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melamori03 [73]
3 years ago
7

100-99+98-97+96-95+...+8-7+6-5+4-3+2-1

Mathematics
2 answers:
lina2011 [118]3 years ago
8 0
100-99+98-97+96-95+...+8-7+6-5+4-3+2-1\\\\=(100-99)+(98-97)+(96-95)+...+(4-3)+(2-1)\\\\=\underbrace{1+1+1+...+1}_{50}=\fbox{50}

sergij07 [2.7K]3 years ago
7 0
x=100-99+98-97+96-95+...+8-7+6-5+4-3+2-1=\\\\=1+1+1+...+1+1+1+1=1\cdot n=n\\\\and\ \ \ 2=100-2\cdot(n-1)\ \ \  \Rightarrow\ \ \ 2n=100\ /:2\ \ \ \Rightarrow\ \ \ n=50\\\\x=50
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\text{Hey there!}

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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]

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

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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

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= 0.2 x 0.4

= 0.08

Part b)

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

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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

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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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