Answer:
Part A: That 6 pounds of rice costs $18
Part B: (1,3) represents the unit price
Part C: 4 pounds of rice, because $12/3 equals 4
Step-by-step explanation:
Part A: The first point, 6, is on the amount of rice axis and the second point, 18, is on the total cost axis.
Part B: The unit price means the price for just 1 of something, so if you go to 1 pound of rice on the graph, you see it's at 3 on the total cost axis. Which means that 1 pound of rice costs $3.
Part C: From Part B you know that 1 pound of rice equals $3. So if you spend $12, then you can divide that by $3 to see how many pounds of rice you bought. 12/3 equals 4, so you bought 4 pounds of rice. Or you can count by 3's until you get to 12: 3, 6, 9, 12. That's 4 times so that means you bought 4 pounds of rice.
Answer:
a=8:12 c=£1.05:0.70 d=30:50:60:40
b=9:15
Step-by-step explanation:
Answer:
True. See explanation below
Step-by-step explanation:
Previous concepts
Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".
The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"
If we assume that we have 
groups and on each group from 
we have individuals on each group we can define the following formulas of variation:
And we have this property
The degrees of freedom for the numerator on this case is given by 
where k represent the number of groups.
The degrees of freedom for the denominator on this case is given by 
.
And the total degrees of freedom would be 
And the we can find the F statistic 
Answer:
As the calculated value of z does not lie in the critical region the null hypothesis is accepted that the GPA mean of the night students is the same as the GPA mean of the day students.
Step-by-step explanation:
Here n= 20
Sample mean GPA = x`= 2.84
Standard mean GPA = u= 2.55
Standard deviation = s= 0.45.
Level of Significance.= ∝ = 0.01
The hypothesis are formulated as
H0: u1=u2 i.e the GPA of night students is same as the mean GPA of day students
against the claim
Ha: u1≠u2
i.e the GPA of night students is different from the mea GPA of day students
For two tailed test the critical value is z ≥ z∝/2= ± 2.58
The test statistic
Z= x`-u/s/√n
z= 2.84-2.55/0.45/√20
z= 0.1441
As the calculated value of z does not lie in the critical region the null hypothesis is accepted that the GPA mean of the night students is the same as the GPA mean of the day students.
Answer:
1 out of 2
Step-by-step explanation:
Sorry if I’m wrong
