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babymother [125]
2 years ago
9

What is the width of the rectangle? 30y 60y4

Mathematics
1 answer:
Bas_tet [7]2 years ago
7 0

Answer:

2 y^3

Step-by-step explanation:

Area of rectangule is length * width

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Solve for X. 6x – 9 = -27​
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A stereo store is offering a special price on a complete set of components (receiver, compact disc player, speakers, turntable).
Juli2301 [7.4K]

Answer:

a) 240 ways

b) 12 ways

c) 108 ways

d) 132 ways

e) i) 0.55

ii) 0.4125

Step-by-step explanation:

Given the components:

Receiver, compound disk player, speakers, turntable.

Then a purcahser is offered a choice of manufacturer for each component:

Receiver: Kenwood, Onkyo, Pioneer, Sony, Sherwood => 5 offers

Compact disc player: Onkyo, Pioneer, Sony, Technics => 4 offers

Speakers: Boston, Infinity, Polk => 3 offers

Turntable: Onkyo, Sony, Teac, Technics => 4 offers

a) The number of ways one component of each type can be selected =

\left(\begin{array}{ccc}5\\1\end{array}\right) \left(\begin{array}{ccc}4\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}4\\1\end{array}\right)

= 5 * 4 * 3 * 4  = 240 ways

b) If both the receiver and compact disk are to be sony.

In the receiver, the purchaser was offered 1 Sony, also in the CD(compact disk) player the purchaser was offered 1 Sony.

Thus, the number of ways components can be selected if both receiver and player are to be Sony is:

\left(\begin{array}{ccc}1\\1\end{array}\right) \left(\begin{array}{ccc}1\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}4\\1\end{array}\right)

= 1 * 1 * 3 * 4 = 12 ways

c) If none is to be Sony.

Let's exclude Sony from each component.

Receiver has 1 sony = 5 - 1 = 4

CD player has 1 Sony = 4 - 1 = 3

Speakers had 0 sony = 3 - 0 = 3

Turntable has 1 sony = 4 - 1 = 3

Therefore, the number of ways can be selected if none is to be sony:

\left(\begin{array}{ccc}4\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right)

= 4 * 3 * 3 * 3 = 108 ways

d) If at least one sony is to be included.

Number of ways can a selection be made if at least one Sony component is to be included =

Total possible selections - possible selections without Sony

= 240 - 108

= 132 ways

e) If someone flips switches on the selection in a completely random fashion.

i) Probability of selecting at least one Sony component=

Possible selections with at least one sony / Total number of possible selections

\frac{132}{240} = 0.55

ii) Probability of selecting exactly one sony component =

Possible selections with exactly one sony / Total number of possible selections.

\frac{\left(\begin{array}{ccc}1\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) + \left(\begin{array}{ccc}4\\1\end{array}\right) \left(\begin{array}{ccc}1\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) + \left(\begin{array}{ccc}4\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}3\\1\end{array}\right) \left(\begin{array}{ccc}1\\1\end{array}\right)}{240}

= \frac{(1*3*3*3)+(4*1*3*3)+(4*3*3*1)}{240}

\frac{27 + 36 + 36}{240} = \frac{99}{240} = 0.4125

5 0
3 years ago
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