Using a number cube and this hat with 5 different-colored marbles, which simulation would help you answer this question? During
the first 70 days of school, Becca’s science class does a lab experiment 2/3 of the time. The class meets Monday through Friday. If you choose a class at random, what are the chances that Becca did a lab experiment on a Friday?
1 answer:
Answer:
see the attachment
Step-by-step explanation:
We assume that the question is interested in the probability that a randomly chosen class is a Friday class with a lab experiment (2/15). That is somewhat different from the probability that a lab experiment is conducted on a Friday (2/3).
Based on our assumption, we want to create a simulation that includes a 1/5 chance of the day being a Friday, along with a 2/3 chance that the class has a lab experiment on whatever day it is.
That simulation can consist of choosing 1 of 5 differently-colored marbles, and rolling a 6-sided die with 2/3 of the numbers being designated as representing a lab-experiment day. (The marble must be replaced and the marbles stirred for the next trial.) For our purpose, we can designate the yellow marble as "Friday", and numbers greater than 2 as "lab-experiment".
The simulation of 70 different choices of a random class is shown in the attachment.
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<em>Comment on the question</em>
IMO, the use of <em>70 trials</em> is coincidentally the same number as the first <em>70 days</em> of school. The calendar is deterministic, so there will be exactly 14 Fridays in that period. If, in 70 draws, you get 16 yellow marbles, you cannot say, "the probability of a Friday is 16/70." You need to be very careful to properly state the question you're trying to answer.
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Refer to the diagram shown below.
Given:
m∠A = 19°
c = 15
By definition,
sin A = a/c
Therefore
a = c*sin A = 15*sin(19°) = 4.8835
cos A = b/c
Therefore
b = c*cos A = 15*cos(19°) =14.1828
Answer:
The lengths are 4.88, 14.18, and 15.00 (nearest hundredth)
Given:
m∠A = 19°
c = 15
By definition,
sin A = a/c
Therefore
a = c*sin A = 15*sin(19°) = 4.8835
cos A = b/c
Therefore
b = c*cos A = 15*cos(19°) =14.1828
Answer:
The lengths are 4.88, 14.18, and 15.00 (nearest hundredth)
The number of bins in the histogram determines the width of the bins and
the frequency (number of members) in a bin.
- Please find attached the graph of the histogram of Gretchen's data created with MS Excel.
Given:
The table from Gretchen's data is presented as follows;
The required number of bins = 8 bins
The steps to create the histogram using MS Excel;
- Enter the 15 dataset for the number of days as presented in the above table in a column on the worksheet.
- Select the dataset entered into the cells from above then select the Insert menu.
- In the Chart section of the insert menu click on the Histogram icon to view the quick options for histograms type charts.
- Select the Histogram option from the drop down menu from the quick options for histograms, to create an histogram chart.
- Right-click the histogram created from the previous step, then select Format Chart Area from the pop up menu.
- In the Format Chart Area task pane, click on the arrow next to Chart Options, then select Horizontal Axis from the options menu.
- Click on the histogram icon under Axis Options, then click on arrow button besides Axis Option sub menu, to display the axis options.
- Click on the Number of bins radio button then enter 8, to create an histogram having 8 bins.
The histogram of Gretchen's data having eight bins created with MS Excel is attached.
Learn more about creating a histogram chart here:
