Answer: Larger the MAD tells us that the ages of swimmer are far from the mean age. Thus means is not a relevant indicator for the data.
Step-by-step explanation:
We know that the mean absolute deviation (MAD) helps to know whether the mean of a data is a worthy indicator for the data values .
The larger the MAD tells us the values are spread out far from the mean .
Also, larger the MAD makes the mean less worthy as an indicator of the data elements within the data set.
Answer:
The correct options are;
B) The mean and median for the security company are both lower than the mean and the median for the collections performed by other companies
B) Since the security company appear to have collected lower revenue than other companies, there is some evidence of stealing by the security company's employees
Step-by-step explanation:
We use the acronym SC for the security company and OC for the other company
SC OC
1.6 1.5
1.8 2.1
1.6 1.9
1.8 2.2
1.7 1.9
1.2 1.7
1.1 2.1
1.2 2.2
1.2 2.2
1.5 1.8
∑x 14.7 19.6
The mean is given by ∑x/n
n = 10
SC OC
∴ Mean 1.47 1.96
The median is the
Hence
SC OC
Median 1.55 2
Therefore, the mean and median for the security company are both lower than the mean and the median for the collections performed by other companies.
b) Since the security company appear to have collected lower revenue than other companies, there is some evidence of stealing by the security company's employees.
Answer:
the LCM would be 8 based on the following set of multiples: Multiples of 2: 2, 4, 6, 8, 10, 12, 14, ... Multiples of 8: 8, 16, 24, 32, 40, 48, 56, ...
Step-by-step explanation:
Answer:
Approximately 3 grams left.
Step-by-step explanation:
We will utilize the standard form of an exponential function, given by:

In the case of half-life, our rate <em>r</em> will be 1/2. This is because 1/2 or 50% will be left after <em>t </em>half-lives.
Our initial amount <em>a </em>is 185 grams.
So, by substitution, we have:

Where <em>f(t)</em> denotes the amount of grams left after <em>t</em> half-lives.
We want to find the amount left after 6 half-lives. Therefore, <em>t </em>= 6. Then using our function, we acquire:

Evaluate:

So, after six half-lives, there will be approximately 3 grams left.
Answer:
28 are 6th
14 and 7th
40% are 8th
Step-by-step explanation: