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Monica [59]
4 years ago
15

If someone flips switches on the selection in a completely random fashion, what is the probability that the system selected cont

ains at least one sony component? exactly one sony component? (round your answer to three decimal places.) at least one sony component correct: your answer is correct. exactly one sony component
Mathematics
1 answer:
DiKsa [7]4 years ago
5 0
 Given that a display allows a customer to hook together any selection of components, one of each type. These are the types:
Receiver: Kenwood, Onkyo, Pioneer, Sony, Sherwood
CD player: Onkyo, Pioneer, Sony, Technics
Speakers: Boston, Infinity, Polk
Cassette: Onkyo, Sony, Teac, Technics:

Part (a):
In how many ways can one component of each type be selected?

The number of ways one type of receiver will be selected is given by 5C1 = 5
The number of ways one type of CD player will be selected is given by 4C1 = 4
The number of ways one type of speakers will be selected is given by 3C1 = 3
The number of ways one type of cassette will be selected is given by 4C1 = 4

Therefore, the number of ways one component of each type can be selected is given by 5 x 4 x 3 x 4 = 240 ways



Part (b):
In how many ways can components be selected if both the receiver and the compact disc player are to be Sony?

The number of ways of selecting a Sony receiver is 1
The number of ways of selecting a Sony CD player is 1
The number of ways one type of speakers will be selected is given by 3C1 = 3
The number of ways one type of cassette will be selected is given by 4C1 = 4

Therefore, the number of ways components can be selected if both the receiver and the compact disc player are to be Sony is given by 1 x 1 x 3 x 4 = 12



Part (c)
In how many ways can components be selected if none of them are Sony?

The number of ways one type of receiver that is not Sony will be selected is given by 4C1 = 4
The number of ways one type of CD player that is not Sony will be selected is given by 3C1 = 3
The number of ways one type of speakers that is not Sony will be selected is given by 3C1 = 3
The number of ways one type of cassette that is not Sony will be selected is given by 3C1 = 3

Therefore, the number of ways that components can be selected if none of them are Sony is given by 4 x 3 x 3 x 3 = 108



Part (d):
In how many ways can a selection be made if at least one Sony component is to be included?

The total number of ways of selecting one component of each type is 240
The number of ways that components can be selected if none of them are Sony is 108

Therefore, the number of ways of selecting at least one Sony component is given by 240 - 108 = 132



Part (e):
If someone flips switches on the selection in a completely random fashion, what is the probability that the system selected contains at least one Sony component?

The total number of ways of selecting one component of each type is 240
The number of ways of selecting at least one Sony component is 132

Therefore, the probability that a system selected at random contains at least one Sony component is given by 132 / 240 = 0.55



Part (f):
If someone flips switches on the selection in a completely random fashion, what is the probability that the system selected contains exactly one Sony component? (Round your answer to three decimal places.)

The number of ways of selecting only a Sony receiver is given by 1 x 3 x 3 x 3 = 27
The number of ways of selecting only a Sony CD player is given by 4 x 1 x 3 x 3 = 36
The number of ways of selecting only a Sony cassette is given by 4 x 3 x 3 x 1 = 36

Thus, the number of ways of selecting exactly one Sony component is given by 27 + 36 + 36 = 99

Therefore, the probability that a system selected at random contains exactly one Sony component is given by 99 / 240 = 0.413
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Suppose a certain airline uses passenger seats that are 16.2 inches wide. Assume that adult men have hip breadths that are norma
Pachacha [2.7K]

Answer:

Each adult male has a 5.05% probability of having a hip width greater than 16.2 inches.

There is a 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by

Z = \frac{X - \mu}{\sigma}

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem

Assume that adult men have hip breadths that are normally distributed with a mean of 14.4 inches and a standard deviation of 1.1 inches. This means that \mu = 14.4, \sigma = 1.1.

What is the probability that any one of those adult male will have a hip width greater than 16.2 inches?

For each one of these adult males, the probability that they have a hip width greater than 16.2 inches is 1 subtracted by the pvalue of Z when X = 16.2. So:

Z = \frac{X - \mu}{\sigma}

Z = \frac{16.2 - 14.4}{1.1}

Z = 1.64

Z = 1.64 has a pvalue of 0.9495.

This means that each male has a 1-0.9495 = 0.0505 = 5.05% probability of having a hip width greater than 16.2 inches.

For the average of the sample

What is the probability that the 110 adult men will have an average hip width greater than 16.2 inches?

Now, we need to find the standard deviation of the sample before using the zscore formula. That is:

s = \frac{\sigma}{\sqrt{110}} = 0.1.

Now

Z = \frac{X - \mu}{\sigma}

Z = \frac{16.2 - 14.4}{0.1}

Z = 18

Z = 18 has a pvalue of 0.9999.

This means that there is a 1-0.9999 = 0.0001 = 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.

7 0
3 years ago
In rectangle FGHI, diagonals FH and GI intersect at E. G F E H I • IE = 3x + 4 • EG = 5x - 6 What is the length of FH ?​
timurjin [86]

Answer:

FH = 38

Step-by-step explanation:

Here, given the point of intersection of the two diagonals at E, we want to calculate the length of FH

The point of intersection of the diagonals is the midpoint of each as the diagonals bisect each other

thus,

3x + 4 = 5x - 6

5x-3x = 6+ 4

2x = 10

x= 10/2

x = 5

but Eg is 5x-6

and FH = 2EG

FH = 2(5x-6)

FH = 10x - 12

Substitute x = 5

FH = 10(5) -12

FH = 50-12

FH = 38

3 0
3 years ago
I got 160 but it is not an answer choice, am I wrong?
Sergio039 [100]
Yes, it is 120 degrees
5 0
3 years ago
4. Last year there were 600 students
AleksAgata [21]

Answer:

There are 552 students

Step-by-step explanation: 600*0.08=48, 600-48=552

6 0
3 years ago
if an area can be washed at a rate of 4,900 cm2/ minute, how many square inches can be washed per hour?
Olenka [21]

Answer : 4.5 × 10⁴ square inches can be washed per hour.

Step-by-step explanation :

As we are given that an area can be washed at a rate of 4,900 cm²/min. Now we have to determine the square inches can be washed per hour.

Given conversions are:

1 cm = 0.39 in    and    1 hr = 60 min

As, 1 cm² = (0.39)² in²    and    1 min = 1/60 hr

So, 1cm^2/min=(0.39)^2\times 60in^2/hr

1cm^2/min=9.126in^2/hr

Now we have to determine the square inches can be washed per hour.

As, 1cm^2/min=9.126in^2/hr

So, 4900cm^2/min=\frac{4900cm^2/min}{1cm^2/min}\times 9.126in^2/hr=44717.4in^2/hr=4.5\times 10^4in^2/hr

Therefore, 4.5 × 10⁴ square inches can be washed per hour.

4 0
3 years ago
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