Question

current attempt in progress
methane is a greenhouse gas implicated as a contributor to global warming. given below is the table of values of ( q = w(t) ), the atmospheric methane level in parts per billion (ppb) ( t ) years after 1980.

( t )	0	5	10	15	20	25	30

(a) evaluate ( \frac{w(10) - w(0)}{10 - 0} ) and ( \frac{w(15) - w(10)}{15 - 10} ).
round your answers to one decimal place.

( \frac{w(10)-w(0)}{10-0} = )
( \frac{w(15)-w(10)}{15-10} = )

(b) compare your answers in (a). what does this tell you about atmospheric methane levels? enter a year in each box to complete the explanation.

this tells us that the methane level rises at a slower average rate from the years to than from the years to .

national oceanic & atmospheric administration... accessed september 1, 2014.

Explanation:

Part (a)

First Expression: $\boldsymbol{\frac{w(10) - w(0)}{10 - 0}}$

Step1: Identify values

From the table, $w(10) = 1715$ and $w(0) = 1575$.

Step2: Substitute and calculate

$\frac{1715 - 1575}{10 - 0} = \frac{140}{10} = 14.0$

Second Expression: $\boldsymbol{\frac{w(15) - w(10)}{15 - 10}}$

Step1: Identify values

From the table, $w(15) = 1750$ and $w(10) = 1715$.

Step2: Substitute and calculate

$\frac{1750 - 1715}{15 - 10} = \frac{35}{5} = 7.0$

Part (b)

The first rate (14.0) is the average rate from $t = 0$ (year 1980) to $t = 10$ (year 1990). The second rate (7.0) is the average rate from $t = 10$ (year 1990) to $t = 15$ (year 1995). Since $7.0 < 14.0$, the methane level rises at a slower average rate from 1990 to 1995 than from 1980 to 1990.

Answer:

s:
(a) $\frac{w(10)-w(0)}{10-0} = \boldsymbol{14.0}$; $\frac{w(15)-w(10)}{15-10} = \boldsymbol{7.0}$
(b) This tells us that the methane level rises at a slower average rate from the years $\boldsymbol{1990}$ to $\boldsymbol{1995}$ than from the years $\boldsymbol{1980}$ to $\boldsymbol{1990}$.