During Distributive property, the number on the outside of the parenthesis basically gets multiplied by the numbers inside the parenthesis. In this case, you would multiply 6 by 5 and 4. When you get the products, then follow the sign that comes next. In this case, you would add the products to get a total sum. I attached a photo for further reference.
Hope this helps!
Each container is 50 pounds, forget about the second sentence it's just there to confuse you. One ton is 2,000 pounds, half of that is 1,000 lbs. 1,000 divided by 20 is 50. 50 pounds each.
There must be More details for the problem
With 10 integers available,
has PMF
We're interested in the statistics of the new random variable
. To do this, we need to know the PMF for
. This isn't too hard to find.
Since the PMF for
gives a value of
whenever
is an integer between 0 and 9, it follows that
must also be an integer for the PMF to give the identical value of
. This means
Now the mean (expectation) is
The variance would be
The standard deviation is the square root of the variance, so you have
Answer:
(a) 91 employees were absent fewer than six days.
(b) 22 employees were absent more than five days.
(c) 20 employees were absent from 6 up to 12 days.
Step-by-step explanation:
The data for the number of days absent during a calendar year by employees of a manufacturing company is given below.
(a)
The number of employees that were absent for fewer than six days is =
![Frequency\ for\ class\ [0\ - \ 3]+Frequency\ for\ class\ [3\ - \ 6]\\=60 +31\\=91](https://tex.z-dn.net/?f=Frequency%5C%20%20for%5C%20%20class%5C%20%5B0%5C%20-%20%5C%203%5D%2BFrequency%5C%20%20for%5C%20%20class%5C%20%5B3%5C%20-%20%5C%206%5D%5C%5C%3D60%20%2B31%5C%5C%3D91)
Thus, there were 91 employees who were absent for fewer than six days.
(b)
The number of employees that were absent for more than 5 days is =
![Frequency\ for\ class\ [6\ -\ 9]+Frequency\ for\ class\ [9\ -\12]+\\Frequency\ for\ class\ [12\ - \15]\\=14+6+2\\=22](https://tex.z-dn.net/?f=Frequency%5C%20%20for%5C%20%20class%5C%20%5B6%5C%20-%5C%209%5D%2BFrequency%5C%20%20for%5C%20%20class%5C%20%5B9%5C%20-%5C12%5D%2B%5C%5CFrequency%5C%20for%5C%20%20class%5C%20%5B12%5C%20-%20%5C15%5D%5C%5C%3D14%2B6%2B2%5C%5C%3D22)
Thus, there were 22 employees who were absent for more than 5 days.
(c)
The number of employees that were absent from 6 up to 12 days is =
![Frequency\ for\ class\ [6\ -\ 9]+Frequency\ for\ class\ [9\ -\12]=14+6\\=20](https://tex.z-dn.net/?f=Frequency%5C%20%20for%5C%20%20class%5C%20%5B6%5C%20-%5C%209%5D%2BFrequency%5C%20%20for%5C%20%20class%5C%20%5B9%5C%20-%5C12%5D%3D14%2B6%5C%5C%3D20)
Thus, there were 20 employees who were absent from 6 up to 12 days.
