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3 years ago

What is the primary goal and secondary goal of the Fried Liver Opening in chess? <br />

1 answer:

3 0

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.

You might be interested in

In the parallelogram shown, find x.

Olin [163]

The side with all the 9s and x is 12 just like its opposite side because this is a parallelogram. We can write the equation to add up the three segments; the ones on the left and right are 9-x:

9-x + x + 9-x = 12

18 - x = 12

x = 6

4 0

3 years ago

Which shows the triangle when it is translated two units left

Naddika [18.5K]

Answer:

I believe it would be -3, 7?

Step-by-step explanation:

IF we move the entire triangle to the left, that's where the coordinates line up.

Hope it helps?

6 0

3 years ago

10. There are nine golf balls numbered from 1 to 9 in a bag. Three balls are randomly selected without replacement to form a 3-d

lbvjy [14]

Answer:

a) 504

b) 56

c) 0.111

Step-by-step explanation:

Data provided in the question:

There are nine golf balls numbered from 1 to 9 in a bag

Three balls are randomly selected without replacement

a) 3-digit numbers that can be formed

= $^nP_r$

n = 9

r = 3

= ⁹P₃

= $\frac{9!}{9!-3!}$

= 9 × 8 × 7

= 504

b) 3-digit numbers start with the digit 1

= _ _ _

in the above 3 blanks first digit is fixed i.e 1

we and we have 8 choices left for the last 2 digits

Thus,

n = 8

r = 2

Therefore,

= 1 × ⁸P₂

= 1 × $\frac{8!}{8!-2!}$

= 1 × 8 × 7

= 56

c) Probability that the 3-digit number formed is less than 200

Now,

The number of 3-digit number formed is less than 200 will be the 3-digit numbers start with the digit 1 i.e part b)

and total 3-digit numbers that can be formed is part a)

therefore,

Probability that the 3-digit number formed is less than 200

= 56 ÷ 504

= 0.111

5 0

3 years ago

1/3 (6x - 9) + 3x - 8

Lena [83]

The answer is 5x-11

You have to distrivute 1/3 into 6x and -9 and after you have your answer for those you can combine like terms.

4 0

3 years ago

A couple intends to have two children, and suppose that approximately 52% of births are male and 48% are female.

Pachacha [2.7K]

a) Probability of both being males is 27%

b) Probability of both being females is 23%

c) Probability of having exactly one male and one female is 50%

Step-by-step explanation:

The probability that the birth is a male can be written as

$p(m) = 0.52$ (which corresponds to 52%)

While the probability that the birth is a female can be written as

$p(f) = 0.48$ (which corresponds to 48%)

Here we want to calculate the probability that over 2 births, both are male. Since the two births are two independent events (the probability of the 2nd to be a male does not depend on the fact that the 1st one is a male), then the probability of both being males is given by the product of the individual probabilities:

$p(mm)=p(m)\cdot p(m)$

And substituting, we find

$p(mm)=0.52\cdot 0.52 = 0.27$

So, 27%.

In this case, we want to find the probability that both children are female, so the probability

$p(ff)$

As in the previous case, the probability of the 2nd child to be a female is independent from whether the 1st one is a male or a female: therefore, we can apply the rule for independent events, and this means that the probability that both children are females is the product of the individual probability of a child being a female:

$p(ff)=p(f)\cdot p(f)$