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Mila [183]
2 years ago
6

Kevin studied the effects of video games on antisocial behavior. Which of these is a response variable?

Mathematics
1 answer:
Flura [38]2 years ago
7 0

Answer:

antisocial behavior

Step-by-step explanation:

its the response variable due to it being caused by video games

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Customers arrive at a grocery store at an average of 2.1 per minute. Assume that the number of arrivals in a minute follows the
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Answer:

a)  P(2)=0.270

b) P(X>3)=0.605

c)  P=0.410

Step-by-step explanation:

We know that customers arrive at a grocery store at an average of 2.1 per minute. We use the  Poisson distribution:

\boxed{P(k)=\frac{\lambda^k \cdot e^{-\lambda}}{k!}}

a)  In this case: \lambda=2.1

P(2)=\frac{2.1^2 \cdot e^{-2.1}}{2}\\\\P(2)=0.270

Therefore, the probability is P(2)=0.270.

b)  In this case: \lambda=2\cdot 2.1=4.2

P(X>3)=1-P(X\leq 3)\\\\P(X>3)=1-\sum_{x=0}^3 \frac{4.2^x \cdot e^{-4.2}}{x!}\\\\P(X>3)=1-0.395\\\\P(X>3)=0.605

Therefore, the probability is P(X>3)=0.605.

c)  We know that two customers came in in the first minute. That is why we calculate the probability of at least 5 customers entering the other 2 minutes.

In this case: \lambda=2\cdot 2.1=4.2

P(X\geq 5)=1-P(X

Therefore, the probability is P=0.410.

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You have $5,000 to invest for ten years. The Bank pays simple interest at an annual rate of 5%. Calculate your balance after 10
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Answer: $7,500

Step-by-step explanation:

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SI = 5000(1 + 0.05[10])

SI = 5000 + 2500

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After 10 years, your balance should be $7,500

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The ​half-life of a radioactive element is 130​ days, but your sample will not be useful to you after​ 80% of the radioactive nu
gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

\frac{dN}{dt}\propto -N

\Rightarrow \frac{dN}{dt} =-\lambda N

\Rightarrow \frac{dN}{N} =-\lambda dt

Integrating both sides

\int \frac{dN}{N} =\int\lambda dt

\Rightarrow ln|N|=-\lambda t+c

When t=0, N=N_0 = initial amount

\Rightarrow ln|N_0|=-\lambda .0+c

\Rightarrow c= ln|N_0|

\therefore ln|N|=-\lambda t+ln|N_0|

\Rightarrow ln|N|-ln|N_0|=-\lambda t

\Rightarrow ln|\frac{N}{N_0}|=-\lambda t.......(1)

                            \frac{N}{N_0}=e^{-\lambda t}.........(2)

Logarithm:

  • ln|\frac mn|= ln|m|-ln|n|
  • ln|ab|=ln|a|+ln|b|
  • ln|e^a|=a
  • ln|a|=b \Rightarrow a=e^b
  • ln|1|=0

130 days is the half-life of the given radioactive element.

For half life,

N=\frac12 N_0,  t=t_\frac12=130 days.

we plug all values in equation (1)

ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130

\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130

\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130

\rightarrow -ln|2|=-\lambda \times 130

\rightarrow \lambda= \frac{-ln|2|}{-130}

\rightarrow \lambda= \frac{ln|2|}{130}

We need to find the time when the sample remains 80% of its original.

N=\frac{80}{100}N_0

\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t

\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t

\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t

\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}

\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}

\Rightarrow t\approx 42

We can use the sample about 42 days.

7 0
3 years ago
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