QuestionA bacteria population grows in a broth at $30^{\circ}$ $\mathrm{C}$ . Given data, find average rates of change for specific time intervals.
(a) From 0 to 4.5 hours: $P(4.5) - P(0) / (4.5 - 0)$ .
(b) From 6.5 to 8 hours: $P(8) - P(6.5) / (8 - 6.5)$ .

Studdy Solution

STEP 1

Assumptions1. The population of bacteria is measured in grams. . The time is measured in hours.
3. The population is a function of time t, denoted as(t).
4. The data provided is accurate and complete.

STEP 2

The average rate of change of a function over an interval [a, b] is given by the formula $Average\, rate\, of\, change = \frac{(b) -(a)}{b - a}$

STEP 3

To find the average rate of change of the population from0 to.5 hours, we can plug in the given values into the formula.
$Average\, rate\, of\, change = \frac{(.5) -(0)}{.5 -0}$

STEP 4

From the table, we know that(0) =0.06 and(4.) =0.19. Substituting these values into the formula, we get $Average\, rate\, of\, change = \frac{0.19 -0.06}{4. -0}$

STEP 5

Calculate the average rate of change.
$Average\, rate\, of\, change = \frac{0.19 -0.06}{4.5} =0.029\, grams/hour$

STEP 6

To find the average rate of change of the population from6.5 to8 hours, we can use the same formula.
$Average\, rate\, of\, change = \frac{(8) -(6.5)}{8 -6.5}$

STEP 7

From the table, we know that(6.5) =0.36 and() =0.49. Substituting these values into the formula, we get $Average\, rate\, of\, change = \frac{0.49 -0.36}{ -6.5}$

STEP 8

Calculate the average rate of change.
$Average\, rate\, of\, change = \frac{0.49 -0.36}{1.5} =0.087\, grams/hour$ On average, the population is increasing at a rate of0.029 gram per hour from0 to4.5 hours and at a rate of0.087 gram per hour from6.5 to8 hours.

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QuestionA bacteria population grows in a broth at $30^{\circ}$ $\mathrm{C}$ . Given data, find average rates of change for specific time intervals.
(a) From 0 to 4.5 hours: $P(4.5) - P(0) / (4.5 - 0)$ .
(b) From 6.5 to 8 hours: $P(8) - P(6.5) / (8 - 6.5)$ .

Studdy Solution

STEP 1

Assumptions1. The population of bacteria is measured in grams. . The time is measured in hours.
3. The population is a function of time t, denoted as(t).
4. The data provided is accurate and complete.

STEP 2

The average rate of change of a function over an interval [a, b] is given by the formula $Average\, rate\, of\, change = \frac{(b) -(a)}{b - a}$

STEP 3

To find the average rate of change of the population from0 to.5 hours, we can plug in the given values into the formula.
$Average\, rate\, of\, change = \frac{(.5) -(0)}{.5 -0}$

STEP 4

From the table, we know that(0) =0.06 and(4.) =0.19. Substituting these values into the formula, we get $Average\, rate\, of\, change = \frac{0.19 -0.06}{4. -0}$

STEP 5

Calculate the average rate of change.
$Average\, rate\, of\, change = \frac{0.19 -0.06}{4.5} =0.029\, grams/hour$

STEP 6

To find the average rate of change of the population from6.5 to8 hours, we can use the same formula.
$Average\, rate\, of\, change = \frac{(8) -(6.5)}{8 -6.5}$

STEP 7

From the table, we know that(6.5) =0.36 and() =0.49. Substituting these values into the formula, we get $Average\, rate\, of\, change = \frac{0.49 -0.36}{ -6.5}$

STEP 8

Calculate the average rate of change.
$Average\, rate\, of\, change = \frac{0.49 -0.36}{1.5} =0.087\, grams/hour$ On average, the population is increasing at a rate of0.029 gram per hour from0 to4.5 hours and at a rate of0.087 gram per hour from6.5 to8 hours.

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Studdy Solution

STEP 1

Assumptions1. The population of bacteria is measured in grams. . The time is measured in hours.3. The population is a function of time t, denoted as(t).4. The data provided is accurate and complete.

STEP 2

The average rate of change of a function over an interval [a, b] is given by the formulaAverage rate of change=(b)−(a)b−aAverage\, rate\, of\, change = \frac{(b) -(a)}{b - a}Averagerateofchange=b−a(b)−(a)​

STEP 3

To find the average rate of change of the population from0 to.5 hours, we can plug in the given values into the formula.Average rate of change=(.5)−(0).5−0Average\, rate\, of\, change = \frac{(.5) -(0)}{.5 -0}Averagerateofchange=.5−0(.5)−(0)​

STEP 4

From the table, we know that(0) =0.06 and(4.) =0.19. Substituting these values into the formula, we getAverage rate of change=0.19−0.064.−0Average\, rate\, of\, change = \frac{0.19 -0.06}{4. -0}Averagerateofchange=4.−00.19−0.06​

STEP 5

Calculate the average rate of change.Average rate of change=0.19−0.064.5=0.029 grams/hourAverage\, rate\, of\, change = \frac{0.19 -0.06}{4.5} =0.029\, grams/hourAveragerateofchange=4.50.19−0.06​=0.029grams/hour

STEP 6

To find the average rate of change of the population from6.5 to8 hours, we can use the same formula.Average rate of change=(8)−(6.5)8−6.5Average\, rate\, of\, change = \frac{(8) -(6.5)}{8 -6.5}Averagerateofchange=8−6.5(8)−(6.5)​

STEP 7

From the table, we know that(6.5) =0.36 and() =0.49. Substituting these values into the formula, we getAverage rate of change=0.49−0.36−6.5Average\, rate\, of\, change = \frac{0.49 -0.36}{ -6.5}Averagerateofchange=−6.50.49−0.36​

STEP 8

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Assumptions1. The population of bacteria is measured in grams. . The time is measured in hours.
3. The population is a function of time t, denoted as(t).
4. The data provided is accurate and complete.

The average rate of change of a function over an interval [a, b] is given by the formula $Average\, rate\, of\, change = \frac{(b) -(a)}{b - a}$

To find the average rate of change of the population from0 to.5 hours, we can plug in the given values into the formula.
$Average\, rate\, of\, change = \frac{(.5) -(0)}{.5 -0}$

From the table, we know that(0) =0.06 and(4.) =0.19. Substituting these values into the formula, we get $Average\, rate\, of\, change = \frac{0.19 -0.06}{4. -0}$

Calculate the average rate of change.
$Average\, rate\, of\, change = \frac{0.19 -0.06}{4.5} =0.029\, grams/hour$

To find the average rate of change of the population from6.5 to8 hours, we can use the same formula.
$Average\, rate\, of\, change = \frac{(8) -(6.5)}{8 -6.5}$

From the table, we know that(6.5) =0.36 and() =0.49. Substituting these values into the formula, we get $Average\, rate\, of\, change = \frac{0.49 -0.36}{ -6.5}$