a) Kết quả bằng 1410.          b) Kết quả bằng 63.

a: Bảng thu gọn:

b: Trung bình mẫu là:

\(\overline{x} = \frac{\sum x_i n_i}{N} = \frac{100 \cdot 4 + 101 \cdot 3 + 102 \cdot 4 + 103 \cdot 3 + 104 \cdot 3 + 105 \cdot 3}{20} = \frac{2048}{20} = 102{,}4\text{ cm}\)

\(\sum x_i^2 n_i = 100^2 \cdot 4 + 101^2 \cdot 3 + 102^2 \cdot 4 + 103^2 \cdot 3 + 104^2 \cdot 3 + 105^2 \cdot 3 = 209762\)

Phương sai của mẫu là:

\(s^2 = \frac{1}{N} \sum x_i^2 n_i - \overline{x}^2 = \frac{209762}{20} - (102{,}4)^2 = 10488{,}1 - 10485{,}76 = 2{,}34\)

Phương sai mẫu hiệu chỉnh là:

\(s^2_{hc} = \frac{N}{N - 1} s^2 = \frac{20}{19} \cdot 2{,}34 \approx 2{,}463\)

a: Bảng thu gọn:

b: Trung bình mẫu là:

\(\overline{x} = \frac{\sum x_i n_i}{N} = \frac{100 \cdot 4 + 101 \cdot 3 + 102 \cdot 4 + 103 \cdot 3 + 104 \cdot 3 + 105 \cdot 3}{20} = \frac{2048}{20} = 102{,}4\text{ cm}\)

\(\sum x_i^2 n_i = 100^2 \cdot 4 + 101^2 \cdot 3 + 102^2 \cdot 4 + 103^2 \cdot 3 + 104^2 \cdot 3 + 105^2 \cdot 3 = 209762\)

Phương sai của mẫu là:

\(s^2 = \frac{1}{N} \sum x_i^2 n_i - \overline{x}^2 = \frac{209762}{20} - (102{,}4)^2 = 10488{,}1 - 10485{,}76 = 2{,}34\)

Phương sai mẫu hiệu chỉnh là:

\(s^2_{hc} = \frac{N}{N - 1} s^2 = \frac{20}{19} \cdot 2{,}34 \approx 2{,}463\)

Trả lời giúp mình các bạn ơi

A=100/99+101/100=10000/9900+9999/9900=19999/9900.                    

B=102/101+103/102=1040/10302+10403/10302=11443/10302        

(-16).125.[(-3).2² ].5³ - 2.10⁵

= (-16) .125. 9 .(-12) . 125 - 2. 100000

= (-4).4.125.9.3.(-4) .125 - 2.100000

= (-4) . 500 .27. (-500) -2.100000

= 27.100000- 2.100000

= 2500000

(-16).125.[(-3).22 ].53 - 2.105

= (-16) .125.(-12) . 125 - 2. 100000

= (-4).4.125.3.(-4) .125 - 2.100000

= (-4) . 500 .3. (-500) -2.100000

= 3.100000- 2.100000

= 100000

A= [(1+101)x101:2]-(102-103)
A= 5151+1
A=5152

B= [1+(-3)]+[4+(-5)]+.......[101+(-103)]+105
B= (-2)+(-2)...........+(-2)+105

=> A>B
B=(-2)x26+105
B=(-56)+105
B= 49

cái => A>B nó nằm ở dưới cùng ấy. Nãy gõ chứ nó bị nhảy phím

\(\frac{3}{7.10}+\frac{3}{10.13}+....+\frac{3}{100.103}\)

\(=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{100}-\frac{1}{103}\)

\(=\frac{1}{7}-\frac{1}{103}\)

\(=\frac{96}{721}\)

\(\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)

\(=\frac{2}{3}\left(\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\frac{2}{3}\left(\frac{1}{7}-\frac{1}{103}\right)\)

\(=\frac{2}{3}.\frac{96}{721}\)

\(=\frac{64}{721}\)

\(A=\)\(\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\)

\(A=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\)

\(A=\frac{1}{7}-\frac{1}{103}\)

\(A=\frac{96}{721}\)

\(B=\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)

\(B=2\left(\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{100.103}\right)\)

\(3B=2.3\left(\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{100.103}\right)\)

\(3B=2\left(\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\right)\)

\(3B=2\left(\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(3B=2\left(\frac{1}{7}-\frac{1}{103}\right)\)

\(3B=2.\frac{96}{721}\)

\(3B=\frac{192}{721}\)

\(\Rightarrow B=\frac{192}{721}:3\)

    \(B=\frac{64}{721}\)

\(A=\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\)

\(A=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\)

\(A=\frac{1}{7}-\frac{1}{103}\)

\(A=\frac{96}{721}\)

Vậy  \(A=\frac{96}{721}\)

\(B=\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)

\(B=\frac{2}{3}.\left(\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\right)\)

\(B=\frac{2}{3}.\left(\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(B=\frac{2}{3}.\left(\frac{1}{7}-\frac{1}{103}\right)\)

\(B=\frac{2}{3}.\frac{96}{721}\)

\(B=\frac{64}{721}\)

Vậy  \(B=\frac{64}{721}\)

_Chúc bạn học tốt_

a)2

b)63

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