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Kobotan [32]
3 years ago
9

Cooper is ordering a cheeseburger for lunch. There are 3 cheese and 5 buns to choose from. How many different hamburger can coop

er order if he makes exactly one selection for each option?
Mathematics
1 answer:
Alik [6]3 years ago
3 0

Answer: 15

Step-by-step explanation: u multiply

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Which of these is an example of a literal equation?
RSB [31]

<u>ANSWER</u>

ax-by=k is an example of literal equation.

<u>EXPLANATION</u>

A literal equation is an equation in which letters or variables are used to represent real values.

A literal equation consists of at least two letters or variables.

The first option consists of two variables but it is not an equation. It is just an expression.


The second option is not a literal equation because it consists of only one variable. This is  just a linear equation in one variable. But a literal equation should have at least two variables or letters.

As for the third option, it does not even contain a variable or letter.

5 0
3 years ago
Read 2 more answers
What is the midpoint between (-1, 1) and (5, -5)
e-lub [12.9K]

Answer:

(2,-2) is your answer

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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
Help me please‼️
aksik [14]

Answer:

a.

500x+0.04=y

400x+0.05=y

for percentages move decimal to the left twice

4 0
3 years ago
198round to the nearest tens and hundreds
Xelga [282]
198 round to the nearest hundreds is 200
198 round to the nearest tens is 200 too
8 0
3 years ago
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