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3 years ago

(15-26)/3 + (-8+19)/2

Mathematics

2 answers:

Gre4nikov [31]3 years ago

7 0

Answer:

11/6

Step-by-step explanation:

(15-26)/3 + (-8+19)/2

Do parentheses:

-11/3 + 11/2

Find the common denominator:

-11(2)/3(2) + 11(3)/2(3)

Multiply:

-22/6 + 33/6

Add the fractions:

11/6

Valentin [98]3 years ago

4 0

Answer is 1.83, with the three repeated.

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An experiment starts and the population of a bacteria culture increases by 10% in the first hour. Assume that the population inc

Ne4ueva [31]

Answer:

a) The doubling time is 7.27 hours.

b) The population is 6 hours will be 3,543.

Step-by-step explanation:

The population of the bacteria is modeled by the following equation:

$P(t) = P(0)e^{rt}$

In which P(0) is the initial population, P(t) is the population after t hours and r is the growth rate.

An experiment starts and the population of a bacteria culture increases by 10% in the first hour.

This means that

$P(1) = 1.1P(0)$

Which helps us find r.

$P(t) = P(0)e^{rt}$

$1.1P(0) = P(0)e^{r}$

$e^{r} = 1.1$

Applying ln to both sides

$\ln{e^{r}} = \ln{1.1}$

$r = 0.0953$

$P(t) = P(0)e^{0.0953t}$

a) What is the doubling time?

This is t when $P(t) = 2P(0)$

$P(t) = P(0)e^{0.0953t}$

$2P(0) = P(0)e^{0.0953t}$

$e^{0.0953t} = 2$

Applying ln to both sides

$\ln{e^{0.0953t}} = \ln{2}$

$0.0953t = \ln{2}$

$t = \frac{\ln{2}}{0.0953}$

$t = 7.27$

The doubling time is 7.27 hours.

b) If the initial population is 2,000, what is the population in 6 hours?

This is P(6) when $P(0) = 2000$ . So

$P(t) = P(0)e^{0.0953t}$

$P(6) = 2000e^{0.0953*6} = 3543$

The population is 6 hours will be 3,543.

3 0

4 years ago

-53 + n = -28 solve for n

frozen [14]

Answer:

n will be equal 25

Step-by-step explanation:

-53 + n = -28

collect like terms, and we will have

n = -28 + 53

n = 25

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