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Dimas [21]
2 years ago
8

Solve each of the following exponential equations using common bases:

Mathematics
1 answer:
pantera1 [17]2 years ago
6 0

Answer:

x=19/6, x=-9/13, x=33/4

Step-by-step explanation:

1) 2^(6x+10)=2^(12x-9), 6x+10=12x-9, x=19/6

2) 3^(4x+6)=3^(-9x-3), 4x+6=-9x-3, x=-9/13

3) 2^(x+6)*2^(3x-27)=2^(6), 4x-27=6, x=33/4

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What is the value of u and v?
Novay_Z [31]

Answer and Explanation:

Since this is an isosceles right triangle, you can use the ratio of side lengths specific to this type of triangle where both legs are the hypotenuse over root 2:

u = v = 164√2 / √2 = 164

so:

u = 164

v = 164

3 0
2 years ago
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What is x-2y=3 in slope intercept form
Gennadij [26K]

So firstly, subtract x on both sides: -2y=-x+3


Next, divide -2 on both sides, and your answer will be y = 1/2x - 3/2 , or the last option.

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3 years ago
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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
Need asap will give brainliest
patriot [66]
Im not sure, but i think it should be
( -3, 1 )
( -3, -2 )
( -6, -3 )
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3 years ago
How can 15/6 mixed number in simplest form
Assoli18 [71]
2 1/2 because you divide the 15 by 6 and 2 R.3   3 is half of 6 so you make it a half leaving you with 2.5 or fraction form:2 1/2
3 0
3 years ago
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