Answer:
Option 1: 0.32
Step-by-step explanation:
Let
P(A) be the experimental probability of getting two
And
P(B) be the experimental probability of getting three
The dice is rolled 360 times.
So the sample space is n(S) = 360
P(A) = n(A)/n(S)
= 54/360
= 0.15
And
P(B) = n(B)/n(S)
= 62/360
= 0.172
As both the events A and B are mutually exclusive,
P(A or B) = P(A) + P(B)
= 0.15 + 0.172
=0.322
Rounding of to one decimal gives us:
0.32
So the probability of rolling a two or three is 0.32 ..
Step-by-step explanation:
please find the attached document
Answer: none
Step-by-step explanation:
(A)
(16÷32/10) ×2 + 0.2×(90)
Using bodmas principle ; solve bracket
(16×10/32)×2 + (2/10×90)
10+18 =28
(B)
{(16÷32/10) × (2+2/10)} ×90
Open brackets
{(16×10/32) × (22/10)} ×90
(5×11/5) ×90
11×90 = 990
(C)
16÷{(32/10×2) + (2/10×8)} +82
Open brackets, solve division first, dolled by addition
16÷(32/5 + 8/5) +82
16÷(40/5) +82
16÷8 +82
2+82= 84
(D)
[16÷(32/10 ×2) + 0.2× (90)]
16÷ (32/5) + 2/10 ×90
Solve division
16×5/32 + 18
5/2 + 18
L.c.m of denominator (2&1) =2
(5+36) / 2 = 41/2
=20.5
I think x = 2 is the answer
There are 60 thirds in three hours
