Answer:
80 chocolates type 'a' and 240 chocolates type 'b'
Step-by-step explanation:
The company selects 25 random packages of each type from which one of type a has an incorrect weight and three of type b have an incorrect weight.
We can say that 
= 4% of the chocolates type 'a' have an incorrect weight while 
= 12% of chocolates type 'b' have an incorrect weight.
Now, if the company selects 2000 random chocolates of each type we ca predict that the amount of chocolates that have an incorrect wright will be:
Type 'a': 2000x0.04 = 80
Type 'b': 2000x0.12 = 240
So the company is expected to find 80 chocolates type 'a' and 240 chocolates type 'b' that have incorrect weight.
Data is collected on the types of vehicles
this is cross sectional data
Time is not relavant for this data and hence cannot be time series data.
b)
from the given information, the variable is type of vehicle which is categorical data
c)
this data is only about the more frequent vehicles the chain cannot ignore the vehicles which are not parked
Area+ length times width.
Make them improper and multiply.
12/5 and 7/3. Then you multiply
Answer:
The value of 
Step-by-step explanation:
We have given 
d=1.7
We have to find 
So, 
from the formula
When n=14
Using the formula:
When n=26
![s_{26}=\frac{26}{2}[2(-18)+(26-1)(1.7)]](https://tex.z-dn.net/?f=s_%7B26%7D%3D%5Cfrac%7B26%7D%7B2%7D%5B2%28-18%29%2B%2826-1%29%281.7%29%5D)
On simplification we get:
Answer:
The decimal 0.345 expressed as a fraction is 69200 .
