Suppose you just received a shipment of seven televisions. Four of the televisions are defective. If two televisions are randomly selected, compute the probability that both televisions work. What is the probability at least one of the two televisions does not work?
The probability that both televisions work is 0.143
(Round to three decimal places as needed.)
The probability that at least one of the two televisions does not work is I
(Round to three decimal places as needed.)
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answer: standard form is ax^2 + bx + c = 0

step-by-step explanation:

standard form is ax^2 + bx + c = 0

(2x + 1)(x - 2) = 0

distribute to get into standard form

2x^2 - 4x + x - 2 = 0

2x^2 - 3x - 2 = 0

a = 2, b= -3, c = -2

a is your answer

x=17

step-by-step explanation:

solve 3x-24=x+10

then that makes it 2x=34 because you subtract an x from both sides to get rid of the one on the right, and then you add 24 to both sides to get rid of the 24 on the left

divide by both sides by 2 to get your answer so that you can get your answer

∠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.

what is the original number? you forgot to add it.