Questions: Multiple Choice 1 point A scientist has discovered that the temperature in Kelvins of a layer in the earth's atmosphere can be given by T=300(2-0.01 ln (l)), where l is the height in kilometers of the layer to the ground. Find the temperature of a point which is 180 kilometers high. Round to the nearest tenth of a K . -652.1 K 472.8 K 584.4K 45.3 K Multiple Choice 1 point Convert 50 feet to meters and centimeters. 14 m ; 11 cm 15 m ; 24 cm 11 m ; 2 cm 18 m ; 36 cm Multiple Choice 1 point In 2009, 80 frogs were introduced into a marsh. By 2015, the population had grown to 180 frogs. Write an algebraic function N(t) representing the population of frogs N over time t.

Multiple Choice 1 point A scientist has discovered that the temperature in Kelvins of a layer in the earth's atmosphere can be given by T=300(2-0.01 ln (l)), where l is the height in kilometers of the layer to the ground. Find the temperature of a point which is 180 kilometers high. Round to the nearest tenth of a K . -652.1 K 472.8 K 584.4K 45.3 K Multiple Choice 1 point Convert 50 feet to meters and centimeters. 14 m ; 11 cm 15 m ; 24 cm 11 m ; 2 cm 18 m ; 36 cm Multiple Choice 1 point In 2009, 80 frogs were introduced into a marsh. By 2015, the population had grown to 180 frogs. Write an algebraic function N(t) representing the population of frogs N over time t.
Transcript text: Multiple Choice 1 point A scientist has discovered that the temperature in Kelvins of a layer in the earth's atmosphere can be given by $T=300(2-0.01 \ln (l))$, where l is the height in kilometers of the layer to the ground. Find the temperature of a point which is 180 kilometers high. Round to the nearest tenth of a K . $-652.1 \mathrm{~K}$ 472.8 K 584.4K 45.3 K Multiple Choice 1 point Convert 50 feet to meters and centimeters. $14 \mathrm{~m} ; 11 \mathrm{~cm}$ $15 \mathrm{~m} ; 24 \mathrm{~cm}$ $11 \mathrm{~m} ; 2 \mathrm{~cm}$ $18 \mathrm{~m} ; 36 \mathrm{~cm}$ Multiple Choice 1 point In 2009, 80 frogs were introduced into a marsh. By 2015, the population had grown to 180 frogs. Write an algebraic function $N(t)$ representing the population of frogs N over time t.
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Solution

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Solution Steps

Solution Approach

  1. Temperature Calculation: To find the temperature at a height of 180 kilometers, substitute \( l = 180 \) into the given formula \( T = 300(2 - 0.01 \ln(l)) \) and compute the result. Round the result to the nearest tenth.

  2. Conversion from Feet to Meters and Centimeters: Use the conversion factor where 1 foot equals 0.3048 meters. Convert 50 feet to meters, then separate the integer part as meters and convert the decimal part to centimeters.

  3. Algebraic Function for Frog Population: Use the given data points (2009, 80) and (2015, 180) to determine a linear function \( N(t) \) that models the population over time. Calculate the slope and use the point-slope form to find the equation.

Step 1: Temperature Calculation

To find the temperature \( T \) at a height of \( l = 180 \) km, we use the formula: \[ T = 300(2 - 0.01 \ln(l)) \] Substituting \( l = 180 \): \[ T = 300(2 - 0.01 \ln(180)) \approx 584.4 \, K \]

Step 2: Conversion from Feet to Meters and Centimeters

To convert \( 50 \) feet to meters, we use the conversion factor \( 1 \, \text{foot} = 0.3048 \, \text{meters} \): \[ \text{meters} = 50 \times 0.3048 \approx 15.24 \, \text{m} \] Separating the integer part gives \( 15 \, \text{m} \) and converting the decimal part to centimeters: \[ \text{centimeters} = (0.24 \, \text{m}) \times 100 \approx 24 \, \text{cm} \]

Step 3: Algebraic Function for Frog Population

Using the data points \( (2009, 80) \) and \( (2015, 180) \), we calculate the slope \( m \): \[ m = \frac{180 - 80}{2015 - 2009} = \frac{100}{6} \approx 16.67 \] The population function \( N(t) \) can be expressed as: \[ N(t) = 16.67(t - 2009) + 80 \] For \( t = 2010 \): \[ N(2010) \approx 96.67 \]

Final Answer

The results are as follows:

  1. Temperature at \( 180 \) km: \( \boxed{584.4 \, K} \)
  2. Conversion of \( 50 \) feet: \( \boxed{15 \, m; \, 24 \, cm} \)
  3. Estimated population in \( 2010 \): \( \boxed{96.67} \) frogs
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