convert metric measurements.
make precise and accurate
measurements with various lab
equipment.

vocabulary:
accuracy
precision
meniscus

practice:
3. list the si units for length,
mass, time and temperature.
4. the following data was
collected for a runner’s speed: 4
mi/hr, 3 mi/hr, 6.9 mi/hr, and 7
mi/hr.
a. is this data precise?
explain.
b. if the runner tends to run
5 mi/hr, is this data an
accurate representation of
the runner’s ability?
explain.
5. convert the following values:
a. 150 cm to mm
b. 1,200 ks to s
6. convert the following values:
a. 20°c to °f
b. 300 k to °c
c. 98°f to k

vocabulary:
conversion factors

practice:
7. use dimensional
analysis to convert
the following values.
a. 191 days to
seconds
b. 32 lbs to
grams
c. 21 mm to
yards
8. write the
following values
in scientific
notation.
a. 787,000
b. 4,990,000
c. 0.0413
d. 0.00000011
9. write the
following values
in standard
notation.
a. 6.9 x 10⁻³
b. 7.008 x 10⁵
c. 1.02 x 10²
d. 5.0553 x 10⁻⁴

© it’s not rocket science® 2017

Question

convert metric measurements.
make precise and accurate
measurements with various lab
equipment.

vocabulary:
accuracy
precision
meniscus

practice:
3. list the si units for length,
   mass, time and temperature.
4. the following data was
   collected for a runner’s speed: 4
   mi/hr, 3 mi/hr, 6.9 mi/hr, and 7
   mi/hr.
   a. is this data precise?
      explain.
   b. if the runner tends to run
      5 mi/hr, is this data an
      accurate representation of
      the runner’s ability?
      explain.
5. convert the following values:
   a. 150 cm to mm
   b. 1,200 ks to s
6. convert the following values:
   a. 20°c to °f
   b. 300 k to °c
   c. 98°f to k

vocabulary:
conversion factors

practice:
7. use dimensional
   analysis to convert
   the following values.
   a. 191 days to
      seconds
   b. 32 lbs to
      grams
   c. 21 mm to
      yards
8. write the
   following values
   in scientific
   notation.
   a. 787,000
   b. 4,990,000
   c. 0.0413
   d. 0.00000011
9. write the
   following values
   in standard
   notation.
   a. 6.9 x 10⁻³
   b. 7.008 x 10⁵
   c. 1.02 x 10²
   d. 5.0553 x 10⁻⁴

Accepted Answer

Let's solve some of these problems. Let's start with problem 5a: Convert 150 cm to mm.

# Explanation:
## Step1: Recall the conversion factor
We know that 1 cm = 10 mm. So to convert centimeters to millimeters, we multiply by 10.
## Step2: Perform the conversion
We have 150 cm. So, $ 150 \, \text{cm} \times \frac{10 \, \text{mm}}{1 \, \text{cm}} = 1500 \, \text{mm} $

# Answer:
1500 mm

Now problem 5b: Convert 1,200 ks to s.

# Explanation:
## Step1: Recall the conversion factor
We know that 1 ks (kilosecond) = 1000 s. So to convert kiloseconds to seconds, we multiply by 1000.
## Step2: Perform the conversion
We have 1200 ks. So, $ 1200 \, \text{ks} \times \frac{1000 \, \text{s}}{1 \, \text{ks}} = 1200000 \, \text{s} $

# Answer:
1200000 s

Problem 6a: Convert 20°C to °F.

# Explanation:
## Step1: Recall the formula for converting Celsius to Fahrenheit
The formula is $ F = \frac{9}{5}C + 32 $
## Step2: Substitute C = 20 into the formula
$ F = \frac{9}{5} \times 20 + 32 $
First, calculate $ \frac{9}{5} \times 20 = 36 $
Then, $ 36 + 32 = 68 $

# Answer:
68°F

Problem 6b: Convert 300 K to °C.

# Explanation:
## Step1: Recall the formula for converting Kelvin to Celsius
The formula is $ C = K - 273.15 $
## Step2: Substitute K = 300 into the formula
$ C = 300 - 273.15 = 26.85 $ (we can also use 273 for simplicity, then $ 300 - 273 = 27 $, but 26.85 is more accurate)

# Answer:
26.85°C (or 27°C if using 273)

Problem 6c: Convert 98°F to K.

# Explanation:
## Step1: First convert Fahrenheit to Celsius using $ C = \frac{5}{9}(F - 32) $
Substitute F = 98: $ C = \frac{5}{9}(98 - 32) = \frac{5}{9} \times 66 = \frac{330}{9} \approx 36.67 $
## Step2: Then convert Celsius to Kelvin using $ K = C + 273.15 $
$ K = 36.67 + 273.15 = 309.82 $

# Answer:
309.82 K

Problem 7a: Convert 191 days to seconds.

# Explanation:
## Step1: Recall the conversion factors
1 day = 24 hours, 1 hour = 60 minutes, 1 minute = 60 seconds. So, 1 day = $ 24 \times 60 \times 60 = 86400 $ seconds.
## Step2: Multiply the number of days by the number of seconds in a day
$ 191 \, \text{days} \times 86400 \, \text{s/day} = 191 \times 86400 $
Calculate $ 191 \times 86400 = 16502400 $

# Answer:
16502400 seconds

Problem 7b: Convert 32 lbs to grams.

# Explanation:
## Step1: Recall the conversion factor
1 lb (pound) = 453.592 grams (approximate)
## Step2: Multiply the number of pounds by the conversion factor
$ 32 \, \text{lbs} \times 453.592 \, \text{g/lb} = 32 \times 453.592 = 14514.944 $ grams

# Answer:
14514.944 grams (or approximately 14515 grams)

Problem 7c: Convert 21 mm to yards.

# Explanation:
## Step1: Recall the conversion factors
1 meter = 1000 mm, 1 yard = 0.9144 meters. So first convert mm to meters, then meters to yards.
## Step2: Convert mm to meters
$ 21 \, \text{mm} \times \frac{1 \, \text{m}}{1000 \, \text{mm}} = 0.021 \, \text{m} $
## Step3: Convert meters to yards
$ 0.021 \, \text{m} \times \frac{1 \, \text{yard}}{0.9144 \, \text{m}} \approx 0.02297 $ yards (approximate)

# Answer:
Approximately 0.023 yards (or 0.02297 yards)

Problem 8a: Write 787,000 in scientific notation.

# Explanation:
## Step1: Recall the form of scientific notation
Scientific notation is $ a \times 10^n $, where $ 1 \leq a < 10 $ and $ n $ is an integer.
## Step2: Move the decimal point to get a number between 1 and 10
For 787000, move the decimal point 5 places to the left: $ 7.87 \times 10^5 $

# Answer:
$ 7.87 \times 10^5 $

Problem 8b: Write 4,990,000 in scientific notation.

# Explanation:
## Step1: Recall the form of scientific notation
Scientific notation is $ a \times 10^n $, where $ 1 \leq a < 10 $ and $ n $ is an integer.
## Step2: Move the decimal point to get a number between 1 and 10
For 4990000, move the decimal point 6 places to the left: $ 4.99 \times 10^6 $

# Answer:
$ 4.99 \times 10^6 $

Problem 8c: Write 0.0413 in scientific notation.

# Explanation:
## Step1: Recall the form of scientific notation
Scientific notation is $ a \times 10^n $, where $ 1 \leq a < 10 $ and $ n $ is an integer.
## Step2: Move the decimal point to get a number between 1 and 10
For 0.0413, move the decimal point 2 places to the right: $ 4.13 \times 10^{-2} $

# Answer:
$ 4.13 \times 10^{-2} $

Problem 8d: Write 0.00000011 in scientific notation.

# Explanation:
## Step1: Recall the form of scientific notation
Scientific notation is $ a \times 10^n $, where $ 1 \leq a < 10 $ and $ n $ is an integer.
## Step2: Move the decimal point to get a number between 1 and 10
For 0.00000011, move the decimal point 7 places to the right: $ 1.1 \times 10^{-7} $

# Answer:
$ 1.1 \times 10^{-7} $

Problem 9a: Write $ 6.9 \times 10^{-3} $ in standard notation.

# Explanation:
## Step1: Recall that $ 10^{-3} = \frac{1}{1000} = 0.001 $
## Step2: Multiply 6.9 by $ 10^{-3} $
$ 6.9 \times 0.001 = 0.0069 $

# Answer:
0.0069

Problem 9b: Write $ 7.008 \times 10^6 $ in standard notation.

# Explanation:
## Step1: Recall that $ 10^6 = 1000000 $
## Step2: Multiply 7.008 by $ 10^6 $
$ 7.008 \times 1000000 = 7008000 $

# Answer:
7008000

Problem 9c: Write $ 1.02 \times 10^2 $ in standard notation.

# Explanation:
## Step1: Recall that $ 10^2 = 100 $
## Step2: Multiply 1.02 by $ 10^2 $
$ 1.02 \times 100 = 102 $

# Answer:
102

Problem 9d: Write $ 5.0553 \times 10^{-4} $ in standard notation.

# Explanation:
## Step1: Recall that $ 10^{-4} = \frac{1}{10000} = 0.0001 $
## Step2: Multiply 5.0553 by $ 10^{-4} $
$ 5.0553 \times 0.0001 = 0.00050553 $

# Answer:
0.00050553

Problem 3: List the SI units for length, mass, time and temperature.

# Brief Explanations:
- Length: The SI unit for length is meter (m).
- Mass: The SI unit for mass is kilogram (kg).
- Time: The SI unit for time is second (s).
- Temperature: The SI unit for temperature is Kelvin (K).

# Answer:
- Length: meter (m)
- Mass: kilogram (kg)
- Time: second (s)
- Temperature: Kelvin (K)

Problem 4a: Is the data (4 mi/hr, 3 mi/hr, 6.9 mi/hr, 7 mi/hr) precise? Explain.

# Brief Explanations:
Precision refers to how close the measured values are to each other. Let's check the differences between the values. The values are 3, 4, 6.9, 7. The differences between consecutive values (or between any two) are relatively large (e.g., 3 to 4 is 1, 4 to 6.9 is 2.9, 6.9 to 7 is 0.1). So the data points are not very close to each other, so the data is not precise.

# Answer:
No, the data is not precise. Precision is about the closeness of measured values to each other. The values (3, 4, 6.9, 7 mi/hr) have relatively large differences between them (e.g., 3 to 4 is 1, 4 to 6.9 is 2.9), so they are not close to each other, indicating low precision.

Problem 4b: If the runner tends to run 5 mi/hr, is this data an accurate representation of the runner’s ability? Explain.

# Brief Explanations:
Accuracy refers to how close the measured values are to the true value (here, 5 mi/hr). Let's check the values: 3, 4, 6.9, 7. The values 3 and 4 are below 5, 6.9 and 7 are above 5. The average of the data: (3 + 4 + 6.9 + 7)/4 = (20.9)/4 = 5.225 mi/hr, which is close to 5 mi/hr. But individual values are spread out. However, accuracy is about how close each measurement is to the true value. Some values (3, 4) are far from 5, while others (6.9, 7) are also far. But the average is close. Depending on the context, if we consider the average, it's somewhat accurate, but individual measurements are not. But generally, accuracy is about the closeness of each measurement to the true value. Since some measurements are far (3, 4, 6.9, 7 are not close to 5), the data is not a very accurate representation, but the average is close.

# Answer:
The data is not a very accurate representation of the runner’s ability (to run 5 mi/hr) for individual measurements, but the average (5.225 mi/hr) is close to 5 mi/hr. Accuracy is about how close measured values are to the true value (5 mi/hr). The individual values (3, 4, 6.9, 7 mi/hr) are not close to 5 mi/hr (e.g., 3 is 2 less, 7 is 2 more), so most measurements are not accurate, though the average is somewhat close.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Recall the conversion factor

Step2: Perform the conversion

Step1: Recall the conversion factor

Step2: Perform the conversion

Step1: Recall the formula for converting Celsius to Fahrenheit

Step2: Substitute C = 20 into the formula

Answer: