按关键词阅读:
Answer: B
Solution:
Let K(n) be the probability that Keiko gets n heads, and let E(n) be the probability that Ephriam gets n heads. E(0)=1/4; E(1)=1/2; E(2)=1/4; K(0)=1/2; K(1)=1/2; K(2)=0;
The probability that Keiko gets 0 heads and Ephriam gets 0 heads is K(0)*E(0)=1/2*1/4=1/8; Similarly for 1 head and 2 heads. Thus, we have:P=K(0)*E(0)+K(1)*E(1)+K(2)*E(2)=1/8+1/2*1/2+0=3/8.Problem 22

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Answer: C
Solution:
The original cube has 6 faces, each with an area of 2*2=4 square units. Thus the original figure had a total surface area of 24 square units.
The new figure has the original surface, with 6 new faces that each have an area of 1 square unit, for a total surface area of 6 additional square units added to it. But 1 square unit of the top of the bigger cube, and 1 square unit on the bottom of smaller cube, is not on the surface, and does not count towards the surface area.
The total surface area is therefore 24+6-1-1=28 square units.
The percent increase in surface area 28-24/24=4/24=17%Problem 23

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Answer: B
Solution:
Let the sum of first 3 numbers is a, the fourth number is x, and the sum of last 3 numbers is c, then :
a+x=5*4;
x+b=8*4;
a+x+b=7*(46/7);
we get x=6 from the above 3 equations.Problem 24

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Answer: D
Solution:
<AFG=(180-<A)/2=80;thus , <BFD=180-<AFG=100;
In Triangle BFD; <B+<D=180-<BFD=180-100=80.Problem 25

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Answer: B
Solution:
To quickly solve this multiple choice problem, make the (not necessarily valid, but very convenient) assumption that ABCD can have any dimension. Give the rectangle dimensions of AB=CD=12 and BC=AD=6, which is the easiest way to avoid fractions. Labelling the right midpoint as M, and the bottom midpoint as N, we know that DN=NC=6; BM=MC=3.
Area(AND)=1/2*6*6=18; Area(MNC)=1/2*6*3=9; Area(ABM)=1/2*12*3=18; Area(AMN)=Area(ABCD)-Area(AND)-Area(MNC)-Area(ABM)=72-18-9-18=27.AMC培训、答疑 , 请联系微信 / 电话:136 1118 1627
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