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2 years ago

What is the mean absolute deviation of 1, 2, 2, 3, 5, 6, 6, 8, 8, 9

Mathematics

1 answer:

nikdorinn [45]2 years ago

6 0

Answer:

Step-by-step explanation:

1+2+2+3+5+6+6+8+8+9/10=5

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A lighthouse keeper has a 10 degrees angle of depression between the horizontal light beam and the line of sight from the ship.

Andrei [34K]

107.75m
This is a trigonometric ratio problem. the angle of depression (10) is opposite the height (19m) and the distance is adjacent to the distance (x). opposite ad adjacent sides to the angle is a tangent problem.
$tan \theta = \frac{opp}{adj}$
plus in values, we know.
$tan (10) = \frac{19}{x}$
switch tan(10) and x using the products property.
$x = \frac{19}{tan (10)}$
plug in to calulator to get answer
x=107.75

4 0

3 years ago

5 1<br> - - - = <br> 8 3<br><br> simplify the expression

tensa zangetsu [6.8K]

The answer is...
7
-
24

8 0

3 years ago

In 2000, the number of fatalities caused by measles was 582,800 worldwide. In 2013, that number was 145,700. Between 2000 and 20

lara31 [8.8K]

Between 2000 and 2013, fatalities from measles decreased by (582,800 - 145,700)/582,800 x 100% = 75%

7 0

3 years ago

Read 2 more answers

In which quadrant of the coordinate plane would the point (6, −8) fall?

True [87]

Answer:

Step-by-step explanation:

8 0

2 years ago

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$

7 0

4 years ago