January . . . $57.85
March . . . . 4 times as much = 4 (57.85) = $231.40
Deposit 78.45 more . . . ($231.40 + 78.45) = <em>$309.85</em> .
Notice that "interest" is never mentioned anywhere in this problem.
In other words, it doesn't matter whether Julie's savings account is
in a bucket in the basement, a mayonnaise jar on the porch, under
her mattress, or in a bank that pays no interest.
Without interest, $309.85 is what she <em><u>does</u></em> have<em><u /></em> in November, which
is about right for savings accounts in banks these days.
What her balance <em><u>should</u></em> be in November is an entirely different subject.
Answer:
0.35
Step-by-step explanation:
Answer:
D. There were no significant effects.
Step-by-step explanation:
The table below shows the representation of the significance level using the two-way between subjects ANOVA.
Source of Variation SS df MS F P-value
Factor A 10 1 10 0.21 0.660
Factor B 50 2 25 0.52 0.6235
A × B 40 2 20 0.42 0.6783
Error 240 5 48 - -
Total 340 10 - - -
From the table above , the SS(B) is determined as follows:
SS(B) = SS(Total)-SS(Error-(A×B)-A)
= 340-(240-40-10)
= 50
A researcher computes the following 2 x 3 between-subjects ANOVA;
k=2
n=3
N(total) = no of participants observed in each group =11
df for Factor A= (k-1)
=(2-1)
=1
df for Factor B = (n-1)
=(3-1)
=2
df for A × B
= 2 × 1
= 2
df factor for total
=(N-1)
=11-1
=10
MS = SS/df
Thus, from the table, the P-Value for all data is greater than 0.05, therefore we fail to reject H₀.
