Answer:
The disadvantages: Convenience samples do not produce representative results. If you need to extrapolate to the target population, convenience samples aren't going to get you there. ... Much larger convenience samples are not more accurate than small probability samples.
Answer:
In 2012, a restaurant served 27,500 customers and in 2014, the restaurant served 29,000 customers.
First determine the increase in the number of customer served each year.
the number of customer increase in each year is, 
therefore, the linear model is;
y = 750x+ 27500 , where y represents the number of customers that the restaurant serves x years after 2012.
Now, to predict the number of customers that the restaurant will serve in 2020 is;
since, x = 2020-2012 = 8 years
or
therefore, the number of customers that the restaurant will serve in 2020 = 33,500
The answer would be
15.9(9.1)= 144.61
Answer:
The answer is : b) a confound
Step-by-step explanation:
While manipulating, is possible that some factors like noise in the hall, can affect learning in one of the groups but not in the other.
This possibility reflects the presence of a confound.
We can define a confounding variable as an external influencing factor which results in bringing changes in the effects of a dependent and independent variable.
This variable changes the outcome of an experiment and produces useless results.
There is no solution ,<span>a+c=-10;b-c=15;a-2b+c=-5 </span>No solution System of Linear Equations entered : [1] 2a+c=-10
[2] b-c=15
[3] a-2b+c=-5
Equations Simplified or Rearranged :<span><span> [1] 2a + c = -10
</span><span> [2] - c + b = 15
</span><span> [3] a + c - 2b = -5
</span></span>Solve by Substitution :
// Solve equation [3] for the variable c
<span> [3] c = -a + 2b - 5
</span>
// Plug this in for variable c in equation [1]
<span><span> [1] 2a + (-a +2?-5) = -10
</span><span> [1] a = -5
</span></span>
// Plug this in for variable c in equation [2]
<span><span> [2] - (-? +2b-5) + b = 15
</span><span> [2] - b = 10
</span></span>
// Solve equation [2] for the variable ?
<span> [2] ? = b + 10
</span>
// Plug this in for variable ? in equation [1]
<span><span> [1] (? +10) = -5
</span><span> [1] 0 = -15 => NO solution
</span></span><span>No solution</span>
