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kupik [55]
2 years ago
5

John bought 3 pencils for $0.75 each and a notebook for $4.25. If the tax rate is 5%, what is the total amount of John's purchas

e including the tax?
Mathematics
1 answer:
SIZIF [17.4K]2 years ago
7 0

Answer:

$6.83

Step-by-step explanation:

First calculate the before-tax total amount:

$0.75 + $0.75 + $0.75 + $4.25, or

3($0.75) + $4.25 = $2.25 + $4.25 = $6.50    (total before tax)

The before-tax total PLUS the 5% tax comes to:

$6.50 + 0.05($6.50) = $6.50 + $0.33  =  $6.83

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Find the perimeter of 24+2×+11 and 6×-8+7×-19+24​
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x+3y

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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
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