Question

Perform each calculation and limit each answer to three significant figures. a) \(67,883 \times 0.004321=?\) b) \(\left(9.67 \times 10^{3}\right) \times 0.0055087=?\)

Short answer

a) 293; b) 53.3

Step-by-step solution

Perform the Multiplication for Part a

To solve the first part, multiply the numbers given: \[67,883 \times 0.004321\]Calculate the product: \[67,883 \times 0.004321 = 293.432563\]

Round the Result for Part a to Three Significant Figures

For the result obtained in Step 1 (293.432563), we must keep only the first three significant digits:- The first three significant digits in 293.432563 are 2, 9, and 3.- The digit following these three is 4, which does not change the last significant figure (3) since it is less than 5.Rounding gives us:\[293\]So, to three significant figures, the answer is 293.

Perform the Multiplication for Part b

Now, solve the second part by multiplying the given numbers: \[(9.67 \times 10^{3}) \times 0.0055087\]First, rewrite \(9.67 \times 10^{3}\) in its expanded form, which is 9670.Calculate the product:\[9670 \times 0.0055087 = 53.267989\]

Round the Result for Part b to Three Significant Figures

For the result from Step 3 (53.267989), limit the value to three significant figures:- The first three significant figures in 53.267989 are 5, 3, and 2.- The digit following the third significant figure is 6, which is 5 or greater, so round up.Rounding gives us:\[53.3\]Therefore, the answer to three significant figures is 53.3.

Key concepts

Rounding

Rounding is a fundamental concept that helps us express numbers more simply by limiting the number of significant figures. This is especially useful when dealing with very large or very small numbers. To round a number:

• Identify how many significant figures you need. For this exercise, it is three.
• Look at the digit immediately after your last significant figure. This digit will tell you whether to round up or stay the same.
• If this digit is 5 or greater, round up the last significant figure by one.
• If it’s less than 5, leave the last significant number as is.

Rounding not only makes numbers easier to read and communicate but also maintains a meaningful level of precision. For example, the number 293.432563 rounded to three significant figures becomes 293. This means you keep the first three significant digits and ignore the rest.

Multiplication

When multiplying numbers, especially those with many digits, it's crucial to follow some steps to ensure accurate results. Here's how you can perform these multiplications:

• First, align the numbers and multiply them as if they were whole numbers.
• Count the total number of decimal places in both numbers being multiplied before starting the multiplication.
• After multiplying, adjust the product by placing the decimal point so that the number of decimal places in the result matches the total counted earlier.

Always multiply carefully to avoid calculation errors. For example, multiplying 67,883 by 0.004321 involves treating 0.004321 as a whole number, and then adjusting the decimal point based on the original figures. Similarly, 9.67 multiplied by 0.0055087 involves first writing 9.67 in expanded form (9670) before multiplying, and then correctly placing the decimal point in the final answer.

Scientific Notation

Scientific notation is a method for expressing very large or small numbers in a more manageable form. It uses powers of ten for simplicity, which makes calculating and comparing such numbers easier. This is particularly useful in scientific contexts.

• To convert a number to scientific notation, move the decimal point so there is one non-zero digit to its left.
• Count how many places you moved the decimal point, which will be the exponent of 10.
• If you moved the decimal to the left, the exponent is positive.
• If you moved the decimal to the right, the exponent is negative.

For instance, the number 9.67 x 10³ is already in scientific notation, representing 9670, which is easier to manipulate when performing calculations. Scientific notation helps with readability and can reduce potential error in large calculations.