Ask question

Ask question

All categories

1 year ago

5

Computer keyboard failures are due to faulty electrical connects (12%) or mechanical defects (88%). Mechanical defects are relat

ed to loose keys (27%) or improper assembly (73%). Electrical connect defects are caused by defective wires (35%), improper connections (13%), or poorly welded wires (52%). a) Find the probability that a failure is due to loose keys. b) Find the probability that a failure is due to improperly connected or poorly welded wires.

2 answers:

777dan777 [17]1 year ago

8 0

Answer:

Probability that a failure is due to loose keys = 0.2376

Probability that a failure is due to improperly connected or poorly welded wires 0.078

Step-by-step explanation:

Given data:

P(A) = percentage of faulty electrical connections is 12% = 0.12

P(B) = mechanical defects is 88% = 0.88

P(C / B ) = percentage of defects that are related to loose keys is 27% = 0.27

P( D/B ) = improper assembly is 73% = 0.73

P(E/A) = defective wires is 35% = 0.35

P( F/A)= improper connections is 13% = 0.13

P( G/A ) = percentage of poorly welded wire is 52% = 0.52

a) Probability of being failure because of loose keys = 0.27\times 0.88 = 0.2376

b) Improperly connected = 0.13\times 0.12 = 0.0156

poorly welded wires= 0.52 \times 0.12 = 0.0624

Probability of being failure because of improperly connected or poorly welded wires = 0.0156 + 0.0624 = 0.078

Margaret [11]1 year ago

6 0

Answer:

a) 23.76%

b) 7.8%

Step-by-step explanation:

a) probability that a failure is due to loose keys.

loose key failure (27%) comes under mechanical failure(88%)

hence, probability that a failure is due to loose keys= 0.27×0.88= 0.2376= 23.76%

b) probability that a failure is due to improperly connected wire which comes under electrical failure = 0.12×0.13

probability that a failure is due to poorly welded wires which comes under electrical failure= 0.52×0.12

now, the probability that a failure is due to improperly connected or poorly welded wires. = 0.12(0.52+0.13)= 0.078= 7.8%

You might be interested in

713.49 written in a standard form

Vikki [24]

Answer: one hundred seven thirteen and forty-nine hundreds

Step-by-step explanation: Standard form is a way of writing down very large or very small numbers easily. 103 = 1000, so 4 × 103 = 4000 . So 4000 can be written as 4 × 10³ . This idea can be used to write even larger numbers down easily in standard form.

7 0

1 year ago

Chandresh is helping his father paint the fence in their back yard. The fence is represented by the shaded part of the diagram b

Lyrx [107]

B. The total area painted is 864; they must buy two cans of paint.

Step-by-step explanation:

Step 1:

A rectangle's area can be calculated by multiplying its length and its width. The wall is made up of 5 different types of rectangular walls. All walls are 8 feet tall but the length varies.

Step 2:

The area of the 20 feet long wall = $(length)(width)= (20)(8) =160,$

The area of the 10 feet long wall = $(length)(width)= (10)(8) =80,$

The area of the 5 feet long wall = $(length)(width)= (5)(8) =40,$

The area of the 4 feet long wall = $(length)(width)= (4)(8) =32,$

The area of the 15 feet long wall = $(length)(width)= (15)(8) =120.$

The area of all the walls = 432 square feet.

Since there are two sides for every wall, total area = $432(2)=864$ square feet.

Step 3:

If one paint can covers 500 square feet,

the number of cans required to paint 864 square feet = $\frac{864}{500} = 1.728$ cans.

so 2 paint cans are needed to paint 864 square feet which is option B.

7 0

2 years ago

If they charge $2.00 per quart, how much money will they make if they sell it all

Tcecarenko [31]

Answer:

Give me more information

Step-by-step explanation:

5 0

2 years ago

17. It takes one minute to fill (3/7) th of a vessel. What is the time taken in

Vinil7 [7]

Answer:

Given that 1minute =3/7of the vessel

let the time taken in minutes to fill the whole of the vessel be' x' minute.

1min =3/7

xmin =whole vessel =1=7/7

=3x/7=1=x=7/3=2.3min=2min.3sec

8 0

1 year ago

The random variable Y is normally distributed with a mean of 470, and the value Y=340 is at the 15th percentile of the distribut

adell [148]

The empirical rule says that about 68% of any normal distribution lies within one standard deviation of the mean. This leaves 32% of the distribution that lies outside this range, with about 16% to either side.

At the 16th percentile, there is a value of $Y=y$ such that

$\mathbb P(Y$

where $\sigma$ is the standard deviation for the distribution, and $Z$ is the random variable corresponding to the standard normal distribution. This value of $y$ would correspond roughly to a z-score of $Z=-1$ .

You're told that $Y=340$ lies at the 15th percentile, so that

$\mathbb P(Y$

Roughly, then, it'd be fair to say that $y\approx340$ . So you have

$\dfrac{340-470}\sigma\approx-1\implies\sigma\approx130$

which falls between (A) and (B). To narrow down the choice, notice that $y$ would have be slightly larger than 340 in order to have $\mathbb P(Y$ . This brings $y$ closer to the mean, and thus suggests the standard deviation for the distribution is actually smaller than our approximation.

This tells us that (A) is the answer.

4 0

1 year ago