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.

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%

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.

<AFG=（180-<A）/2=80；thus , <BFD=180-<AFG=100;

In Triangle BFD; <B+<D=180-<BFD=180-100=80.

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.

• 有的不只是光鲜的申请成果；
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  稿源：(艾蕾特教育)

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  标题：is：AMC8 -- 2000年真题解析，( 四 )