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Sagot :
### Part (a)
First, let's calculate the natural logarithm ([tex]\(\ln t\)[/tex]) for each value of [tex]\(t\)[/tex] using the given time intervals in minutes.
[tex]\[ \begin{aligned} &\ln 5 \approx 1.6094 \\ &\ln 15 \approx 2.7081 \\ &\ln 30 \approx 3.4012 \\ &\ln 60 \approx 4.0943 \\ &\ln 120 \approx 4.7875 \\ &\ln 240 \approx 5.4806 \\ &\ln 480 \approx 6.1738 \\ &\ln 720 \approx 6.5793 \\ &\ln 1440 \approx 7.2724 \\ &\ln 2880 \approx 7.9655 \\ &\ln 5760 \approx 8.6587 \\ &\ln 10080 \approx 9.2183 \\ \end{aligned} \][/tex]
After converting all [tex]\(t\)[/tex] values to [tex]\(\ln t\)[/tex] values, we proceed with linear regression to estimate the coefficients [tex]\(a\)[/tex] and [tex]\(b\)[/tex] for the linear model [tex]\(P = a \ln t + b\)[/tex].
From the linear regression, we find:
[tex]\[ a \approx -7.7867 \][/tex]
[tex]\[ b \approx 86.2831 \][/tex]
So, the values are:
[tex]\[ a = -7.7867 \][/tex]
[tex]\[ b = 86.2831 \][/tex]
### Part (b)
We need to find the time [tex]\(t\)[/tex] when [tex]\(P = 0\)[/tex]. We solve the equation [tex]\(0 = a \ln t + b\)[/tex]:
[tex]\[ 0 = -7.7867 \ln t + 86.2831 \][/tex]
Solving for [tex]\(\ln t\)[/tex]:
[tex]\[ \ln t = \frac{-b}{a} = \frac{-86.2831}{-7.7867} \approx 11.0796 \][/tex]
Now, converting [tex]\(\ln t\)[/tex] back to [tex]\(t\)[/tex]:
[tex]\[ t = e^{11.0796} \approx 64318.3 \text{ minutes} \][/tex]
To convert this time into days:
[tex]\[ \text{Days} = \frac{64318.3}{1440} \approx 45.0822 \][/tex]
Therefore, the model predicts that the subjects will recognize no words in about [tex]\(45.0822\)[/tex] days.
### Part (c)
To find the time [tex]\(t\)[/tex] when [tex]\(P = 100\)[/tex], solve the equation [tex]\(100 = a \ln t + b\)[/tex]:
[tex]\[ 100 = -7.7867 \ln t + 86.2831 \][/tex]
Solving for [tex]\(\ln t\)[/tex]:
[tex]\[ \ln t = \frac{100 - b}{a} = \frac{100 - 86.2831}{-7.7867} \approx -1.7645 \][/tex]
Now, converting [tex]\(\ln t\)[/tex] back to [tex]\(t\)[/tex]:
[tex]\[ t = e^{-1.7645} \approx 0.1718 \text{ minutes} \][/tex]
To convert this time into seconds:
[tex]\[ \text{Seconds} = 0.1718 \times 60 \approx 10.3062 \][/tex]
Therefore, the model predicts that the subjects will recognize all words in about [tex]\(10.3062\)[/tex] seconds.
First, let's calculate the natural logarithm ([tex]\(\ln t\)[/tex]) for each value of [tex]\(t\)[/tex] using the given time intervals in minutes.
[tex]\[ \begin{aligned} &\ln 5 \approx 1.6094 \\ &\ln 15 \approx 2.7081 \\ &\ln 30 \approx 3.4012 \\ &\ln 60 \approx 4.0943 \\ &\ln 120 \approx 4.7875 \\ &\ln 240 \approx 5.4806 \\ &\ln 480 \approx 6.1738 \\ &\ln 720 \approx 6.5793 \\ &\ln 1440 \approx 7.2724 \\ &\ln 2880 \approx 7.9655 \\ &\ln 5760 \approx 8.6587 \\ &\ln 10080 \approx 9.2183 \\ \end{aligned} \][/tex]
After converting all [tex]\(t\)[/tex] values to [tex]\(\ln t\)[/tex] values, we proceed with linear regression to estimate the coefficients [tex]\(a\)[/tex] and [tex]\(b\)[/tex] for the linear model [tex]\(P = a \ln t + b\)[/tex].
From the linear regression, we find:
[tex]\[ a \approx -7.7867 \][/tex]
[tex]\[ b \approx 86.2831 \][/tex]
So, the values are:
[tex]\[ a = -7.7867 \][/tex]
[tex]\[ b = 86.2831 \][/tex]
### Part (b)
We need to find the time [tex]\(t\)[/tex] when [tex]\(P = 0\)[/tex]. We solve the equation [tex]\(0 = a \ln t + b\)[/tex]:
[tex]\[ 0 = -7.7867 \ln t + 86.2831 \][/tex]
Solving for [tex]\(\ln t\)[/tex]:
[tex]\[ \ln t = \frac{-b}{a} = \frac{-86.2831}{-7.7867} \approx 11.0796 \][/tex]
Now, converting [tex]\(\ln t\)[/tex] back to [tex]\(t\)[/tex]:
[tex]\[ t = e^{11.0796} \approx 64318.3 \text{ minutes} \][/tex]
To convert this time into days:
[tex]\[ \text{Days} = \frac{64318.3}{1440} \approx 45.0822 \][/tex]
Therefore, the model predicts that the subjects will recognize no words in about [tex]\(45.0822\)[/tex] days.
### Part (c)
To find the time [tex]\(t\)[/tex] when [tex]\(P = 100\)[/tex], solve the equation [tex]\(100 = a \ln t + b\)[/tex]:
[tex]\[ 100 = -7.7867 \ln t + 86.2831 \][/tex]
Solving for [tex]\(\ln t\)[/tex]:
[tex]\[ \ln t = \frac{100 - b}{a} = \frac{100 - 86.2831}{-7.7867} \approx -1.7645 \][/tex]
Now, converting [tex]\(\ln t\)[/tex] back to [tex]\(t\)[/tex]:
[tex]\[ t = e^{-1.7645} \approx 0.1718 \text{ minutes} \][/tex]
To convert this time into seconds:
[tex]\[ \text{Seconds} = 0.1718 \times 60 \approx 10.3062 \][/tex]
Therefore, the model predicts that the subjects will recognize all words in about [tex]\(10.3062\)[/tex] seconds.
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