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Anna35 [415]
3 years ago
8

Of 256 consumers polled, some like only thriller movies, some prefer just comedies, and some like both. If 78 people like only t

hrillers, and 116 like both, how many people prefer only comedies
Mathematics
2 answers:
rodikova [14]3 years ago
7 0
ANSWER

The number of people who like only comedies is

n(C \cup \: T ' ) = 62

EXPLANATION

The total number of consumers polled is

n(C \cup \: T) = 256

The number of consumers who like both Comedies and Thrillers is

n(C \cap \: T) = 116

The number of people who like thrillers is

n(T) = 116 + 78 = 194

The number of people who like comedies can be calculated using the formula,

n(C \cup \: T) = n(C ) + n(T) - n(C \cap \: T)
We substitute the above values to obtain,

256= n(C ) +194- 116

256= n(C ) +78

256 - 78= n(C )

n(C ) = 178

The number of people who like only comedies

n(C \cup \: T ' ) = 178 - 116

n(C \cup \: T ' ) = 62
Margaret [11]3 years ago
5 0

Answer:

62.

Step-by-step explanation:

256-78-116=62

Simple subtraction.

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During optimal conditions, the rate of change of the population of a certain organism is proportional to the population at time
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Answer:

The population is of 500 after 10.22 hours.

Step-by-step explanation:

The rate of change of the population of a certain organism is proportional to the population at time t, in hours.

This means that the population can be modeled by the following differential equation:

\frac{dP}{dt} = Pr

In which r is the growth rate.

Solving by separation of variables, then integrating both sides, we have that:

\frac{dP}{P} = r dt

\int \frac{dP}{P} = \int r dt

\ln{P} = rt + K

Applying the exponential to both sides:

P(t) = Ke^{rt}

In which K is the initial population.

At time t = 0 hours, the population is 300.

This means that K = 300. So

P(t) = 300e^{rt}

At time t = 24 hours, the population is 1000.

This means that P(24) = 1000. We use this to find the growth rate. So

P(t) = 300e^{rt}

1000 = 300e^{24r}

e^{24r} = \frac{1000}{300}

e^{24r} = \frac{10}{3}

\ln{e^{24r}} = \ln{\frac{10}{3}}

24r = \ln{\frac{10}{3}}

r = \frac{\ln{\frac{10}{3}}}{24}

r = 0.05

So

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At what time t is the population 500?

This is t for which P(t) = 500. So

P(t) = 300e^{0.05t}

500 = 300e^{0.05t}

e^{0.05t} = \frac{500}{300}

e^{0.05t} = \frac{5}{3}

\ln{e^{0.05t}} = \ln{\frac{5}{3}}

0.05t = \ln{\frac{5}{3}}

t = \frac{\ln{\frac{5}{3}}}{0.05}

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