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Olegator [25]
3 years ago
7

Theo made a donation to a charity. His grandfather agreed to add $4.00 to Theo's donation amount and then donate half of that su

m. Theo's grandfather donated $4.25. Write and solve an equation to find the amount of Theo's donation
Mathematics
2 answers:
miss Akunina [59]3 years ago
8 0

So you have to solve for "donation" - so let's call that X. And the things we know are what the grandfather donated, as a function - meaning based upon - Theo's donation. We'll call that Y.

So Y is equal to 1/2 of $4 plus Theo's donation...... 0.5*(x+4).

So... 4.25 = 0.5(X+4), solving for X

4.25=0.5x + 2

-2            -2

2.25=0.5x - divide each side by 0.5 to get x alone.

x = 4.50

Plugging back in:

(4.50 + 4.00)*0.5 = 4.25!

ValentinkaMS [17]3 years ago
8 0

Answer:i don't understand

Step-by-step explanation:

I need an equation plz.

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Answer:

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Step-by-step explanation:

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3 years ago
Makenzie leaves her home at 10am. She has to be at her grandmother's house, which is 42 miles away. The
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no

Step-by-step explanation:

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the scale of a map is 1 cm : 72km. What is the actual distance between two towns that are 4 cm apart on the map
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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

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3 years ago
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If it was just the splitting It would be
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