There were 200,000 animals of a certain species in 1980. Since then, this number has decreased by 4.5% each year. Approximately how many animals of this species will be left in 2025?

Approximately 25187 animals of this species will be left in 2025

Step-by-step explanation:

We would apply the formula for exponential decay which is expressed as

y = b(1 - r)^x

Where

y represents the population of animals after x years.

x represents the number of years.

b represents the initial population of animals.

r represents rate of decay.

From the information given,

b = 200000

r = 4.5% = 4.5/100 = 0.045

x = 2025 - 1980 = 45 years

Therefore,

y = 200000(1 - 0.045)^45

y = 200000(0.955)^45

y = 25187

There'll be approximately 25187.3059 animals of this species in 2025.

Step-by-step explanation:

In this case we have a compounded interest problem, but the interest rate is negative, since the number will be decreasing. To solve it we can use the compound interest formula shown bellow:

M = C*(1 + r)^(t)

Where M is the final amount, C is the initial amount, r is the interest rate and t is the time elapsed.

In this case the time elapsed was from 1980 to 2025, so 45 years. Applying the data to the formula gives us:

M = 200000*(1 + (-0.045))^(45)

M = 200000*(1 - 0.045)^(45)

M = 200000*(0.955)^(45)

M = 25187.3059

There'll be approximately 25187.3059 animals of this species in 2025.

$53.50

step-by-step explanation:

you have to subtract the amount of money chuck earned ($72.50) by the cost of the gas for the mower ($11.25) and the new string for the trimmer he bought ($7.75).

72.50-11.25=61.25

and then subtract $61.25 by the second thing he bought (the new string for the trimmer)

61.25-7.75=53.5

so in total it's $53.50 (53 dollars and 50 cents)

exactly you go it

step-by-step explanation: