Why is my brainly account not getting deleted even after a waiting long time?

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

3 years ago

Whats the correct answer <br> 41/8928

chubhunter [2.5K]

The answer to 41 divided by 8,928 would be 0.0046

6 0

3 years ago

Find the nth term of 67,56,45,34

kvv77 [185]

Answer:

an = 67 -11(n -1)

Step-by-step explanation:

The terms of your arithmetic sequence have a common difference of -11. The first term is 67.

These values can be used in the generic formula for the n-th term:

an = a1 +d(n -1) . . . . . . first term a1, common difference d

an = 67 -11(n -1) . . . . . formula for the n-th term

7 0

2 years ago

I need the answers plz

garik1379 [7]

Answer:

These problems are an example of equations with two unknowns. The way these equations are solved is that we write these equations one under the another.

If both equations have, such is the case here, same parts, we can simply cancel the same parts out and subtract the rest of equatuons. That way, we are left with only one unknown (the other one was eliminated), which makes it easy to solve.

After we have found the value of an unknown, we just plug it back into any of the starting equations and solve for the second unknown.

2. Adult ticket costs $12 and child ticket costs $14.

3. Adult ticket costs $10 and child ticket costs $5.

4. One daylily costs $9 and one bush of ornamental grass costs $2.

5. A van can carry 15 and a bus can carry 56 students.

Step-by-step explanation:

2. If we mark the price of one adult ticket with x and the price of one child ticket with y, we get that:

- first day: 7x + 12y = $252

- second day: 7x + 10y = $224

Now, we can make a system:

7x + 12y = 252

7x + 10y = 224

We can now subtract these two equations and 7x will cancel out, so we get:

12y - 10y = 252 - 224

2y = 28

y = 14

Now, we can plug the value of y into any of the two equations:

7x + 10y = 224

7x + 140 = 224

7x = 84

x = 12

3. Similarly, if we mark the price of one adult ticket with x and the price of one child tickey with y, we'll get a system:

x + 12y = 70

x + 9y = 55

Again, if we subtract these two, x will cancel out, so we have:

12y - 9y = 70 - 55

3y = 15

y = 5

Now, we plug the value of y into any of the two equations, and we get:

x + 9y = 55

x + 45 = 55

x = 10

4. Using the same principle, we can mark the price of one daylily with x and the price of one bunch of ornamental grass with y, we'll get a system:

12x + 11y = 130

12x + 12y = 132

Again, we subtract so that 12x cancel out and we get:

11y - 12y = 130 - 132

-y = -2

If we get minuses on both sides, we can simply multiply both sides with -1 and we get:

y = 2

Again, we plug y:

12x + 12y = 132

12x + 24 = 132

12x = 108

x = 9

5. If we mark number of students in a van with x and the number of students in a bus with y, we get a system:

2x + 12y = 702

2x + y = 86

As you've probably already noticed the pattern, we subtract equations and cancel 2x out to get:

12y - y = 702 - 86

11y = 616

y = 56

Once again, we plug the value of y into any equation:

2x + y = 86

2x + 56 = 86

2x = 30

x = 15

4 0

3 years ago

What is the absolute value of -0.58?

cluponka [151]

.58. Absolute value gets rid of negative.

7 0

2 years ago

Read 2 more answers