A batch of 40 parts contains six defects. If two parts are drawn randomly one at a time without replacement, what is the probability that both parts are defective? b. If this experiment is repeated, with replacement, what is the probability that both parts are defective?

a) Therefore, the probability is P=1/52.

b) Therefore, the probability is P=9/400.

Step-by-step explanation:

We know that a  batch of 40 parts contains six defects.

a) We calculate the probability that the both parts are defective, if two parts are drawn randomly one at a time without replacement.

The probability for the first is 6/40.

The probability for the second is 5/39.  

We get:

Therefore, the probability is P=1/52.

b) We calculate the probability that the both parts are defective, if two parts are drawn randomly one at a time with replacement.

The probability for the first is 6/40.

The probability for the second is 6/40.  

We get:

Therefore, the probability is P=9/400.

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15