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Vikentia [17]
3 years ago
12

Express 16 = 2x as a logarithmic equation.

Mathematics
1 answer:
Dmitrij [34]3 years ago
8 0
If you meant
16=2ˣ then
remember
y=a^x translates to log_ay=x

so
16=2^x translates to log_216=x
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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
These are the values in Ariel’s data set. (1, 67), (3, 88), (5,97), (6, 101), (8, 115) Ariel determines the equation of a linear
notsponge [240]
(3,5)
(1,-3)
(5,1)
(6,-2)
(8,-1)
5 0
3 years ago
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bulgar [2K]

Answer:

1. 11/12

2. 1/2

3. 2/3

4. 3/4

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6. 4/5

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Step-by-step explanation:

1. 11/12

2. 5/10= 1/2

3. 4/6=2/3

4. 6/8=3/4

5. 3/3=1

6. 4/5

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jok3333 [9.3K]
P(get a package of cupcakes) = 11/12


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Arlecino [84]

Answer:

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4 0
2 years ago
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