Prove theorems about triangles

Mathematics

1 answer:

Aleks04 [339]3 years ago

3 0

Answer & step-by-step explanation:

a) impossible

b) the sum of the angles in a triangle is 180°

63° + 59° + 57° = 179°

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Shalnov [3]

$8.05 \cdot 10^8 - 9.16 \cdot 10^6 \\\\ =10^6(8.05 \cdot 10^2-9.16) \\\\ =10^6(805-9.16) \\\\ =10^6(795.84) \\\\ =795.84 \cdot 10^6 \\\\ = 7.96 \cdot 10^8\ kilograms$

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3 years ago

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Kevin Horn is the national sales manager for National Textbooks Inc. He has a sales staff of 40 who visit college professors all

Oduvanchick [21]

Answer:

Following are the solution to the given points:

Step-by-step explanation:

Given values:

$38, 40, 41, 45, 48, 48, 50, 50, 51, 51, 52, 52, 53, \\\\$ $54, 55, 55, 55, 56, 56, 57, 59, 59, 59, 62, 62, 62, \\ 63, 64, 65,66, 66, 67, 67, 69, 69, 71, 77, 78, 79, 79$

Total staff=40

In point a:

To calculate the median number first we arrange the value into ascending order and then collect the even numbers of calls that were also made. Its average of the middle terms is thus the median.

The midterms =55 and 59 so,

Median = $\frac{55+59}{2} = \frac{114}{2}=57$

In point b:

First-quarter $Q_1 = \frac{1}{4} \text{number of 8th call}$

$=\frac{1}{4} \times 30 th \\\\= 7.5 th$

The first quarterlies are $7.5th \ \ that \ is = (7+0.5)th\ \ term$

therefore the multiply of 0.5 by calculating the difference of the 7th and 8th term are:

$=0.5 \times 0= 0 \\\\\to Q_1 = 50+0=50 \\\\\to Q_3=\frac{3}{4} \text{number of 8th call} \\\\=\frac{3}{4} \times 30 th \\\\=22.5 \ th = (22+0.5)th \ \ term$

therefore the it is multiply by the 0.5 for the difference of the 22nd and 23rd term:

$= 0.5 \times 1=0.5 \\\\\to Q_3= 63+0.5=63.5$

In point c:

First decile $D_1 = \frac{1}{18} \text{number of 8th call}$

$= \frac{1}{10} \times 30 th \\\\= 3rd \ \ term\\\\\to D_1= 41 \\\\\to D_9= \frac{9}{10} \times 8\ th\ call\\\\=\frac{9}{10} \times 30 th \\\\=27th \ \ term\\\\\to D9 = 67$