Mean and median of 58,57,42,44,51,54

Mean and median of 58,57,42,44,51,54

The median was 22.5

17, 53, 53, 58, 50, 53, 17. In this group of data, what is the median? What is the average? Who answered the first? I will choose water,

The median is 53 and the average is 43

The median of a set of data 63, 67, 58, x, 56 is 61, so what is x?

x=61.

The median of 68, 43, 68, 76, 58, 68, 105 is ()

The definition of median is as follows:
Arrange 68, 43, 68, 76, 58, 68, 105 from small to large, 43, 58, 68.68.68, 76105, the number in the middle is 68
The median was 68

As shown in the figure, FC ∥ ab ∥ De, and ∥ A: ∥ D: ∥ B = 2:3:4, calculate the degree of ∥ a, ∥ D, ∥ B

A refers to α bar. Let ∠ 3 = 2K, then ∠ d = 3k, ∠ B = 4K, because FC / / AB / / De, so ∠ 2 = 180 °～ B = 180 °～ 4K ∠ 1 = 180 °～ d = 180 °～ 3k, because ∠ 1 + ∠ 2 + ∠ 3 = 180 °, so 2K + 180-3k + 180-4k = 180, so k = 36, so ∠ 3 = 72 °, so ∠ B = 144 °, and ∠ d = 108 °---

As shown in the figure, point C is on the line AB, Da ⊥ AB, EB ⊥ AB, FC ⊥ AB, and Da = BC, EB = AC, FC = AB, ∠ AFB = 51 ° to calculate the degree of ∠ DFE

Prove: as shown in the figure, connect BD, AE, ∨ Da ⊥ AB, FC ⊥ AB, ∨ ad ∥ CF, ∨ DAB = ∨ BCF = 90 ° and ∫ Da = BC, FC = AB, ≌ DAB ≌ BCF (SAS), ∨ BD = BF, ∨ BDF = ∨ BFD, and ∨ ad ∥ CF, ∨ ADF = ∨ CFD, ∨ ABF = ∨ DFB + ≌ ADF = ∨ BFC + 2 ∨ CFD

As shown in the figure, point C is on the line AB, Da ⊥ AB, EB ⊥ AB, FC ⊥ AB, and Da = BC, EB = AC, FC = AB, ∠ AFB = 51 ° to calculate the degree of ∠ DFE

As shown in the figure, connect BD, AE, \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\baf = 129 °, ■ ∠ BFC + 2 The degree of ∠ DFE is 39 degrees

As shown in the figure, point C is on the line AB, Da ⊥ AB, EB ⊥ AB, FC ⊥ AB, and Da = BC, EB = AC, FC = AB, ∠ AFB = 51 ° to calculate the degree of ∠ DFE

Prove: as shown in the figure, connect BD, AE, ∨ Da ⊥ AB, FC ⊥ AB, ∨ ad ∥ CF, ∨ DAB = ∨ BCF = 90 ° and ∫ Da = BC, FC = AB, ≌ DAB ≌ BCF (SAS), ∨ BD = BF, ∨ BDF = ∨ BFD, and ∨ ad ∥ CF, ∨ ADF = ∨ CFD, ∨ ABF = ∨ DFB + ≌ ADF = ∨ BFC + 2 ∨ CFD

As shown in the figure, point C is on the line AB, Da ⊥ AB, EB ⊥ AB, FC ⊥ AB, and Da = BC, EB = AC, FC = AB, ∠ AFB = 51 ° to calculate the degree of ∠ DFE

As shown in the figure, connect BD, AE, \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\baf = 129 °, ■ ∠ BFC + 2 The degree of ∠ DFE is 39 degrees

As shown in the figure, AB is parallel to CD, ad and BC intersect at point E, angle B = 50 ° and the degree of angle c is calculated

AB / / CD, angle B = angle c (equal internal stagger angle), so it's 50 degrees