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DochEvi [55]
3 years ago
9

A bag contains 5 red, 5 blue, 4 green, and 2 yellow marbles. Find the probability of randomly choosing: a green then yellow (ind

ependent)
a.3/8
b.1/16
c.1/30
d.1/36
Mathematics
2 answers:
Snowcat [4.5K]3 years ago
6 0

Answer:

1/16

Step-by-step explanation:

you need to add it to have a correct answer

Nata [24]3 years ago
5 0
1/16 is the correct answer
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Please help me complete this quiz lol
melomori [17]
I believe the first answer is correct
7 0
3 years ago
Is (-5,-5) a solution of y \geq -2x +4y≥−2x+4 ?
FromTheMoon [43]

Answer:

  No

Step-by-step explanation:

Substitute the values and see.

  -5 ≥ -2(-5) +4

  -5 ≥ 14 . . . . . . . . FALSE

The point (-5, -5) is <em>not a solution</em> to the inequality

  y ≥ -2x +4

___

A graph can quickly show you this.

7 0
3 years ago
What's the "Build Time" and "Paint/Assembly Time"
erik [133]

5x + 2y=50

or

5x + 2y<=50 (less than or equal to)

then

x + 2y=30

or

x + 2y<=30(less than or equal to)

5 0
3 years ago
let t : r2 →r2 be the linear transformation that reflects vectors over the y−axis. a) geometrically (that is without computing a
tangare [24]

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b)  two eigen values of 'k' = 1, -1

for k = 1, eigen vector is \left[\begin{array}{c}0\\1\end{array}\right]

for k = -1 eigen vector is \left[\begin{array}{c}1\\0\end{array}\right]

See the figure for the graph:

(a) for any (x, y) ∈ R² the reflection of (x, y) over the y - axis is ( -x, y )

∴ x → -x hence '-1' is the eigen value.

∴ y → y hence '1' is the eigen value.

also, ( 1, 0 ) → -1 ( 1, 0 ) so ( 1, 0 ) is the eigen vector for '-1'.

( 0, 1 ) → 1 ( 0, 1 ) so ( 0, 1 ) is the eigen vector for '1'.

(b) ∵ T(x, y) = (-x, y)

T(x) = -x = (-1)(x) + 0(y)

T(y) =  y = 0(x) + 1(y)

Matrix Representation of T = \left[\begin{array}{cc}-1&0\\0&1\end{array}\right]

now, eigen value of 'T'

T - kI =  \left[\begin{array}{cc}-1-k&0\\0&1-k\end{array}\right]

after solving the determinant,

we get two eigen values of 'k' = 1, -1

for k = 1, eigen vector is \left[\begin{array}{c}0\\1\end{array}\right]

for k = -1 eigen vector is \left[\begin{array}{c}1\\0\end{array}\right]

Hence,

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b)  two eigen values of 'k' = 1, -1

for k = 1, eigen vector is \left[\begin{array}{c}0\\1\end{array}\right]

for k = -1 eigen vector is \left[\begin{array}{c}1\\0\end{array}\right]

Learn more about " Matrix and Eigen Values, Vector " from here: brainly.com/question/13050052

#SPJ4

6 0
1 year ago
If 1-cosA = 1/2 , Then find the value of sinA.
zimovet [89]

Step-by-step explanation:

Given :

1- cosA = 1/2

or, CosA = 1 -1/2

Therefore ; CosA = 1/2 = b/h

According to the Pythagoras theorem,

P = root under h^2 - b^2

= root under (2)^2 - (1)^2

= root under 4 -1

= root 3

Again,

SinA = P/h

= root 3 / 2

8 0
2 years ago
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