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Nady [450]
2 years ago
10

How many triangles can be formed by joining the vertices of an octagon?need answer and solution pls​

Mathematics
1 answer:
Scrat [10]2 years ago
7 0

Answer:

6 triangles

Solution: In a regular octagon, by joining one vertex to the remaining non-adjacent vertices, 6 triangles can be formed.Step-by-step explanation:

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Select the function that represents the following transformations of f(x) = | x |:
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g

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Which symbol correctly compares the fractions below?<br> 1/8 ? 1/2
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<

Step-by-step explanation:

1/2 = 4/8

=> 1/8 < 4/8

=> 1/8 < 1/2

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1467

Step-by-step explanation:

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2 years ago
On a particular day, the wind added 4 miles per hour to Alfonso’s rate when he was cycling with the wind and subtracted 4 miles
Greeley [361]

d = r · t ⇒ t = d/r

going

t = 54/(r + 4)

coming

t = 30/(r - 4)

54/(r + 4) = 30/(r - 4) times are equal

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8 0
3 years ago
Military radar and missile detection systems are designed to warn a coutry of an enemy attack. Assume that a particular detectio
Tom [10]

Answer:

0.96 = 96% probability that at least one of them detect an enemy attack.

Step-by-step explanation:

For each radar, there are only two possible outcomes. Either it detects the attack, or it does not. The missiles are operated independently, which means that we use the binomial probability distribution 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.

Assume that a particular detection system has a 0.80 probability of detecting a missile attack.

This means that p = 0.8

If two military radars are installed in two different areas and they operate independently, the probability that at least one of them detect an enemy attack is

This is P(X \geq 1) when n = 2. So

P(X \geq 1) = 1 - P(X = 0)

In which

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

P(X = 0) = C_{2,0}.(0.8)^{0}.(0.2)^{2} = 0.04

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.04 = 0.96

0.96 = 96% probability that at least one of them detect an enemy attack.

6 0
3 years ago
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