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hammer [34]
2 years ago
6

A metal bar weighs 7 Kg. 93% of the bar is Silver. How many Kgs of silver is there in the bar?​

Mathematics
2 answers:
Sindrei [870]2 years ago
8 0
<h3>Solution:</h3>

<u>From the given data</u>

<u>93% of the bar is silver :</u>

\small\bold{ → 93\%    \: \: of \: \:   7kg}

\small \bold{→\frac{93}{100}  \times 7 }

\small \bold{→ \frac{651}{100}  }

\small \bold{→ 6.51 \: kg  }

<u>Therefore, 6.51 Kg of Silver is there in the bar.</u>

Wittaler [7]2 years ago
3 0
<h2><u>PERCENT</u></h2>

<h3>\mathbb{QUESTION:}</h3>

» A metal bar weighs 7 Kg. 93% of the bar is Silver. How many Kgs of silver is there in the bar?

<h3>\mathbb{ANSWER:}</h3>
  • 6.51 kilogram (kg) of Silver is there in the bar.

— — — — — — —

<h3>\mathbb{SOLUTION:}</h3>

  • \bold{  93\%  \: of  \: 7 \: kg   \:  \: so, }\\  \\  \tt \:   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  = 93/100*7   \\ \tt =  \frac{651}{100}  \\ \:  \:  \:    \:  \:  \:  \:  \:  \:   \:  \tt= \underline{ \boxed{ \purple{ \tt 6.51  \: kg}} }\\

Therefore, 6.51 kg of Silver is there in the bar.

_______________❦︎_______________

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Mrac [35]

Answer: 0.58 m^3

Step-by-step explanation:

V= 2600 / 1500

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5 0
3 years ago
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

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Alex787 [66]
6+2(x+8)+3x+11+x
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quester [9]
I think x equals 54


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Vesnalui [34]
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7 0
4 years ago
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