Substituting the values given, we get
(2)*(3) + 4 ;
Using BODMAS
We get
6 + 4
= 10
V=448 in^3
l=8 in
w=12
h=?
V=l•w•h/3
448=8•12•h/3 multiply both sides by 3
3•448=96•h
1344=96•h
h=1344/96
h=14 in
V=l•w•h/3
h=4in
l=3in
w=2.5 in
V=?
V=4•3•2.5/3
V=30/3
V=10 in^3
Answer:
D
Step-by-step explanation:
A P E X
Answer: 0.0241
Step-by-step explanation:
This is solved using the probability distribution formula for random variables where the combination formula for selection is used to determine the probability of these random variables occurring. This formula is denoted by:
P(X=r) = nCr × p^r × q^n-r
Where:
n = number of sampled variable which in this case = 21
r = variable outcome being determined which in this case = 5
p = probability of success of the variable which in this case = 0.31
q= 1- p = 1 - 0.31 = 0.69
P(X=5) = 21C5 × 0.31^5 × 0.69^16
P(X=5) = 0.0241
<u>Answer- </u>
In tossing four fair dice, the probability of getting at most one 3 is 0.86.
<u>Solution-</u>
The probability of getting at most one 3 is, either getting zero 3 or only one 3.
( ∵ xxxx )
( ∵ 3xxx, x3xx, xx3x, xxx3 )
P(Atmost one 3) = P(A) + P(B) = 0.48 + 0.38 = 0.86
