Currently, the best estimate for the age of the universe is 12.72 giga-years. How did you do?

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Hello, can I get some help with question 6 please?
**Text for an Educational Website:**

---

**Understanding the Age of the Universe through Hubble’s Constant**

6. The units of Hubble's Constant (H) are km/(sec Mpc). Meanwhile, the units of the reciprocal of Hubble's Constant (1/H) are (sec Mpc)/km. Watch the units when you multiply (1/H) by the appropriate conversions...

**Using the conversions from Questions #4, 5:**

\[
\frac{1}{H} \left(\frac{\text{sec Mpc}}{\text{km}}\right) \times \frac{3.08 \times 10^{19} \text{ km}}{1 \text{ Mpc}} \times \frac{1 \text{ Gyr}}{3.15 \times 10^{16} \text{ sec}} = \frac{1}{H} \times \frac{3.08 \times 10^{19}}{3.15 \times 10^{16}} \text{ Gyr}
\]

(Note that the strikeouts showing cancellations have 1, 2, or 3 lines in them to help you see the exact unit cancellations at work.)

Therefore, if you multiply 1/H by the conversion factor given by the second line of the equation above, you get a period of time in giga-years! This is the approximate age of the universe. Calculate the age of the universe, using your value of H.

**Current Estimate:**

Currently, the best estimate for the age of the universe is 12.72 giga-years. How did you do?

---

**Explanation of Diagram:**

The diagram shows a step-by-step conversion of units to calculate the age of the universe. It starts with the reciprocal of Hubble's Constant and uses conversions with different units such as kilometers, mega parsecs, gigayears, and seconds. The result of these conversions is an expression that allows the calculation of the age of the universe in gigayears, demonstrating the importance of unit conversion in astronomical calculations.
Transcribed Image Text:**Text for an Educational Website:** --- **Understanding the Age of the Universe through Hubble’s Constant** 6. The units of Hubble's Constant (H) are km/(sec Mpc). Meanwhile, the units of the reciprocal of Hubble's Constant (1/H) are (sec Mpc)/km. Watch the units when you multiply (1/H) by the appropriate conversions... **Using the conversions from Questions #4, 5:** \[ \frac{1}{H} \left(\frac{\text{sec Mpc}}{\text{km}}\right) \times \frac{3.08 \times 10^{19} \text{ km}}{1 \text{ Mpc}} \times \frac{1 \text{ Gyr}}{3.15 \times 10^{16} \text{ sec}} = \frac{1}{H} \times \frac{3.08 \times 10^{19}}{3.15 \times 10^{16}} \text{ Gyr} \] (Note that the strikeouts showing cancellations have 1, 2, or 3 lines in them to help you see the exact unit cancellations at work.) Therefore, if you multiply 1/H by the conversion factor given by the second line of the equation above, you get a period of time in giga-years! This is the approximate age of the universe. Calculate the age of the universe, using your value of H. **Current Estimate:** Currently, the best estimate for the age of the universe is 12.72 giga-years. How did you do? --- **Explanation of Diagram:** The diagram shows a step-by-step conversion of units to calculate the age of the universe. It starts with the reciprocal of Hubble's Constant and uses conversions with different units such as kilometers, mega parsecs, gigayears, and seconds. The result of these conversions is an expression that allows the calculation of the age of the universe in gigayears, demonstrating the importance of unit conversion in astronomical calculations.
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