Given the conditional relative frequency table below which was generated by
column using frequency table data comparing the number of calories in a meal to
whether the meal was prepared at home or at a restaurant.
Number of Calories and Location of Meal Preparation.
Home Restaurant Total
≥ 500 calories 0.15 0.55 0.28
< 500 calories 0.85 0.45 0.72
Total 1.0 1.0 1.0
To determine whether there is an association between where food is prepared and the number of carories the food contain, we recall that an "association" exists between two categorical variables if the column conditional relative frequencies are different
for the columns of the table. The bigger the differences in
the conditional relative frequencies, the stronger the association
between the variables. If the conditional relative frequencies are
nearly equal for all categories, there may be no association between the
variables. Such variables are said to be <span>independent.
For the given conditional relative freduency, we can see that there is a significant difference between the columns of the table.
i.e. 0.15 is significantly different from 0.55 and 0.85 is significantly different from 0.45
Therefore, we can conclude from the given answer options that t</span><span>here is an association because the value 0.15 is not similar to the value 0.55</span>
The answer is 1 in the series
Answer: 250 mi
Step-by-step explanation:
Here we can think in a triangle rectangle:
The distance from Birmingham to Atlanta is roughly 150 mi, and this is one of the cathetus.
And the distance from Birmingham to Nashville is roughly 200 mi, this is the other cathetus of the triangle.
Now, the distance from Atlanta to Nashville will be the hypotenuse of this triangle rectangle.
Now we can apply the Pythagorean's theorem:
A^2 + B^2 = H^2
Where A and B are the cathetus, and H is the hypotenuse:
Then:
H = √(A^2 + B^2)
H = √(150^2 + 200^2) mi = √(62,500) mi = 250 mi
Then the estimated distance from Atlanta to Nashville is 250 mi
It is given here that there is 1/3 probability of professional baseball player will get a hit. Hence if at least three hits are gained out of 5 attempts, the calculation goes: 5C3* (1/3)^3*(2/3)^2 + 5C4* (1/3)^4*(2/3)^1 +5C5 *<span>(1/3)^5*(2/3)^0 equal to 0.21. </span>
