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konstantin123 [22]
2 years ago
11

Equilateral Triangle. 6in 6in 6in 5in

Mathematics
1 answer:
OLEGan [10]2 years ago
8 0
Triangle area formula is A = 1/2bh
The base is 6 inches
The height is 5 inches
A=1/2bh
A=1/2(6)(5)
A=3(5)
A=15 inches ^2
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A stereo store is offering a special price on a complete set ofcomponents (receiver, compact disc player, speakers, cassette dec
Korvikt [17]

Answer:

Step-by-step explanation:

(a)

The number of receivers is 5.

The number of CD players is 4.

The number of speakers is 3.

The number of cassettes is 4.

Select one receiver out of 5 receivers in 5C_1 ways.

Select one CD player out of 4 CD players in 4C_1 ways.

Select one speaker out of 3 speakers in 3C_1 ways.

Select one cassette out of 4 cassettes in 4C_1 ways.

Find the number of ways can one component of each type be selected.

By the multiplication rule, the number of possible ways can one component of each type be selected is,

The number of ways can one component of each type be selected is

=5C_1*4C_1*3C_1*4C_1\\\\=5*4*3*4\\\\=240

Part a

Therefore, the number of possible ways can one component of each type be selected is 240.

(b)

The number of Sony receivers is 1.

The number of Sony CD players is 1.

The number of speakers is 3.

The number of cassettes is 4.

Select one Sony receiver out of 1 Sony receivers in ways.

Select one Sony CD player out of 1 Sony CD players in ways.

Select one speaker out of 3 speakers in ways.

Select one cassette out of 4 cassettes in 4C_1 ways.

Find the number of ways can components be selected if both the receiver and the CD player are to be Sony.

By the multiplication rule, the number of possible ways can components be selected if both the receiver and the CD player are to be Sony is,

Number of ways can one components of each type be selected

=1C_1*1C_1*3C_1*4C_1\\\\=1*1*3*4\\\\=12

Therefore, the number of possible ways can components be selected if both the receiver and the CD player are to be Sony is 12.

(c)

The number of receivers without Sony is 4.

The number of CD players without Sony is 3.

The number of speakers without Sony is 3.

The number of cassettes without Sony is 3.

Select one receiver out of 4 receivers in 4C_1 ways.

Select one CD player out of 3 CD players in 3C_1 ways.

Select one speaker out of 3 speakers in 3C_1 ways.

Select one cassette out of 3 cassettes in 3C_1 ways.

Find the number of ways can components be selected if none is to be Sony.

By the multiplication rule, the number of ways can components be selected if none is to be Sony is,

=4C_1*3C_1*3C_1*3C_1\\\\=108

[excluding sony from each of the component]

Therefore, the number of ways can components be selected if none is to be Sony is 108.

(d)

The number of ways can a selection be made if at least one Sony component is to be included is,

= Total possible selections -Total possible selections without Sony

= 240-108

= 132  

Therefore, the number of ways can a selection be made if at least one Sony component is to be included is 132.

(e)

If someone flips the switches on the selection in a completely random fashion, the probability that the system selected contains at least one Sony component is,

= \text {Total possible selections with at least one Sony} /\text {Total possible selections}

= 132  / 240

= 0.55

The probability that the system selected contains exactly one Sony component is,

= \text {Total possible selections with exactly one Sony} /\text {Total possible selections}\frac{1C_1*3C_1*3C_1*3C_1+4C_11C_13C_13C_1+4C_13C_13C_13C_1}{240} \\\\=\frac{99}{240} \\\\=0.4125

Therefore, if someone flips the switches on the selection in a completely random fashion, then is the probability that the system selected contains at least one Sony component is 0.55.

If someone flips the switches on the selection in a completely random fashion, then is the probability that the system selected contains exactly one Sony component is 0.4125.

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Would be the 3rd one I believe

Step-by-step explanation:

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One paper clip has alength of 1/10 foot. all of sam's paper clips lined up endto end measure 5 feet long.how many paper clips do
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Sam has 50 paper clips

If one paper clip has a length of 1/10 foot than 10 paper clips would make 1 foot.
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Gina Chuez has considered starting her own custom greeting card business. With an initial start up cost of $1500, she figures it
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1500 + 0.45x < 1.70x
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4 years ago
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