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kiruha [24]
4 years ago
6

Points J and K lie in plane H. Plane H contains two points. Point J is to the lower left and point K is to the uppercase right.

How many lines can be drawn through points J and K? 0 1 2 3
Mathematics
2 answers:
FrozenT [24]4 years ago
9 0

Answer:

The correct option is B) 1

Step-by-step explanation:

Consider the provided information.

It is given that there is a plane H on which two pints J and K are located.

We need to find how many lines can be drawn through the points J and K.

The Axiom of Euclid's geometry says that "Through any given two points X and Y, only one line can be drawn "

Therefore by Axiom in Euclid's geometry we can concluded that for the given points J and K in plane H , only one line can be drawn through points J and K.

Hence, the correct option is B) 1

xxMikexx [17]4 years ago
4 0

Answer:

1

Step-by-step explanation:

There's a theorem that says "through 2 points it only cross one and only one line". Having this as our start point, we can guarantee that there's only one line that can be drawn through those points

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11) Forty percent of the homes constructed in the Quail Creek area include a security system. Three homes are selected at random
Alekssandra [29.7K]

Answer:

1) Pr(all three) = 0.064

2) Pr(none) = 0.216

3) Independent event

Step-by-step explanation:

Let Probability of homes constructed in the Quail Creek area with a security system = Pr (security)

Where Pr = probability

This is a binomial distribution problem. There ate only two outcomes in this distribution: a success and a failure

Where p = success, q = failure

For n trials,

Pr(X = x) = n!/[(n-x)!x!] × p^x × q^(n-x)

Pr (security) = 40% = 0.4

p = 0.4

q = 1-p = 1-0.4 = 0.6

Numbers selected at random = n = 3

See attachment for more details on the workings.

1) Probability all three of the selected homes have a security system = Pr(all three)

Pr(all three) = 1× (0.4)³ (0.6)^0 = 0.4³

= 0.4×0.4×0.4

Pr(all three) = 0.064

2) Probability none of the three selected homes have a security system

= Pr(none)

Pr(none) = 1 × (0.4)^0 × (0.6)³

= 0.6³ = 0.6×0.6×0.6

Pr(none) = 0.216

3) The events were assumed to be independent as the selection of one house doesn't affect the outcome of the selection of another.

6 0
3 years ago
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