Add them all up.
A triangle has 3 sides, and you said each side is 3. So:
3+3+3 = 9
Answer:
:5:5&&'4&4&4&:&;4"33*2**223$&4&4&55&
Play usually continues 7.Qf3+ Ke6 8.Nc3 (see diagram). Black will play 8...Nb4 or 8...Ne7 and follow up with c6, bolstering his pinned knight on d5. If Black plays 8...Nb4, White can force the b4 knight to abandon protection of the d5 knight with 9.a3?! Nxc2+ 10.Kd1 Nxa1 11.Nxd5, sacrificing a rook, but current analysis suggests that the alternatives 9.Qe4, 9.Bb3 and 9.O-O are stronger. White has a strong attack, but it has not been proven yet to be decisive.
Because defence is harder to play than attack in this variation when given short time limits, the Fried Liver is dangerous for Black in over-the-board play, if using a short time control. It is also especially effective against weaker players who may not be able to find the correct defences. Sometimes Black invites White to play the Fried Liver Attack in correspondence chess or in over-the-board games with longer time limits (or no time limit), as the relaxed pace affords Black a better opportunity to refute the White sacrifice.
Answer:
Step-by-step explanation:
You are to make 5 assemblies.
Each assembly requires the use of 1 Type A bolt.
To make the 5 assemblies, you need 5 Type A bolts.
The container of bolts has a total of 60 bolts.
The focus - Type A bolts - is 20 out of this 60.
The probability of obtaining a Type A bolt at all, is 20/60, which is = 1/3
(A) What is the probability of taking the exact number of Type A bolts you need for your 5 assemblies, if you randomly take 10 bolts from the container?
- The exact number of Type A bolts you need for the 5 assemblies is 5
1/3 × 5/10 = 5/30 = 1/6 = 0.167
(B) What is the probability of taking/having less than 5 Type A bolts out of the randomly selected 10 bolts? The solution is to sum up the following:
1/3 × 4/10 = 0.133
1/3 × 3/10 = 0.1
1/3 × 2/10 = 0.067
1/3 × 1/10 = 0.033
1/3 × 0/10 = 0
TOTAL = 0.333
Answer:
c
Step-by-step explanation:
