Question #5d92a

3 Answers
Jun 23, 2017

Answer:

#4 s.f#

#1,150 xx 10^6#

Explanation:

#((29-20.2) xx (1.79xx10^5))/(1.37)#

Solution

#(8.8 xx 1.79 xx 10^5)/1.37#

#(1,575,200)/1.37#

#1,149.781.022#

#1,150,000 = 4 s.f#

#1,150 xx 10^6#

Jul 13, 2017

Answer:

#1.15 xx 10^6# to 3 significant figures.

Explanation:

If you use the decimals as they are given and calculate the answer, you will get an exact result:

#((29-20.2) xx 1.79xx10^5)/(1.37)#

#= 1,149,781.022#

This answer has #10# significant figures.

However, such as answer is not practical. There are usually guidelines given as to the accuracy of the answer.
This also varies from subject to subject.

The greatest level of accuracy for any of the numbers is to #2# decimal places. But in comparison to #1.79 xx 10^5,# 2 decimal places has no significance.

#= 1,149,781.022 " "# is given to #10# sig figs.

#= 1,149,781" "#is given to the nearest integer.

#= 1,150,000 " "# is given to #4# sig figs (#3# would be the same)
This would be my choice as it is of a similar size to the biggest value.

The most appropriate answer for the situation should be given.

The answer should be in the same format as the question, so we can write it in scientific notation as #1.15 xx 10^6#

Jul 15, 2017

Answer:

#1.1xx10^6#
Two

Explanation:

As we observe the expression has three factors.

  1. #(29-20.2)#
    It consists of two numbers
    (a) #29-># has two significant figures and
    (b) #20.2-># has three significant figures.
    Since both are stated within parenthesis their final result has Two Significant Figures.

  2. #1.79xx10^5#
    Once we write a number in scientific notation, the number of significant figures is taken from the non-exponent number part.
    We see that in this number it is
    #1.79-># has three significant figures.

  3. #1.37#
    This has three significant figures.

As such, the final result must be stated in scientific notation with two significant figures (least of all).

The given expression #=1149781.02#

In scientific notation with accuracy of two significant figures final result is #=1.1xx10^6#