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Ugo [173]
3 years ago
8

Find an explicit rule for the nth term of the sequence. 9, 36, 144, 576, ...

Mathematics
2 answers:
Sphinxa [80]3 years ago
7 0
Something like (n4)
N representing the number before. For example, if n=9 then (9*4)=36.
Hope it makes since for you. I know there might be more to it, but that's what I know off the top of my head.
andreyandreev [35.5K]3 years ago
4 0

Answer:

an= 9* 4^n-1

Step-by-step explanation:

an= a1 r^ n-1

a1=9

r=4

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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
diamong [38]

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%.

4 0
3 years ago
Given f(x) = 4x - 6, evaluate f(3).<br><br> A. 2<br><br> B. 1<br><br> C.12<br><br> D. 6
Dima020 [189]

Answer:

6

Step-by-step explanation:

7 0
3 years ago
Please help
mafiozo [28]

Answer:

Easy it's a right hand child

Step-by-step explanation:

Because if both husband and wife both have right hands. And it's 10% chance the child will be left handed. Then it's possible that the child is going to be a right hand.

4 0
3 years ago
Anybody got some answers?
kiruha [24]
I don’t know if this is right but this is what I figured.
37.245508982
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3 years ago
Solve:<br>4c + 22 = 66<br>C = ​
Amanda [17]
Answer:

C= 11

Explanation:
8 0
3 years ago
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