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 $2\sqrt{21}-3\sqrt{3}$.