Answered

IDNLearn.com makes it easy to find answers and share knowledge with others. Join our community to access reliable and comprehensive responses to your questions from experienced professionals.

The number of species in a sample is given by [tex]\( S(n)=a \ln \left(1+\frac{n}{a}\right) \)[/tex]. Here, [tex]\( n \)[/tex] is the number of individuals in the sample, and [tex]\( a \)[/tex] is a constant that indicates the diversity of species in the community. If [tex]\( a=0.48 \)[/tex], find [tex]\( S(n) \)[/tex] for each value of [tex]\( n \)[/tex].

1. Find [tex]\( S(n) \)[/tex] for [tex]\( n=100 \)[/tex].
[tex]\[
S(100)= \square \quad \text{(Round to the nearest whole number as needed.)}
\][/tex]

2. Find [tex]\( S(n) \)[/tex] for [tex]\( n=200 \)[/tex].
[tex]\[
S(200)= \square \quad \text{(Round to the nearest whole number as needed.)}
\][/tex]

3. Find [tex]\( S(n) \)[/tex] for [tex]\( n=150 \)[/tex].
[tex]\[
S(150)= \square \quad \text{(Round to the nearest whole number as needed.)}
\][/tex]

4. Find [tex]\( S(n) \)[/tex] for [tex]\( n=10 \)[/tex].
[tex]\[
S(10)= \square \quad \text{(Round to the nearest whole number as needed.)}
\][/tex]

Sagot :

GinnyAnsweridn
To solve for the number of species in a sample using the given equation [tex]\( S(n) = a \ln \left( 1 + \frac{n}{a} \right) \)[/tex] where [tex]\(a = 0.48\)[/tex], let's calculate [tex]\( S(n) \)[/tex] for the specified values of [tex]\( n \)[/tex] step by step.

### Step-by-Step Calculations

#### Find [tex]\( S(100) \)[/tex]

1. Substitute [tex]\( n = 100 \)[/tex] and [tex]\( a = 0.48 \)[/tex] into the equation:
[tex]\[ S(100) = 0.48 \ln \left( 1 + \frac{100}{0.48} \right) \][/tex]
2. Calculate [tex]\( \frac{100}{0.48} \)[/tex]:
[tex]\[ \frac{100}{0.48} \approx 208.33 \][/tex]
3. Add 1 to the result:
[tex]\[ 1 + 208.33 \approx 209.33 \][/tex]
4. Calculate the natural logarithm of 209.33:
[tex]\[ \ln(209.33) \approx 5.34 \][/tex]
5. Multiply by the constant [tex]\( 0.48 \)[/tex]:
[tex]\[ 0.48 \times 5.34 \approx 2.56 \][/tex]
6. Round to the nearest whole number:
[tex]\[ S(100) \approx 3 \][/tex]

#### Find [tex]\( S(200) \)[/tex]

1. Substitute [tex]\( n = 200 \)[/tex] and [tex]\( a = 0.48 \)[/tex] into the equation:
[tex]\[ S(200) = 0.48 \ln \left( 1 + \frac{200}{0.48} \right) \][/tex]
2. Calculate [tex]\( \frac{200}{0.48} \)[/tex]:
[tex]\[ \frac{200}{0.48} \approx 416.67 \][/tex]
3. Add 1 to the result:
[tex]\[ 1 + 416.67 \approx 417.67 \][/tex]
4. Calculate the natural logarithm of 417.67:
[tex]\[ \ln(417.67) \approx 6.03 \][/tex]
5. Multiply by the constant [tex]\( 0.48 \)[/tex]:
[tex]\[ 0.48 \times 6.03 \approx 2.89 \][/tex]
6. Round to the nearest whole number:
[tex]\[ S(200) \approx 3 \][/tex]

#### Find [tex]\( S(150) \)[/tex]

1. Substitute [tex]\( n = 150 \)[/tex] and [tex]\( a = 0.48 \)[/tex] into the equation:
[tex]\[ S(150) = 0.48 \ln \left( 1 + \frac{150}{0.48} \right) \][/tex]
2. Calculate [tex]\( \frac{150}{0.48} \)[/tex]:
[tex]\[ \frac{150}{0.48} \approx 312.5 \][/tex]
3. Add 1 to the result:
[tex]\[ 1 + 312.5 \approx 313.5 \][/tex]
4. Calculate the natural logarithm of 313.5:
[tex]\[ \ln(313.5) \approx 5.75 \][/tex]
5. Multiply by the constant [tex]\( 0.48 \)[/tex]:
[tex]\[ 0.48 \times 5.75 \approx 2.76 \][/tex]
6. Round to the nearest whole number:
[tex]\[ S(150) \approx 3 \][/tex]

#### Find [tex]\( S(10) \)[/tex]

1. Substitute [tex]\( n = 10 \)[/tex] and [tex]\( a = 0.48 \)[/tex] into the equation:
[tex]\[ S(10) = 0.48 \ln \left( 1 + \frac{10}{0.48} \right) \][/tex]
2. Calculate [tex]\( \frac{10}{0.48} \)[/tex]:
[tex]\[ \frac{10}{0.48} \approx 20.83 \][/tex]
3. Add 1 to the result:
[tex]\[ 1 + 20.83 \approx 21.83 \][/tex]
4. Calculate the natural logarithm of 21.83:
[tex]\[ \ln(21.83) \approx 3.08 \][/tex]
5. Multiply by the constant [tex]\( 0.48 \)[/tex]:
[tex]\[ 0.48 \times 3.08 \approx 1.48 \][/tex]
6. Round to the nearest whole number:
[tex]\[ S(10) \approx 1 \][/tex]

### Final Results

- [tex]\( S(100) = 3 \)[/tex]
- [tex]\( S(200) = 3 \)[/tex]
- [tex]\( S(150) = 3 \)[/tex]
- [tex]\( S(10) = 1 \)[/tex]
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. IDNLearn.com is your reliable source for accurate answers. Thank you for visiting, and we hope to assist you again.