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answers of the practical exam
1.
5002-802=243600.
2.a)
$1995=3\times 5\times 7\times 19$;$S(1995)=4\times 6\times 8\times 20=3840<2\times 1995$; 1995 is deficient;

b)
1997 is a prime number. 1997 is deficient;
c)
1999 is a prime number. 1999 is deficient.

3.
$\frac{6^{30}-6^{10}}{5}.$
4.
P(A)=4/36=1/9, P(B)=20/36=5/9, $P(A\cap B)=2/36=1/18,$P(A|B)=2/20=1/10, $P(A\cap C)=4/36=1/9$, P(A|C)=4/27 and P(B|C)=15/27=5/9.

5-6.a)
${}_8C_{5}(\frac{3}{5})^5(\frac{2}{5})^3;$
b)
${}_9C_{6}(\frac{3}{5})^6(\frac{2}{5})^3+
{}_9C_{7}(\frac{3}{5})^7(\frac{2}{5})^2+
{}_9C_{8}(\frac{3}{5})^8\frac{2}{5}+
(\frac{3}{5})^9;$
c)
65 and 25;
d)
${}_{10}C_{7}(\frac{3}{5})^7(\frac{2}{5})^3;$
e)
${}_{10}C_{6}(\frac{3}{5})^6(\frac{2}{5})^4+
{}_{10}C_{7}(\frac{3}{5})^7(\frac{2}{5})^3+
{}_{10}C_{8}(\frac{3}{5})^8(\frac{2}{5})^2.$

7.a)
18!;
b)
18C3.
8.a)
mean =5, the average deviation=40/7 and the standard deviation $\sigma=\sqrt{\frac{8^2+5^2+3^2+10^2+2^2+5^2+7^2}{6}}=\sqrt{46};$
b)
five-number:-5,-2,7,10,13;

9.a)
$d((-1,2,5,0),(-3,6,8,6))=\sqrt{2^2+4^2+3^2+6^2}=\sqrt{65}\gt 8$;(-1,2,5,0) is outside of the hypersphere;

b)
$d((1,9,9,7),(-3,6,8,6))=\sqrt{4^2+3^2+1^2+1^2}=\sqrt{27}<8$;(1,9,9,7) is inside of the hypersphere;

c)
$d((1,9,9,8),(-3,6,8,6))=\sqrt{4^2+3^2+1^2+2^2}=\sqrt{30}<8$;(1,9,9,8) is inside of the hypersphere;

10.
$d((1,9,9,8),(2,0,5,0))=\sqrt{1^2+9^2+4^2+8^2}=\sqrt{162}$;the equation is (x-1)2+(y-9)2+(z-9)2+(t-8)2=162;

11.
$r_1=\sqrt{(2-0)^2+(-2-0)^2+(2-0)^2}=2\sqrt{3};$$r_2=\sqrt{(-6-(-5))^2+(-4-(-5))^2+(6-5)^2}=\sqrt{3};$$d=\sqrt{(-6-2)^2+(-4-(-2))^2+(6-2)^2}=\sqrt{84}=2\sqrt{21}$; Thus the maximum possible distance is $2\sqrt{21}+3\sqrt{3}$ and the minimum possible distance is $2\sqrt{21}-3\sqrt{3}$.



 

Jie Wu
12/8/1997