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erma4kov [3.2K]

3 years ago

12

A bag contains 8 blue marbles, 3 red marbles, 6 purple marbles, and 7 green marbles. A marble is chosen at random from the bag.

What is the probability of choosing a green or blue marble from the bag?

2 answers:

olga55 [171]3 years ago

5 0

Answer:

62.5%

Step-by-step explanation:

Total marbles: 8+3+6+7 = 24

Green or blue marbles: 7+8=15

15/24=0.625

sammy [17]3 years ago

3 0

You would probably have a higher chance of getting the green because there are more of them?

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a ship leaves port on a bearing of 44.0 degrees, and travels 11.4 mi. the ship turns east and travels 6.2 mi. How far is the shi

Rudik [331]

Based on the information provided, $\overrightarrow{r}$ = at a bearing angle of = 59.88° .

We are requested to calculate the overall displacement of the ship, both in magnitude and direction, once it departs the harbor under the stated parameters. First, let's define a bearing.

<h3>What is bearing in the context given above?</h3>

Angles are generally measured anticlockwise from the positive x-axis, whereas bearing angles are evaluated clockwise from the positive y-axis. A bearing is NOT a standard angle measuring tool.

Therefore, a bearing of 44.0° indicates that this is an angle 90.0° - 44.0° = 46°. This is the angle to be used in the calculations.

We're given that the first displacement is 11.4° at an angle of 46° (which was computed earlier).

Step 1 - Split the above into components

x1 = 11.4 Cos 46°

= 7.92m

y1 = 11.4Sin 46.0°

= 11.4 * 0.71933980033

= 8.2m

The second displacement is a simple 6.2mi due east, that is, the positive

x-direction. The components are thus:

Δx = x1 + x2

= 7.92 + 6.2

= 14.12mi

Δy = 1 + y2

= 8.2 + 0

= 8.2

r = √(x total)² + (y total)²

=√[(14.21)²+(8.2)²]

= √(201.9241 +67.24)

= √269.1641

= 16.4mi

The direction of the displacement vector is given by:

tan Ф = (Δy)/(Δx)

= arctan (8.2/14.12)

= arctan (0.5807365439)

= 30.12°

Recall that we were asked for the bearing angle. The bearing angle is what we get when we subtract 30.12° from 90°.

That is:
= 90 - 30.12

= 59.88°

Learn more about bearing angle:
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7 0

2 years ago

8 tins holds 42-2-3 liter of oil. How many liters can 1such tin hold

GarryVolchara [31]

Answer = 4.625

Solution =42-2-3=37

37 divided by 8 = 4.625

Holley Gus’s it’s the answer

8 0

3 years ago

yarga [219]

Answer:

The first one

Step-by-step explanation:

5 0

3 years ago

What is the equation, in point-slope form, for a line that goes through (8, −4) and has a slope of −5/6 ?

horrorfan [7]

Point-slope form, y-y1 = m(x-x1)

(y-(-4)) = -5/6 (x-8)

y+4 = -5x/6+20/3

y = -5x/6+20/3 - 4

y = -5x/6 + 8/3

6 0

3 years ago

Read 2 more answers

Can someone please help me with these 7 questions please?

yarga [219]

$(1)\ (-xy)^3(xz)$

Expand

$(-xy)^3(xz) = (-x)^3* y^3*(xz)$

$(-xy)^3(xz) = -x^3* y^3*xz$

Rewrite as:

$(-xy)^3(xz) = -x^3*x* y^3*z$

Apply law of indices

$(-xy)^3(xz) = -x^4y^3z$

$(2)\ (\frac{1}{3}mn^{-4})^2$

Expand

$(\frac{1}{3}mn^{-4})^2 =(\frac{1}{3})^2m^2n^{-4*2}$

$(\frac{1}{3}mn^{-4})^2 =\frac{1}{9}m^2n^{-8$

$(3)\ (\frac{1}{5x^4})^{-2}$

Apply negative power rule of indices

$(\frac{1}{5x^4})^{-2}= (5x^4)^2$

Expand

$(\frac{1}{5x^4})^{-2}= 5^2x^{4*2}$

$(\frac{1}{5x^4})^{-2}= 25x^{8$

$(4)\ -x(2x^2 - 4x) - 6x^2$

Expand

$-x(2x^2 - 4x) - 6x^2 = -2x^3 + 4x^2 - 6x^2$

Evaluate like terms

$-x(2x^2 - 4x) - 6x^2 = -2x^3 -2x^2$

Factor out x^2

$-x(2x^2 - 4x) - 6x^2 = (-2x-2)x^2$

Factor out -2

$-x(2x^2 - 4x) - 6x^2 = -2(x+1)x^2$

$(5)\ \sqrt{\frac{4y}{3y^2}}$

Divide by y

$\sqrt{\frac{4y}{3y^2}} = \sqrt{\frac{4}{3y}}$

Split

$\sqrt{\frac{4y}{3y^2}} = \frac{\sqrt{4}}{\sqrt{3y}}$

$\sqrt{\frac{4y}{3y^2}} = \frac{2}{\sqrt{3y}}$

Rationalize

$\sqrt{\frac{4y}{3y^2}} = \frac{2}{\sqrt{3y}} * \frac{\sqrt{3y}}{\sqrt{3y}}$

$\sqrt{\frac{4y}{3y^2}} = \frac{2\sqrt{3y}}{3y}$

$(6)\ \frac{8}{3 + \sqrt 3}$

Rationalize

$\frac{8}{3 + \sqrt 3} = \frac{3 - \sqrt 3}{3 - \sqrt 3}$

$\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{(3 + \sqrt 3)(3 - \sqrt 3)}$

Apply different of two squares to the denominator

$\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{3^2 - (\sqrt 3)^2}$

$\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{9 - 3}$

$\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{6}$

Simplify

$\frac{8}{3 + \sqrt 3} = \frac{4(3 - \sqrt 3)}{3}$

$(7)\ \sqrt{40} - \sqrt{10} + \sqrt{90}$

Expand

$\sqrt{40} - \sqrt{10} + \sqrt{90} =\sqrt{4*10} - \sqrt{10} + \sqrt{9*10}$

Split

$\sqrt{40} - \sqrt{10} + \sqrt{90} =\sqrt{4}*\sqrt{10} - \sqrt{10} + \sqrt{9}*\sqrt{10}$

Evaluate all roots

$\sqrt{40} - \sqrt{10} + \sqrt{90} =2*\sqrt{10} - \sqrt{10} + 3*\sqrt{10}$

$\sqrt{40} - \sqrt{10} + \sqrt{90} =2\sqrt{10} - \sqrt{10} + 3\sqrt{10}$

$\sqrt{40} - \sqrt{10} + \sqrt{90} =4\sqrt{10}$

$(8)\ \frac{r^2 + r - 6}{r^2 + 4r -12}$

Expand

$\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{r^2 + 3r-2r - 6}{r^2 + 6r-2r -12}$

Factorize each

$\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{r(r + 3)-2(r + 3)}{r(r + 6)-2(r +6)}$

Factor out (r+3) in the numerator and (r + 6) in the denominator

$\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{(r -2)(r + 3)}{(r - 2)(r +6)}$

Cancel out r - 2

$\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{r + 3}{r +6}$

$(9)\ \frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14}$

Cancel out x

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{x^2 - 5x - 14}$

Expand the numerator of the 2nd fraction

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{x^2 - 7x+2x - 14}$

Factorize

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{x(x - 7)+2(x - 7)}$

Factor out x - 7

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{(x + 2)(x - 7)}$

Factor out 4 from 4x + 8

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4(x + 2)}{x} \cdot \frac{1}{(x + 2)(x - 7)}$

Cancel out x + 2

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4}{x} \cdot \frac{1}{(x - 7)}$

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4}{x(x - 7)}$

$(10)\ (3x^3 + 15x^2 -21x) \div 3x$

Factorize

$(3x^3 + 15x^2 -21x) \div 3x = 3x(x^2 + 5x -7) \div 3x$

Cancel out 3x

$(3x^3 + 15x^2 -21x) \div 3x = x^2 + 5x -7$

$(11)\ \frac{m}{6m + 6} - \frac{1}{m+1}$

Take LCM

$\frac{m}{6m + 6} - \frac{1}{m+1} = \frac{m(m + 1) - 1(6m + 6)}{(6m + 6)(m + 1)}$

Expand

$\frac{m}{6m + 6} - \frac{1}{m+1} = \frac{m^2 + m- 6m - 6}{(6m + 6)(m + 1)}$

$\frac{m}{6m + 6} - \frac{1}{m+1} = \frac{m^2 - 5m - 6}{(6m + 6)(m + 1)}$

$(12)\ \frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}}$

Rewrite as:

$\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} \div \frac{2}{y^2 - 9}$

Express as multiplication

$\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} * \frac{y^2 - 9}{2}$

Express y^2 - 9 as y^2 - 3^2

$\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} * \frac{y^2 - 3^2}{2}$

Express as difference of two squares

$\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} * \frac{(y - 3)(y+3)}{2}$

$\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{1} * \frac{(y+3)}{2}$

$\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{y+3}{2}$

Read more at:

brainly.com/question/4372544

3 0

3 years ago