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3 years ago

Please help me with my homework

Mathematics

1 answer:

wariber [46]3 years ago

6 0

Answer:

steps below

Step-by-step explanation:

5-1) Center of dilation: intersect of AA' and BB' (or CC' ...) (M in the graph)

5-2) scale factor = A'B' / AB = 6/4 = 1.5

5-3) ratio of all corresponding sides are equal = 1.5 and parallel, it also means all corresponding angles are similar.

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

erica [24]

Answer:

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

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

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

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

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

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

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

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