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Juli2301 [7.4K]
3 years ago
12

Gina went to a restaurant for lunch. On her bill the sales tax equals $2.34. If the tax is 8.5%, what was the subtotal (without

tax) of her meal?
Mathematics
1 answer:
bagirrra123 [75]3 years ago
4 0

Answer:

$2.5389

Step-by-step explanation:

Given data

Cost of lunch = $2.34

Tax= 8.5%

Let us compute the cost of the tax

=8.5/100*2.34

=0.085*2.34

= $0.1989

Hence the total is = 0.1989+2.34

=$2.5389

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Answer:

0.166666667, to round up it is 0.17

Step-by-step explanation:

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How do i write 11 5/8 as an improper fraction
FinnZ [79.3K]
11 5/8 as a improper fraction 

1) Simplify Fraction 
⇒ Multiply denominator with whole number
8 x 11 = 88

⇒ Answer of multiplication and add the numerator 
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93/8 is improper Fraction. 


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What is the answer to 41x72.94
Alex Ar [27]

Answer:

2990.54

Step-by-step explanation:

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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
Answerrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr pls
Gelneren [198K]

Answer:

the answer is b.11

Step-by-step explanation:

4 0
3 years ago
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