Between 1866 and 2006, carbon dioxide (co2) concentration in the atmosphere rose from roughly 262 parts per million to 377 parts per million. assume that this growth can be modeled with an exponential function q=q0x(1+r)^t (the “0”after the “q” is lower than the other parts of the equation)

a. by experimenting with various values of the growth rate r, find an exponential function that fits the data for 1866 and 2006

b. use this exponential model to predict when the co2 concentration will double its 1866 level

A. The growth rate of the CO² concentration in the atmosphere, from 1866 to 2006 was 0.26% annually

B. The CO² concentration in the atmosphere will double its 1866 level after 266 years and 5 months approximately.

Step-by-step explanation:

1. Let's review the information provided to us to answer the question correctly:

Parts per million of CO² concentration in 1866 = 262

Parts per million of CO² concentration in 2006 = 377

Duration of the model = 140 years (2006 - 1866)

2. Let's find the growth rate (r) of this model after 140 years, using the following exponential function:

FV = PV * (1 + r) ⁿ

PV = Parts per million of CO² concentration in 1866 = 262

FV = Parts per million of CO² concentration in 2006 = 377

number of periods (n) = 140 (140 years compounded annually)

Growth rate (r) = r

Replacing with the real values, we have:

377 = 262 * (1 + r) ¹⁴⁰

377/262 = (1 + r) ¹⁴⁰

¹⁴⁰√ 377/262 = 1 + r

r = ¹⁴⁰√ 377/262 - 1

r = 1.002602672   - 1

r = 0.002602672 = 0.26% (Rounding to two decimal places)

The growth rate from 1866 to 2006 was 0.26% annually

3. Use this exponential model to predict when the CO² concentration will double its 1866 level

FV = PV * (1 + r) ⁿ

PV = Parts per million of CO² concentration in 1866 = 262

FV = Parts per million of CO² concentration in ? = 262 * 2 = 524

number of periods (n) = n

Growth rate (r) = 0.26% = 0.0026

Replacing with the real values, we have:

524 = 262 * (1 + 0.0026) ⁿ

524/262 =  (1 + 0.0026) ⁿ

2 = 1.0026ⁿ

n = ㏒ 2/ ㏒ 1.0026

n = 0.30103/0.00113

n = 266.4 (Rounding to the next tenth)

0.4 of a year = 0.4 * 12 = 5 months (Rounding to the next whole)

The CO² concentration in the atmosphere will double its 1866 level after 266 years and 5 months approximately.

d) 30 units^2

area of rectangle = l × b

we know,

length = 10 unit

breadth = 3 unit

applying the formula,

= l × b

= 10 × 3

= 30 unit^2

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a

step-by-step explanation: