The population of a parish is growing with an annual percentage rate compounded continuously. The population reaches 1.1 times its previous size in 2 years. Find the annual percentage rate according to the exponential growth function

Growth rate = ln(1.1)/2

Step-by-step explanation:

Given:

P1 = 1.1 P0

Number of year n = 2

Computation:

P1 = P0[eⁿˣ]

x = growth rate

So,

1.1 P0 = P0[eⁿˣ]

1.1 = eⁿˣ

1.1 = e²ˣ

By taking log

Growth rate = ln(1.1)/2

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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