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Anika [276]
3 years ago
12

A bag contains 5 red marbles, 6 blue, and 4 green. If three marbles are drawn what is the probability to the nearest 10th all th

ree are blue
Mathematics
1 answer:
fredd [130]3 years ago
8 0

Answer: 4.4%

Step-by-step explanation:

Probability of drawing blue first is 6 in 15

Assuming you get that one, the second is 5 in 14 (as you've taken a blue one out)

If that's successful you have 5 in 13 chances of pulling another blue

As you want the first AND second AND third to happen, you multiply them…

(6/15)*(5/14)*(4/13) = 0.04395

So as a percentage, to the nearest tenth is 4.4%

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When was the last time that texas tried to become its own state
Rudik [331]

Answer:

The Texas Declaration of Independence was signed in 1836 and the Texas Republic ratified its own constitution shortly after. In 1837, the U.S. formally recognized Texas as a nation, but the new country struggled economically without any of the resources of its larger neighbors, leading most (but not all) residents to welcome statehood by the time it came through in 1845. Only a little more than a decade passed, however, before the state tried to secede from its national government again, joining with the Confederacy in 1861.

Step-by-step explanation:

3 0
4 years ago
find two complementary angles such that the measure of the first angle is X, and the measure of the second angle is ( 3X -2 )
mojhsa [17]

Step-by-step explanation:

<em>The </em><em>given </em><em>angles </em><em>are </em><em>complementary</em>

<em>Hence </em><em>sum </em><em>of </em><em>these </em><em>are </em><em>equal </em><em>to </em><em>9</em><em>0</em><em>°</em>

<em>X </em><em>+</em><em> </em><em>(</em><em> </em><em>3</em><em>X</em><em> </em><em>-</em><em> </em><em>2</em><em>)</em><em> </em><em>=</em><em> </em><em>9</em><em>0</em>

<em>X </em><em>+</em><em> </em><em>3</em><em>X</em><em> </em><em>-</em><em> </em><em>2</em><em> </em><em>=</em><em> </em><em>9</em><em>0</em>

<em>4</em><em>X</em><em> </em><em>-</em><em> </em><em>2</em><em> </em><em>=</em><em> </em><em>9</em><em>0</em>

<em>4</em><em>X</em><em> </em><em>=</em><em> </em><em>9</em><em>0</em><em> </em><em>+</em><em> </em><em>2</em>

<em>4</em><em>X</em><em> </em><em>=</em><em> </em><em>9</em><em>2</em>

<em>X </em><em>=</em><em> </em><em>9</em><em>2</em><em>/</em><em>4</em>

<em>X </em><em>=</em><em> </em><em> </em><em>2</em><em>3</em>

<em>First </em><em>angle </em><em>(</em><em>X)</em><em> </em><em>=</em><em> </em><em>2</em><em>3</em>

<em>Second </em><em>angle </em><em>(</em><em> </em><em>3</em><em>X</em><em> </em><em>-</em><em> </em><em>2</em><em> </em><em>)</em><em>=</em><em> </em><em>3</em><em>(</em><em>2</em><em>3</em><em>)</em><em> </em><em>-</em><em> </em><em>2</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>6</em><em>7</em>

4 0
3 years ago
If line segment RU is considered the base of parallelogram RSTU, what is the corresponding height of the parallelogram?
klasskru [66]
Given that line segment RU with vertices R(1, 1) and U(4, 5) is considered the base of parallelogram RSTU.

Then, the line segment ST with vertices S(7, 0) and T(10, 4) is the top of the parallelogram.

The corresponding height of the parallelogram is the length of a line with endponts at RU and ST and perpendicular to both RU and ST.

The equation of the line segment RU is given by
\frac{y-1}{x-1} = \frac{5-1}{4-1} = \frac{4}{3}  \\  \\ 3(y-1)=4(x-1) \\  \\ 3y-3=4x-4 \\  \\ 3y=4x-1 \\  \\ y= \frac{4}{3} x- \frac{1}{3}

Recall that given that two lines are perpendicular, the product of the slope of the two lines is -1.
Let the slope of the line perpendicular to line RU be m, then
\frac{4}{3} m=-1 \\  \\ m=- \frac{3}{4}

Thus, the equation of the line perpendicular to RU passing through point (1, 1) is given by
y-1=- \frac{3}{4} (x-1) \\  \\ 4(y-1)=-3(x-1) \\  \\ 4y-4=-3x+3 \\  \\ 4y=-3x+7 \\  \\ y=- \frac{3}{4} x+ \frac{7}{4}

The equation of the line segment ST is given by
\frac{y-0}{x-7} = \frac{4-0}{10-7} = \frac{4}{3}  \\  \\ 3y=4(x-7)=4x-28 \\  \\ y= \frac{4}{3} x- \frac{28}{3}

The line perpendicular to line segment RU intersected line segment ST at the point given by
- \frac{3}{4} x+ \frac{7}{4}=\frac{4}{3} x- \frac{28}{3} \\  \\ \frac{4}{3} x+\frac{3}{4} x=\frac{7}{4}+\frac{28}{3} \\  \\  \frac{25}{12} x= \frac{133}{12}  \\  \\ x= \frac{133}{25}  \\  \\ y=\frac{4}{3} \left(\frac{133}{25}\right)- \frac{28}{3}= -\frac{56}{25}

Thus, the corresponding height of the parallelogram is the line with endpoints
(1,1) \ and \ \left(\frac{133}{25},-\frac{56}{25}\right)

Recall that the length of a line passing through points
(x_1,y_1) \ and \ (x_2,y_2)
is given by
l= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Thus, the length of the line passing through points
(1,1) \ and \ \left(\frac{133}{25},-\frac{56}{25}\right)
is given by
l= \sqrt{\left(\frac{133}{25}-1\right)^2+\left(-\frac{56}{25}-1\right)^2}  \\  \\ = \sqrt{\left( \frac{108}{25}\right)^2+\left(- \frac{81}{25} \right)^2}= \sqrt{ \frac{11,664}{625} + \frac{6,561}{625} }  \\  \\ = \sqrt{ \frac{729}{25} } = \frac{27}{5} =5.4

Therefore, <span>the corresponding height of the given parallelogram is 5.4 units</span>
8 0
4 years ago
Read 2 more answers
Help me with the work please if you can
timurjin [86]

Answer:

1/3

Step-by-step explanation:

48+32+16=96

numerator : 32

denominator : 96

32/96=1/3

6 0
3 years ago
Savannah has a 22 ounce soda. She drinks 13 ounces. Enter the percentage of ounces Savannah has left of her soda. Round your ans
Kaylis [27]

Answer:

59%

Step-by-step explanation:

13/22=.59 or 59%

Take the amount she drank and divide by the total size of soda.

4 0
4 years ago
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