A quantity with an initial value of 7600 decays exponentially at a rate of 55% every 7 days. What is the value of the quantity after 2 weeks, to the nearest hundredth?

1539

Step-by-step explanation:

We solve the above question using the Exponential decay formula

= A(t) = Ao(1 - r) ^t

Ao = Initial Amount invested = 7600

r = Decay rate = 55% = 0.54

t = time in weeks = 2

Hence:

A(t) = 7600(1 - 0.55)²

A(t) = 7600 × (0.45)²

A(t) = 1539

Therefore, the value of the quantity after 2 weeks is 1539

step-by-step explanation:

a

step-by-step explanation:

when a number is between two straight lines it means the absolute value.

to find absolute value you need to know how many numbers is that number away from 0.

when finding absolute value the number will always come out positive

so absolute value of -3 is 3

absolute value of 3/2 is 3/2

absolute value of -2 is 2

absolute value of 3.5 is 3.5

absolute value of -1 is 1

now put these numbers in order:

a)   |-1|, |3/2|, |-2|, |-3|, |3.5|