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Fofino [41]
3 years ago
14

Algunos números dibisibles entre 3 son divisibles entre 4​

Mathematics
1 answer:
Drupady [299]3 years ago
4 0

Answer:

si

Step-by-step explanation:

por ejemplo 12

12÷3=4. 12÷4= 3.

al hacer la división el resultado es exacto o sea que no te sobra nada y no llegas a un número decimal.

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Answer quick for my hw pls
Readme [11.4K]

Answer:

day 14 , 450 dollars

Step-by-step explanation:

We called x is the number of days

Zippy Rent-a-car: 25x + 100

Speedy Rent-a-car: 30x + 30

25x + 100 = 30x + 30

25x -30x = 30 - 100

-5x = -70

x = 14

5 0
3 years ago
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Point E is located at (2, −3), and point F is located at (−2, −1). Find the point that is 3 over 4 the distance from point E to
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Your answer will be -2+2+4= 4
-3-1+4=0
so it will be 4,0
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According to the scores on the last math test, 80%, or 20, of the students in the class received an A. Find the number of studen
Sladkaya [172]
80% is the same as 4/5, so 4/5 of the class= 20. That means that if there was a picture, 4 parts would be shaded, and 1 would not. 1/4 of 20 is 5, so 1 part= 5 people. 5 x 5 = 25. 25 is the answer. :)
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3 years ago
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The king comes from a family of 2 children. what is the probability that the other child is his sister?
stira [4]
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4 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
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