$\tan \angle A = \frac{3}{6} \\ \\ \therefore \: \tan \angle A = 0.5 \\ \\ \therefore \: \angle A = { \tan}^{ - 1} 0.5 \\ \\ \therefore \: \angle A = { \tan}^{ - 1} 0.5 \\ \\ \therefore \: \angle A = { \tan}^{ - 1} tan(26.565051177 \degree) \\ \\ \therefore \: \angle A = 26.57 \degree$

7 0

3 years ago

Read 2 more answers

4/? = ?/ 9

svet-max [94.6K]

Set ? as x, and cross-multiply.

so x^2= 36

x = 6 or x = -6

3 0

3 years ago

Find the mean and the standard deviation for each set of values. Show your work. 15,17,19,20,14,23,12

vekshin1

1. Mean.

$\overline{x}=\dfrac{15+17+19+20+14+23+12}{7}=\dfrac{120}{7}\approx\boxed{17.14}$

2. <span>Standard deviation.

First we have to square each value and calculate the mean of that squares:

$\overline{x^2}=\dfrac{15^2+17^2+19^2+20^2+14^2+23^2+12^2}{7}=\boxed{\dfrac{2144}{7}}$

so the variance:

$s^2=\overline{x^2}-\overline{x}^2=\dfrac{2144}{7}-\left(\dfrac{120}{7}\right)^2\approx\boxed{12.41}$

and the standard deviation:

$s=\sqrt{s^2}=\sqrt{12.41}\approx\boxed{3.52}$
</span>

3 0

3 years ago