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3 years ago

Please help ?!!!!!!!

Mathematics

1 answer:

HACTEHA [7]3 years ago

3 0

Answer: N is the transversal.

Step-by-step explanation:

its the transversal because lines L and M are parallel to each other and line N intersects them both which means line N is the transversal. a transversal line means that it is a line that crosses two or more lines on a plane is a transversal. a transversal intersecting two parallel lines creates two different types of angled pairs.

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There are 196 people in

lianna [129]

Answer:

they need 4 buses

60+60+60=180 that does not get you to 196 so you would add 1 more bus

which equals 4 buses

8 0

3 years ago

The probability that a single radar station will detect an enemy plane is 0.65.

taurus [48]

Answer:

a) We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

b) If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

Step-by-step explanation:

For each station, there are two two possible outcomes. Either they detected the enemy plane, or they do not. This means that we can solve this problem using concepts of the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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

And p is the probability of X happening.

In this problem, we have that:

The probability that a single radar station will detect an enemy plane is 0.65. This means that $n = 0.65$ .

(a) How many such stations are required to be 98% certain that an enemy plane flying over will be detected by at least one station?

This is the value of n for which $P(X = 0) \leq 0.02$ .

n = 1.

$P(X = 0) = C_{1,0}.(0.65)^{0}.(0.35)^{1} = 0.35$

n = 2

$P(X = 0) = C_{2,0}.(0.65)^{0}.(0.35)^{2} = 0.1225$

n = 3

$P(X = 0) = C_{3,0}.(0.65)^{0}.(0.35)^{3} = 0.0429$

n = 4

$P(X = 0) = C_{4,0}.(0.65)^{0}.(0.35)^{4} = 0.015$

We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

(b) If seven stations are in use, what is the expected number of stations that will detect an enemy plane?

The expected number of sucesses of a binomial variable is given by:

$E(x) = np$

So when $n = 7$

$E(x) = 7*(0.65) = 4.55$

If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

6 0

3 years ago

20 + 30 x 0 + 1 = what ?

natta225 [31]

21 is the real answer
because of BODMAS

6 0

3 years ago

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Given that a line has a slope

zheka24 [161]

Answer: the last option (:

Step-by-step explanation:

You always put the slope in front of the x intercept in the equation.

5 0

4 years ago

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Nina has prepared the following two-column proof below. She is given that ∠OLN ≅ ∠LNO and she is trying to prove that OL ≅ ON. S

Anit [1.1K]

Step number 3 should really be step number 7, it should be placed after step 4, 5, and 6. The reason is because we won't know that ∠LEO ≅ ∠NEO until after we learn that LE ≅ EN. Because of this, step 3 is in the wrong spot (mistake number one). The secod mistake is that step 7, triangle OLE ≅ triangle ONE is congruent by Angle-Side-Angle (ASA) Postulate, not Side-Angle-Side (SAS) Postulate. It is congruent by ASA because we know that both triangles have equal angles N and L. We also know that the perpendicular bisector creates a 90° angle. So m∠LEO = 90° and ∠NEO = 90°. Therefore, we already have 2 congruent angles in both of the triangles. We also learn that line LE ≅ EN based on the definition of a perpendicular bisector, so we have know one that one side of each triangle is congruent. It is ASA and not AAS, because the ASA Postulate states that two angles and one included side of one triangle are congruent to two angles and one included side of another triangle.

7 0

4 years ago