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QveST [7]
3 years ago
5

The apartment you like rents for $700 a month. The landlord requires two months’ rent for the security deposit. How much is the

security deposit?
Mathematics
1 answer:
Vlada [557]3 years ago
6 0

Answer: 1400.00

Step-by-step explanation:

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The angle measures 22°, 62°, 118°, and 158° are written on slips of paper. You choose two slips of paper at random.
mylen [45]
Answer: 1/3

There are C(4,2) = 6 ways to choose a pair of numbers.  Only 2 of those pairs are supplementary measures {22, 158} and {62, 118}. The probability is 2/6 = 1/3.


3 0
3 years ago
A square field has an area of 4823m²(show your work! ) ​
8_murik_8 [283]

Answer:

69.45 m to the nearest hundredth.

Step-by-step explanation:

I guees you want the length of one side.

A square has 4 sides of equal length and the area = s^2  where s = length of each side.

So the length of a side of a square of area 4823

= √(4823)

= 69.4478 m.

7 0
1 year ago
Image attached on the bottom
julia-pushkina [17]

Answer:

c

Step-by-step explanation:

i just took it

5 0
3 years ago
Read 2 more answers
I think I know the answer but I just want to double check.
Levart [38]
D would be the correct answer
8 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
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