We know that
<span>1. The number of all 4-letter words written using the alphabet of 20 symbols (from A to T) is
20</span><span>^4=160,000
</span><span>
2. The number of all 4-letter words written using the alphabet of 19 symbols (from B to T) is
19</span><span>^4=130,321
</span><span>
3. The difference represents exactly the number of all words of the inhabitants of the island of Jumble
difference=</span>(160,000-130,321)= 29679.
the answer is
29679
Answer:
57
Step-by-step explanation:
Let c represent the number of children ($1.75 each) and a represent the number of adults ( $2.00 each).
We know that there were 340 people total, so c + a = 340. This implies that a = 340 - c
We also know that $1.75 c + $2.00 a = $609.25
By substituting a with 340 -c we have $1.75 c + $2.00 (340 -c) = $609.25
Use the distributive property to obtain $1.75 c + $680 - $2.00 c = $609.25
Subtract $680 from both sides and combine like terms to get - $0.25 c = -
$70.75
Now, divide both sides by -$0.25 to get c = 283, the number of children.
The number of adults is 340 - c or 340 - 283 = 57
Triangle ABC is a right angle triangle. The point of circumcentre and the point of orthocentre is the same point, as shown in the diagram below. The coordinate of the point is (2.5, 2.5)
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.
