<span>2/15 if drawn without replacement.
1/9 if drawn with replacement.
Assuming that the chips are drawn without replacement, there are 6 * 5 different possibilities. And that's a low enough number to exhaustively enumerate them. So they are:
1,2 : 1,3 : 1,4 : 1,5 : 1,6
2,1 : 2,3 : 2,4 : 2,5 : 2,6
3,1 : 3,2 : 3.4 : 3,5 : 3,6
4,1 : 4,2 : 4.3 : 4,5 : 4,6
5,1 : 5,2 : 5.3 : 5,4 : 5,6
6,1 : 6,2 : 6.3 : 6,4 : 6,5
Of the above 30 possible draws, there are 4 that add up to 5. So the probability is 4/30 = 2/15
If the draw is done with replacement, then there are 36 possible draws. Once again, small enough to exhaustively list, they are:
1,1 : 1,2 : 1,3 : 1,4 : 1,5 : 1,6
2,1 : 2,2 : 2,3 : 2,4 : 2,5 : 2,6
3,1 : 3,2 : 3,3 : 3.4 : 3,5 : 3,6
4,1 : 4,2 : 4.3 : 4,4 : 4,5 : 4,6
5,1 : 5,2 : 5.3 : 5,4 : 5,5 : 5,6
6,1 : 6,2 : 6.3 : 6,4 : 6,5 : 6,6
And of the above 36 possibilities, exactly 4 add up to 5. So you have 4/36 = 1/9</span>
The y - intercept of an equation is the value of y when x is substituted with zero. From the given equation,
y = 25x + 225
Substituting,
y = (25)(0) + 225
y = 225
Thus, the y - intercept of the given equation is 225.
Answer:
The first digit of a two digit number can be any of digits 1 - 9. It cannot be 0 though. Therefore there are 9 possible digits for the first place.
There are 5 possible digits for the second position. The two digit number has to be odd and therefore the final digit must be 1,3,5,7 or 9
Therefore for each and every one of the nine first digits there are 5 digits that the second can be.
Therefore ANSWER = 9 * 5 = 45 possible permutations.
2. The largest two digit number = 99
Subtract 57 and you get 42
ANSWER = 42
The forty two numbers are 58 ,59, 60, 61......98, 99
Answer:
Therefore the value of k = 6.
Step-by-step explanation:
Given:
LN = m
NM = l
OM = k
NO = 4
LO = 8
LM = 8 + k and
Δ LNM ,Δ LON and Δ MON are right Triangle.
To Find :
Om = k = ?
Solution:
In Right angle Triangle By Pythagoras Theorem we have,
So, In Right angle Triangle Δ LON we have,
LN² = ON² + OL²
m² = 4² + 8²
m² = 80 ............( 1 )
Now in Right angle Triangle Δ MON we have,
MN² = ON² + MO²
l² = 4² + k² ....................( 2 )
Now In Right angle Triangle Δ LNM we have,
LM² = LN² + MN²
(8 + k)² = m² + l² .................( 3 )
Substituting equation 1 and equation 2 in equation 3
(8+k)² = 80 + 4² + k²
Applying (A+B)² = A² +2AB + B² we get
Therefore the value of k = 6.
