{P(B)} = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|a)P(a)}

where

P(B|A) is probability of event B given event A

P(B|a) is probability of event B not given event A  

and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.

For this problem,

Let P(B1) = Probability of machine B1 = 0.3

P(B2) = Probability of machine B2 = 0.2

P(B3) = Probability of machine B3 = 0.5

Let P(D) = Probability of a defective product

P(N) = Probability of a Non-defective product

P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003

P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006

P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010

Likewise,

P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997

P(N|B2) be probability of a non-defective product produced by machine 2  = 1 - P(D|B2) = 1 - 0.006 = 0.994

P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990

For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)

P(B1|N) =\frac{P(N|B1)P(B1)}{P(N|B1)P(B1) + P(N|B2)P(B2) + (P(N|B3)P(B3)} = \frac{(0.297)(0.3)}{(0.297)(0.3) + (0.994)(0.2) + (0.990)(0.5)} = 0.1138

Hence the probability that a non-defective product is produced by machine B1 is 11.38%.

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