E.1 lesson 14 word problems with scientific notation | PDF - Free Printable

Let's go through each problem in Lesson 14: Word Problems in Scientific Notation step by step and verify the solutions. I'll explain the reasoning behind each.

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Problem 1:

> A corporation has $4.8 \times 10^5$ employees. The average annual salary for each of its employees is \$65,100. Approximately how much is the corporation paying its employees each year?

#### Step-by-step solution:

We are given:
- Number of employees: $4.8 \times 10^5$
- Average salary per employee: \$65,100 = $6.51 \times 10^4$

Total cost = (Number of employees) × (Average salary)

$$
(4.8 \times 10^5) \times (6.51 \times 10^4)
$$

Multiply the coefficients:
$$
4.8 \times 6.51 = 31.248
$$

Add the exponents:
$$
10^5 \times 10^4 = 10^{9}
$$

So,
$$
31.248 \times 10^9
$$

Now convert to proper scientific notation:
$$
3.1248 \times 10^{10}
$$

Answer: Approximately $\boxed{3.1 \times 10^{10}}$ dollars per year.

✔ Your answer $3.1 \times 10^{10}$ is correct!

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Problem 2:

> If 446 billion telephone calls were placed by 135 million United States telephone subscribers, what was the average number of calls per subscriber? (Write your answer in scientific notation.)

#### Step-by-step solution:

Given:
- Total calls: 446 billion = $4.46 \times 10^{11}$
- Subscribers: 135 million = $1.35 \times 10^8$

Average calls per subscriber:
$$
\frac{4.46 \times 10^{11}}{1.35 \times 10^8}
$$

Divide coefficients:
$$
\frac{4.46}{1.35} \approx 3.3037
$$

Subtract exponents:
$$
10^{11 - 8} = 10^3
$$

So,
$$
3.3037 \times 10^3 \approx 3.3 \times 10^3
$$

Answer: $\boxed{3.3 \times 10^3}$ calls per subscriber.

✔ Your answer $3.3 \times 10^3$ is correct!

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Problem 3:

> A light year is the distance light travels in a year. One light year is about $5.88 \times 10^{12}$ miles. The sun is about $1.6 \times 10^8$ light years from Proxima Centauri. About how many miles is the sun from that star?

Wait — this seems incorrect. The Sun is not $1.6 \times 10^8$ light years from Proxima Centauri. That’s way too far. Actually, Proxima Centauri is about 4.24 light years away from Earth (and the Sun).

But let’s assume the problem says:

> "The sun is about $1.6 \times 10^8$ light years from Proxima Centauri" — which is likely a typo or fictional scenario.

But let’s follow the math as written.

We are to compute:
$$
\text{Distance} = (\text{light years}) \times (\text{miles per light year})
= (1.6 \times 10^8) \times (5.88 \times 10^{12})
$$

Multiply coefficients:
$$
1.6 \times 5.88 = 9.408
$$

Add exponents:
$$
10^8 \times 10^{12} = 10^{20}
$$

So,
$$
9.408 \times 10^{20}
$$

Rounded to two significant figures:
$$
\boxed{9.4 \times 10^{20}} \text{ miles}
$$

But your answer shows $9.4 \times 10^{20}$ — so you're correct if the distance is indeed $1.6 \times 10^8$ light years.

⚠️ However, this value is astronomically wrong in reality — Proxima Centauri is only ~4.24 light years away. But since this is a math problem, we accept the numbers as given.

✔ Your answer $9.4 \times 10^{20}$ is correct based on the given values.

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Problem 4:

> The density of silver is $1.05 \times 10^4$. The density of calcium is $1.5 \times 10^3$. How many times greater is the density of silver than of calcium?

#### Step-by-step solution:

We divide the density of silver by the density of calcium:

$$
\frac{1.05 \times 10^4}{1.5 \times 10^3}
$$

Divide coefficients:
$$
\frac{1.05}{1.5} = 0.7
$$

Subtract exponents:
$$
10^{4 - 3} = 10^1
$$

So:
$$
0.7 \times 10^1 = 7
$$

Thus, silver is 7 times denser than calcium.

Answer: $\boxed{7}$

✔ Your answer $7$ is correct!

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Problem 5:

> The distance from Earth to the imaginary planet Melo is $1.7 \times 10^9$ miles. If a spaceship is capable of traveling $3.408 \times 10^4$ miles per hour, how many days will it take the spaceship to reach the planet Melo? Round your answer to the nearest day.

#### Step-by-step solution:

We use:
$$
\text{Time} = \frac{\text{Distance}}{\text{Speed}}
$$

$$
= \frac{1.7 \times 10^9}{3.408 \times 10^4}
$$

Divide coefficients:
$$
\frac{1.7}{3.408} \approx 0.499
$$

Subtract exponents:
$$
10^{9 - 4} = 10^5
$$

So:
$$
0.499 \times 10^5 = 4.99 \times 10^4 \text{ hours}
$$

Now convert hours to days:
$$
\frac{4.99 \times 10^4}{24} = \frac{49,900}{24} \approx 2079.17 \text{ days}
$$

Rounded to the nearest day: $\boxed{2079}$ days

But wait — your work shows:

$$
\frac{1.7 \times 10^9}{3.408 \times 10^4} = 0.499 \times 10^5 = 4.99 \times 10^4 \text{ hours}
$$

Then:
$$
\frac{4.99 \times 10^4}{24} = \frac{49,900}{24} = 2079.17 \Rightarrow \boxed{2079} \text{ days}
$$

But you wrote 499 days — which is incorrect.

✘ Your final answer 499 days is wrong.

You forgot to divide by 24 (hours per day). You computed total hours, but the question asks for days.

Let’s fix it:

$$
\text{Time in days} = \frac{4.99 \times 10^4}{24} = \frac{49,900}{24} \approx 2079.17 \approx \boxed{2079} \text{ days}
$$

✔ Corrected answer: $\boxed{2079}$ days

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✔ Final Answers Summary:

1. $3.1 \times 10^{10}$ → ✔️ Correct
2. $3.3 \times 10^3$ → ✔️ Correct
3. $9.4 \times 10^{20}$ → ✔️ Correct (given data)
4. $7$ → ✔️ Correct
5. $499$ → ✘ Incorrect; should be 2079 days

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🔧 Correction for Problem 5:

Corrected Work:

$$
\text{Time (hours)} = \frac{1.7 \times 10^9}{3.408 \times 10^4} = 4.99 \times 10^4 \text{ hours}
$$

$$
\text{Time (days)} = \frac{4.99 \times 10^4}{24} = 2079.17 \approx \boxed{2079} \text{ days}
$$

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✔ Final Answer for Problem 5: 2079 days

Let me know if you'd like this formatted neatly for submission!