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gregori [183]
3 years ago
7

A ray of light is reflected off a mirror such that the reflected ray is perpendicular to the original ray, as shown in the diagr

am. What is the equation of the reflected ray?

Mathematics
2 answers:
PSYCHO15rus [73]3 years ago
6 0

answer could be

y+x = 6

(-2,-8)

Paladinen [302]3 years ago
3 0
The equation of the reflected ray is y+x = 6. Since based on the equation of the original ray is y-x=4 and the equation of the normal line y=5, the point where the normal line, the original ray and the reflected ray intersect is at point (1,5). By definition, substituting the normal line (y=5) and the point (1,5) to the given equations, y+x=6 satisfies the equality and therefore is the equation of the reflected ray
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Mike’s mom made a batch of soup that was 5 4/10 cups. Each serving was 6/10 cups. How many servings of soup did Mike’s mom make?
Mars2501 [29]

change 5 4/10 to an improper fraction  (10*5 +4)/10 = 54/10

54/10 divided by 6/10

copy dot flip

54/10* 10/6

54/6

9

There are 9 servings of soup


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Rate : every minute 7 pies are eaten 7:1
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2x+ 5+ 256 whats the answer and also JOIN HERE IF YOUR ARE BORED :)
Brums [2.3K]

Answer:

x = - 130.5

Step-by-step explanation:

2x + 5 + 256

2x = -261

x = - 130.5

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Two corporate baseball teams are scheduled to play a game together. They agree that if both teams attend or if neither team atte
sp2606 [1]

Answer:

2.99% probability that the cost will be paid by only one team

Step-by-step explanation:

The binomial probability distribution is used to solve this question.

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 combinations 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.

Probability that a team plays:

A team plays if it has at most 2 injured players out of 11.

11 players, so n = 11

Each player with a 5% probability of injury, so p = 0.05

Then

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

In which

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

P(X = 0) = C_{11,0}.(0.05)^{0}.(0.95)^{11} = 0.5688

P(X = 1) = C_{11,1}.(0.05)^{1}.(0.95)^{10} = 0.3293

P(X = 2) = C_{11,2}.(0.05)^{2}.(0.95)^{9} = 0.0867

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5688 + 0.3293 + 0.0867 = 0.9848

Each team has a 0.9848 probability of showing up to play.

What is the probability that the cost will be paid by only one team?

This happens if one team shows up and the other do not.

2 teams, so n = 2

Each team has a 0.9848 probability of showing up to play, so p = 0.9848.

This probability is P(X = 1).

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

P(X = 1) = C_{2,1}.(0.9848)^{1}.(0.0152)^{1} = 0.0299

2.99% probability that the cost will be paid by only one team

7 0
4 years ago
a bag has 5red,3blue and 7green marbles. what is the probability of randomly selecting a blue marble from the bag?
pickupchik [31]
Probability = yes/total
P(blue)=3/15= 1/5
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3 years ago
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