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kolbaska11 [484]
3 years ago
14

A gross is equal to 144 items. A baker's dozen is equal to

Mathematics
1 answer:
yawa3891 [41]3 years ago
8 0

Answer:

Approximately 11  baker's dozens equal 1 gross.

Step-by-step explanation:

The number of items in a baker's dozen = 13

The gross =  144 items

So, the number of baker's dozen in a gross

=  \frac{\textrm{Total number of items in a gross}}{\textrm{Total number of items on a baker's dozen}}

= \frac{144}{13}  = 11.07

Hence, approximately 11  baker's dozens equal 1 gross.

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50 points
Leya [2.2K]

Answer:

\frac{2}{4} inch

Step-by-step explanation:

Step 1

List the length of the wingspan of the five butterflies

\frac{1}{4}, \frac{1}{4}, 1, \frac{2}{4}, \frac{3}{4}

Step 2:

The two butterflies with the shortest wingspan has a wing length of \frac{1}{4} inches each.

Step 3:

Total length of the wingspan of the two butterflies with the shortest wingspan

= =\frac{1}{4} +\frac{1}{4} \\
       = \frac{2}{4}

Final answer:

= \frac{2}{4} inch

5 0
2 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
Which property is shown by –8 • (4 + 3) = –8 • 4 + –8 • 3?
larisa86 [58]
<span> –8 • (4 + 3) = –8 • 4 + –8 • 3

answer

</span><span>• Distributive Property</span>
6 0
3 years ago
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HELP ME PLEASE! AGH!
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I'm a third grader!!!!!!! Not a highschooler
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3 years ago
Where common multiples between three and five and one and 50
Arlecino [84]
300 is a multiple of 3,5,1, and 50
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