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zhenek [66]
3 years ago
10

What is the solution to 0.25t = 0.25 - t ?

Mathematics
1 answer:
Ede4ka [16]3 years ago
8 0

i think it's t=0.2    Hope it helps:)

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Find the value of x.
Ainat [17]

Answer:

\textsf{x=22.5}

Step-by-step explanation:

\textsf{According to intersecting tangent - secant theorem,}

\textsf{x=1/2[(4x+5)-50]}

\textsf{2(x)=4x+5-50}

\textsf{2x=4x-45}

\textsf{2x-4x=-45}

\textsf{-2x=-45}

\textsf{x= -45/-2}

\textsf{x=22.5}

\textsf{OAmalOHopeO}

5 0
3 years ago
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
PLEASE HELP ME<br><br> rename 8 ten thousands 4 hundreds
Naddik [55]
80,400 or eighty thousand four hundred
6 0
3 years ago
Read 2 more answers
SOMEONE HELP ME PLZ ASAP!!!!​
Elanso [62]

A=22cm

B=21cm

C=23cm

Good luck

4 0
3 years ago
Select the correct answer.
Semenov [28]

Answer:

The last option: 3 x (7 + 2) = (3 x 7) + (3 x 2)

Step-by-step explanation:

There is addition on the inside, and you are distributing the multiplication from the outside.

3 0
3 years ago
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