Please help me Please!!!!!!!!!! :((((

You have received an order of 100 robotic resistance spot welders which contains 5 defective welders. You randomly select 15 wel

erica [24]

Answer:

a)

$P(X = 0) = h(0,100,15,5) = \frac{C_{5,0}*C_{95,15}}{C_{100,15}} = 0.4357$

$P(X = 1) = h(1,100,15,5) = \frac{C_{5,1}*C_{95,14}}{C_{100,15}} = 0.4034$

$P(X = 2) = h(2,100,15,5) = \frac{C_{5,2}*C_{95,13}}{C_{100,15}} = 0.1377$

$P(X = 3) = h(3,100,15,5) = \frac{C_{5,3}*C_{95,12}}{C_{100,15}} = 0.0216$

$P(X = 4) = h(4,100,15,5) = \frac{C_{5,4}*C_{95,11}}{C_{100,15}} = 0.0015$

$P(X = 5) = h(5,100,15,5) = \frac{C_{5,5}*C_{95,10}}{C_{100,15}} = 0.00004$

b) 0.154% probability that there are at least 4 defective welders in the sample

Step-by-step explanation:

The welders are chosen without replacement, so the hypergeometric distribution is used.

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

100 welders, so $N = 100$

Sample of 15, so $n = 15$

In total, 5 defective, so $k = 5$

(a) Determine the PMF of the number of defective welders in your sample?

There are 5 defective, so this is P(X = 0) to P(X = 5). Then

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 0) = h(0,100,15,5) = \frac{C_{5,0}*C_{95,15}}{C_{100,15}} = 0.4357$

$P(X = 1) = h(1,100,15,5) = \frac{C_{5,1}*C_{95,14}}{C_{100,15}} = 0.4034$

$P(X = 2) = h(2,100,15,5) = \frac{C_{5,2}*C_{95,13}}{C_{100,15}} = 0.1377$

$P(X = 3) = h(3,100,15,5) = \frac{C_{5,3}*C_{95,12}}{C_{100,15}} = 0.0216$

$P(X = 4) = h(4,100,15,5) = \frac{C_{5,4}*C_{95,11}}{C_{100,15}} = 0.0015$

$P(X = 5) = h(5,100,15,5) = \frac{C_{5,5}*C_{95,10}}{C_{100,15}} = 0.00004$

(b) Determine the probability that there are at least 4 defective welders in the sample?

$P(X \geq 4) = P(X = 4) + P(X = 5) = 0.0015 + 0.00004 = 0.00154$

0.154% probability that there are at least 4 defective welders in the sample

5 0

3 years ago

Use the numbers 0 through 9 to complete each equation so the statement about it's solution is true

belka [17]

For the first equation, to get an equation that only has 1 solution your two missing numbers could be 3x and 6. This is because when you solve it, you get 7x + 2 = 3x + 6
- 3x
4x + 2 = 6
- 2
4x = 4
x = 1

For the second equation, your two numbers could be 7x and 4, because it means that x has to be different as 7 multiplied by one number and added to 2 cannot equal 7 multiplied by the same number but added to 4.

I hope this helps!

3 0

3 years ago

Solve by factoring 12x^3+12x^2=0

stepladder [879]

Factor out the GCF
12x^2(x + 1) = 0
Now you can do two parts
12x^2 = 0, x = 0
x + 1 = 0, x = -1

Solution: x = 0, x = -1

7 0

3 years ago

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babymother [125]

Answer: $y-0 = \frac{5}{6}(x+8)$