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A product is made up of components A, B, C, D, E, F, G, H, I, and J. Components A, B, C, and F have a 1/10,000 chance of failure

lesantik [10]

Answer:

The overall reliability is 99.7402 %

Step-by-step explanation:

The overall reliability of the product is calculated as the product of the working probability of the components.

For components A,B,C and F we have :

$P(failure)=\frac{1}{10000}$

⇒

$P(Work)=1-\frac{1}{10000}=\frac{9999}{10000}=0.9999$

For components D,E,G and H we have :

$P(failure)=\frac{3}{10000}$

⇒

$P(Work)=1-\frac{3}{10000}=\frac{9997}{10000}=0.9997$

Finally, for components I and J :

$P(failure)=\frac{5}{10000}$

⇒

$P(Work)=1-\frac{5}{10000}=\frac{9995}{10000}=0.9995$

Now we multiply all the working probabilities. We mustn't forget that we have got ten components in this case :

Components A,B,C and F with a working probability of 0.9999

Components D,E,G and H with a working probability of 0.9997

Components I and J with a working probability of 0.9995

Overall reliability = $(0.9999)^{4}(0.9997)^{4}(0.9995)^{2}=0.997402$

0.997402 = 99.7402 %

4 0

3 years ago

50 Points!

Lesechka [4]

Answer:

D 1/4

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Jackson bought 555 ounces of raisins for \$4$4dollar sign, 4. How much do raisins cost per ounce?

stepan [7]

Answer:

$0.80/oz

Step-by-step explanation:

$4 for 5 ounces

unit cost = price/weight = $4/(5 oz.) = $0.80/oz

8 0

3 years ago

If 9a+5b+5c=9 what is -10c-18a-10b?

Rus_ich [418]

9a + 5b + 5c = 9

Multiplying the whole equation by -2.

-18a - 10b - 10c = -18.

Hence, the answer is -18.

6 0

3 years ago

A slitter assembly contains 48 blades. Five blades are selected at random and evaluated each day of sharpness. If any dull blade

Alex73 [517]

Answer:

Part a

The probability that assembly is replaced the first day is 0.7069.

Part b

The probability that assembly is replaced no replaced until the third day of evaluation is 0.0607.

Part c

The probability that the assembly is not replaced until the third day of evaluation is 0.2811.

Step-by-step explanation:

Hypergeometric Distribution: A random variable x that represents number of success of the n trails without replacement and M represents number of success of the N trails without replacement is termed as the hypergeometric distribution. Moreover, it consists of fixed number of trails and also the two possible outcomes for each trail.

It occurs when there is finite population and samples are taken without replacement.

The probability distribution of the hyper geometric is,

$P(x,N,n,M)=\frac{(\limits^M_x)(\imits^{N-M}_{n-x})}{(\limits^N_n)}$

Here x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.

Probability that at least one of the trail is succeed is,

$P(x\geq1)=1-P(x$

(a)

Compute the probability that the assembly is replaced the first day.

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48.

Number of blades selected at random from the assembly n= 5

Number of blades in an assembly dull is M = 10.

The probability mass function is,

$P(X=x)=\frac{[\limits^M_x][\limits^{N-M}_{n-x}]}{[\limits^N_n]};x=0,1,2,...,n\\\\=\frac{[\limits^{10}_x][\limits^{48-10}_{5-x}]}{[\limits^{48}_5]}$

The probability that assembly is replaced the first day means the probability that at least one blade is dull is,

$P(x\geq 1)=1- P(x$

(b)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48

Number of blades selected at random from the assembly N = 5

Number of blades in an assembly dull is M = 10

From the information,

The probability that assembly is replaced (P) is 0.7069.

The probability that assembly is not replaced is (Q) is,

$q=1-p\\= 1-0.7069= 0.2931$

The geometric probability mass function is,

$P(X = x)= q^{x-1} p; x =1,2,....=(0.2931)^{x-1}(0.7069)$

The probability that assembly is replaced no replaced until the third day of evaluation is,

$P(X = 3)=(0.2931)^{3-1}(0.7069)\\=(0.2931)^2(0.7069)= 0.0607$

(c)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48

Number of blades selected at random from the assembly n = 5

Suppose that on the first day of the evaluation two of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M = 2.

$P(x=0)=\frac{(\limits^2_0)(\limits^{48-2}_{5-0})}{\limits^{48}_5}\\\\=\frac{(\limits^{46}_5)}{(\limits^{48}_5)}\\\\= 0.8005$

Suppose that on the second day of the evaluation six of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M = 6.

$P(x=0)=\frac{(\limits^6_0)(\limits^{48-6}_{5-0})}{(\limits^{48}_5)}\\\\=\frac{(\limits^{42}_5}{(\limits^{48}_5)}\\\\= 0.4968$

Suppose that on the third day of the evaluation ten of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M

= 10.

$P(x\geq 1)=1- P(x$

The probability that the assembly is not replaced until the third day of evaluation is,

P(The assembly is not replaced until the third day)=P(The assembly is not replaced first day) x P(The assembly is not replaced second day) x P(The assembly is replaced third day)

=(0.8005)(0.4968)(0.7069)= 0.2811

5 0

4 years ago