Someone , can’t seem to get the problems

A) 525,500

B) decreasing by 0.995% per year

C) 430,243

D) After 20 years, the population can be expected to be about 20% smaller.

E) 2009

Step-by-step explanation:

A) t=0 represents the year 2000, so put 0 where t is in the expression and evaluate it. Of course, e^0 = 1, so the y-value is 525.5 thousand, or 525,500.

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B) Each year, the population is multiplied by e^-0.01 ≈ 0.99004983, or about 1 - 0.995%. That is, the population is decreasing by 0.995% per year.

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C) t represents the number of years since 2000, so the year 2020 is represented by t=20. Put that value in the equation and do the arithmetic.

y = 525.5·e^(-0.01·20) = 525.5·e^-0.2 ≈ 430.243 . . . . thousands

The population in 2020 is predicted to be 430,243.

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D) The decrease is about 1% per year, so a rough estimate of the decrease over 20 years is 20%. The population of about 500,000 will decrease by about 100,000 in that time period, so will be about 400,000. The value we calculated is in that ballpark. (The actual decrease is about 18.13%; or about 95.2 thousand.)

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E) Your working shows the general idea, but you need to remember the numbers in the equation are thousands:

480 = 525.5·e^(-0.01t)

0.913416 = e^(-0.01t) . . . . divide by 525.5

ln(0.913416) = -0.01·t . . . . take the natural log

-100ln(0.913416) = t ≈ 9.06

The population will be 480 thousand after 9 .06 years, in the year 2009.

b. x + 1

step-by-step explanation:

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