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slavikrds [6]
3 years ago
11

Simplify y / 3 - x / 4 / x / 6 + y / 8​

Mathematics
1 answer:
s2008m [1.1K]3 years ago
4 0
Wait is there any other information?
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The power supplied to a typical black-andwhite television set is 48.1 W when the set is connected to 72 V. How much electric ene
Mnenie [13.5K]

Answer:

Energy consumed = 1.73× 10⁵ J

Step-by-step explanation:

Given:

Power P = 48.1 W, Time = 1 Hrs= 3600 S

Formula of Power = Energy / Time

⇒ Energy = Power × Time

Energy = 48.1 W × 3600 S = 173,160 Joule = 1.73× 10⁵ J

8 0
4 years ago
Order the fractions from least to greatest.<br> 1<br> -1.25, 0.125,<br> 1<br> 4
liberstina [14]

Answer:

-1.25, -1/4, -1/8, 0.125

Step-by-step explanation:

5 0
3 years ago
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Choose the correct answer to 1,889/36.<br> 42 R17<br> 51 R53<br> 52 R15<br> 52 R17
ozzi

Answer:

52 R17

Step-by-step explanation:

52 * 36 = 1872

1889 - 1872 = 17

--> 52 R17

3 0
3 years ago
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The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes
saul85 [17]

Answer:

a)0.08  , b)0.4  , C) i)0.84  , ii)0.56

Step-by-step explanation:

Given data

P(A) =  professor arrives on time

P(A) = 0.8

P(B) =  Student aarive on time

P(B) = 0.6

According to the question A & B are Independent  

P(A∩B) = P(A) . P(B)

Therefore  

{A}' & {B}' is also independent

{A}' = 1-0.8 = 0.2

{B}' = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B')  (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

P(\frac{B'}{A}) = \frac{P(B'\cap A)}{P(A)} = \frac{0.4\times 0.8}{0.8} = 0.4

Part c)

Assume the events are not independent

Given Data

P(\frac{{A}'}{{B}'}) = 0.4

=\frac{P({A}'\cap {B}')}{P({B}')} = 0.4

P({A}'\cap {B}') = 0.4 x P({B}')

= 0.4 x 0.4 = 0.16

P({A}'\cap {B}') = 0.16

i)

The probability that at least one of them is on time

P(A\cup B) = 1- P({A}'\cap {B}')  

=  1 - 0.16 = 0.84

ii)The probability that they are both on time

P(A\cap  B) = 1 - P({A}'\cup {B}') = 1 - [P({A}')+P({B}') - P({A}'\cap {B}')]

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

6 0
3 years ago
Plz help ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
Deffense [45]
Hi I figured it out for you all you need to do is plug in the X factor .number 3 is y=-1
5 0
3 years ago
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