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sdas [7]
2 years ago
9

Nick has 3 shirts: a white one, a black one, and a blue one. He also has two pairs of pants, one blue and one tan. What is the p

robability, if Nick gets dressed in the dark, that he winds up wearing the white shirt and tan pants? Show your work.
Mathematics
2 answers:
castortr0y [4]2 years ago
6 0
Well if he put the black shirt on you won’t be able to see him and that’s a fact
Vikki [24]2 years ago
5 0
There is a 1 out of 12, you could have a white shirt and blue pants and on and on with them all
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Can someone please help.
yaroslaw [1]

9514 1404 393

Answer:

  (C)  0.27 × 9 = 2.43

Step-by-step explanation:

Each colored section is 27 small squares, so is 27/100 = 0.27 of the whole. There are 9 colored sections, together totalling 2 wholes plus 43 small squares, or 2.43.

The equation this represents could be ...

  0.27 × 9 = 2.43 . . . . . matches choice C

7 0
3 years ago
Y=5x translate down 3 units​
LekaFEV [45]

Answer:

y = 5x - 3.

Step-by-step explanation:

The value of y will go down by 3 units so it is

y = 5x - 3.

8 0
4 years ago
What is 290 square root
Vladimir79 [104]
The square root of 290 is 17.0294.
6 0
3 years ago
Read 2 more answers
What is the volume of the prism given below?
Y_Kistochka [10]

Answer:

Answer A

Step-by-step explanation:

because formula of a trapezium = (hxbxl) so

(6x8x19) = 912units^3

5 0
3 years ago
Read 2 more answers
Use the following vectors to answer parts​ (a) and​ (b). v1equals=[Start 3 By 1 Matrix 1st Row 1st Column 1 2nd Row 1st Column n
erik [133]

Answer:

(1) No matter what's the value of h, \vec{v}_3 is never in the span of \vec{v}_1 and \vec{v}_2.

(2) The three vectors \vec{v}_1, \vec{v}_2, and \vec{v}_3 are always linearly dependent for all real h.

Step-by-step explanation:

<h3>(a)</h3>

If \vec{v}_3 is in the span of \vec{v}_1 and \vec{v}_2, there need to exist real a and b such that

a\; \vec{v}_1 + b\; \vec{v}_2 = \vec{v}_3.

Assume that such a and b do exist.

In other words,

\displaystyle a \left[\begin{array}{c}{1 \\ -4\\2} \end{array}\right] + b\left[\begin{array}{c}{-4 \\ 16\\-8}\end{array}\right] = \left[\begin{array}{c}5 \\7 \\ h\end{array}\right].

\displaystyle \left[\begin{array}{c}{a \\ -4a\\2a} \end{array}\right] + \left[\begin{array}{c}{-4b \\ 16b\\-8b}\end{array}\right] = \left[\begin{array}{c}5 \\7 \\ h\end{array}\right].

\left\{\begin{array}{rrcr} a & - 4b &=& 5\\ -4a & + 16b &= &7\\2a & -8b & =&h\end{aligned}\right..

Rewrite as an augmented matrix and row-reduce:

\displaystyle \left[ \begin{array}{cc|c} 1 & -4 & 5 \\ -4 & 16 & 7 \\ 2 & -8 & h\end{array}\right].

(Add four times row one to row two and -2 times row one to row three.)

\displaystyle \sim \left[ \begin{array}{cc|c} 1 & -4 & 5 \\ 0 & 0 & 27 \\ 0 & 0 & h - 10\end{array}\right].

Note that in row two,

  • Left-hand side: 0;
  • Right-hand side: 27\neq 0.

In other words, this system is inconsistent. There's no real a and b that would satisfy the condition

a\; \vec{v}_1 + b\; \vec{v}_2 = \vec{v}_3.

Hence \forall h \in \mathbb{R}, \quad \vec{v}_3\not \in \text{Span}\{\vec{v}_1, \vec{v}_2\}.

There's no real h that allows h, \vec{v}_3 to be part of the span of \vec{v}_1 and \vec{v}_2.

<h3>(b)</h3>

If the three vectors are linearly dependent, at least one of them can be expressed as the linear combination of the other two.

Note that

\vec{v}_2 = (-4)\vec{v}_1 + 0 \; \vec{v}_3. In other words, \vec{v}_2 can be written as the linear combination of the other two vectors. Additionally, since the coefficient in front of \vec{v}_3 is zero, neither the exact value of \vec{v}_3 nor the value of h will make a difference. Therefore, for all h \in \mathbb{R}, the three vectors \vec{v}_1, \vec{v}_2, and \vec{v}_3 are linearly dependent.

8 0
3 years ago
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