3 : 3,6,9,12,15,18,21,24,27,30
4 : 4,8,12,16,20,24,28,32,36
7: 7,14,21,28,35,42,49,56,63
9: 9,18,27,36,45,54,63,72,81
In a throw of 2 fair dice, there are 6*6=36 equiprobability outcomes.
To get a sum of 5, there are 4 ways, (1,4),(2,3),(3,2),(4,1) with probability of 4/36=1/9
To get at least one 5, there are 6+6-1=11 outcomes (note (5,5) has been counted in both, so subtracted from sum). The probability is 11/36
Since the two events are mutually exclusive (once we have a five, the sum can no longer be 5), we can add the probabilities to get the probability of one event or the other.
P(sum of 5 OR at least one 5)=1/9+11/36=4/36+11/36=15/36=5/9
Answer:
Solution
verified
Verified by Toppr
m
2
−3m−1=0
m
2
−3m=1 → (1)
Third term =(
2
1
coeeficientofm)
2
(
2
1
×(−3))
2
=(
2
−3
)
2
=
4
9
Adding
4
9
to both sides of equation (1), we get
m
2
−3m+
4
9
=1+
4
9
∴ m
2
−3m+
4
9
=
4
4+9
∴ (m−
2
3
)
2
=
4
13
Taking square roots on both sides
∴ m−
2
3
=±
2
13
∴ m=
2
3
+
2
13
or m=
2
3
−
2
13
m=
2
3+
13
,
2
3−
13
are the roots of the given quadratic equation.
Answer:
$17/ $25/$41
Step-by-step explanation:
brainliest pls
90=l+l+w+w
90=28+28+w+w
90=56+2w
34=2w
17=w
The width of the playground is 17.
