Questions: The number of years N(r) since two independently evolving languages split off from a common ancestral language is approximated by N(r)=-5000 ln r Where r is the proportion of the words from the ancestral language that are common to both languages now. a. N(0.9)=530 (Round to the nearest tens place.) b. N(0.5)=3500 (Round to the nearest hundreds place.) c. N(0.3)= (Round to the nearest thousands place.)

The number of years N(r) since two independently evolving languages split off from a common ancestral language is approximated by
N(r)=-5000 ln r

Where r is the proportion of the words from the ancestral language that are common to both languages now.
a. N(0.9)=530 (Round to the nearest tens place.)
b. N(0.5)=3500 (Round to the nearest hundreds place.)
c. N(0.3)= (Round to the nearest thousands place.)

Transcript text: The number of years $N(r)$ since two independently evolving languages split off from a common ancestral language is approximated by \[ N(r)=-5000 \ln r \] Where $r$ is the proportion of the words from the ancestral language that are common to both languages now. a. $\mathrm{N}(0,9)=530$ (Round to the nearest tens place.) b. $\mathrm{N}(0.5)=3500$ (Round to the nearest hundreds place.) c. $N(0.3)=$ $\square$ (Round to the nearest thousands place.)

Solution

Solution Steps

Step 1: Substitute \( r = 0.9 \) into the formula \( N(r) = -5000 \ln r \)

\[ N(0.9) = -5000 \ln(0.9) \]

Step 2: Calculate \( \ln(0.9) \)

\[ \ln(0.9) \approx -0.1053605 \]

Step 3: Multiply by \(-5000\) and round to the nearest tens place

\[ N(0.9) = -5000 \times (-0.1053605) \approx 526.8025 \] \[ N(0.9) \approx 530 \quad \text{(rounded to the nearest tens place)} \]

Step 4: Substitute \( r = 0.5 \) into the formula \( N(r) = -5000 \ln r \)

\[ N(0.5) = -5000 \ln(0.5) \]

Step 5: Calculate \( \ln(0.5) \)

\[ \ln(0.5) \approx -0.693147 \]

Step 6: Multiply by \(-5000\) and round to the nearest hundreds place

\[ N(0.5) = -5000 \times (-0.693147) \approx 3465.735 \] \[ N(0.5) \approx 3500 \quad \text{(rounded to the nearest hundreds place)} \]

Step 7: Substitute \( r = 0.3 \) into the formula \( N(r) = -5000 \ln r \)

\[ N(0.3) = -5000 \ln(0.3) \]

Step 8: Calculate \( \ln(0.3) \)

\[ \ln(0.3) \approx -1.203973 \]

Step 9: Multiply by \(-5000\) and round to the nearest thousands place

\[ N(0.3) = -5000 \times (-1.203973) \approx 6019.865 \] \[ N(0.3) \approx 6000 \quad \text{(rounded to the nearest thousands place)} \]

Final Answer

a. \( \boxed{530} \)
b. \( \boxed{3500} \)
c. \( \boxed{6000} \)

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