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Pepsi [2]
3 years ago
15

Simplify this complex fraction 4/3/2/5

Mathematics
2 answers:
choli [55]3 years ago
4 0

Answer:

If the question is (4/3) / (2/5), then the answer is 10/3 or 3 and 1/3

Step-by-step explanation:

olga2289 [7]3 years ago
3 0

Answer:

3 1/3

Step-by-step explanation:

You might be interested in
What is the value of 2/3^-4
Reptile [31]

Answer:

\frac{16}{49}

Step-by-step explanation:

To solve problems like this, we need to multiply the base ,\frac{2}{3} , the amount of times as the exponent, 4.

Essentially, the equation is \frac{2}{3} x \frac{2}{3}  x \frac{2}{3}  x \frac{2}{3} .

The product would be \frac{16}{49}.

Now we can't forget that the exponent is a negative. But because the exponent is an even number, we don't need to worry about that.

Hope this helped :)

8 0
3 years ago
7.9 - 4.012. I really don't know the answer
Lunna [17]
3.888 :))))))))))))))))))))))))))
7 0
3 years ago
Use cubes. Draw to show how you make ten. find the sum.
Mila [183]
If you're asking for the model for the second problem (I didn't see any unsolved), here it is!

7 0
3 years ago
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
Help pleeeeeeeeeease ?
cupoosta [38]
The answer is choice D.
Cause it can't be calculate.
5 0
3 years ago
Read 2 more answers
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