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statuscvo [17]
3 years ago
11

State whether the given measurements determine zero, one, or two triangles. C = 37°, a = 16, c = 14

Mathematics
2 answers:
notka56 [123]3 years ago
8 0

Answer:

sides are     a = 16                   c = 14             b =22.96

angles are   A = 43.39°           B = 99.61°        C = 37°

Step-by-step explanation:

using sine law of triangle,

\frac{sin A}{a} = \frac{Sin B}{b}= \frac{sinC}{c}

using,

\frac{sin A}{a} = \frac{sinC}{c}

given that,

C= 37°                 a = 16                   c = 14

now,

\frac{sin A}{16} = \frac{sin37^o}{14}

sin A = 0.687

A = 43.39°

hence B = 180° - (43.39° + 37°)

           B = 99.61°

\frac{sin 43.39^o}{16} = \frac{sin99.61^o}{b}

b = 22.96

Hence, there will be only one triangle which will form and there

sides are     a = 16                   c = 14             b =22.96

angles are   A = 43.39°           B = 99.61°        C = 37°

Sveta_85 [38]3 years ago
5 0

Two triangles: b ≈ 2.62 or 22.94

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a) P(X \leq 4) = 0.6289

P(X < 4) = 0.4335

b) P(4 \leq X \leq 8) = 0.5452

c) P(X \geq 8) = 0.0511

d) 0.2605

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

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In which

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e = 2.71828 is the Euler number

\mu is the mean in the given time interval.

It is important to know that the variance has the same value as the mean in a Poisson distribution. The standard deviation is the square root of the variance.

In this problem, we have that:

\mu = 4, \sigma = \sqrt{4} = 2.

To help our solution, i am going to find each of P(X = x) from x = 0 to 8[/tex]

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-4}*4^{0}}{(0)!} = 0.0183

P(X = 1) = \frac{e^{-4}*4^{1}}{(1)!} = 0.0733

P(X = 2) = \frac{e^{-4}*4^{2}}{(2)!} = 0.1465

P(X = 3) = \frac{e^{-4}*4^{3}}{(3)!} = 0.1954

P(X = 4) = \frac{e^{-4}*4^{4}}{(4)!} = 0.1954

P(X = 5) = \frac{e^{-4}*4^{5}}{(5)!} = 0.1563

P(X = 6) = \frac{e^{-4}*4^{6}}{(6)!} = 0.1042

P(X = 7) = \frac{e^{-4}*4^{7}}{(7)!} = 0.0595

P(X = 8) = \frac{e^{-4}*4^{8}}{(8)!} = 0.0298

(a) Compute both P(X ≤ 4) and P(X < 4)

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0183 + 0.0733 + 0.1465 + 0.1954 + 0.1954 = 0.6289

---------

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0183 + 0.0733 + 0.1465 + 0.1954 = 0.4335

(b) Compute P(4 ≤ X ≤ 8).

P(4 \leq X \leq 8) = P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.1954 + 0.1563 + 0.1042 + 0.0595 + 0.0298 = 0.5452

(c) Compute P(8 ≤ X)

That is P(X \geq 8).

The sum of the probabilities must be 1 in decimal. Either X is greater or equal to 8, or X is lesser than 8.

So

P(X < 8) + P(X \geq 8) = 1

P(X \geq 8) = 1 - P(X < 8)

From what we have in a) and b)

P(X < 8) = P(X < 4) + P(4 \leq X < 8) = 0.4335 + 0.1954 + 0.1563 + 0.1042 + 0.0595 = 0.9489

So

P(X \geq 8) = 1 - P(X < 8) = 1 - 0.9489 = 0.0511

(d) What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation

One standard deviation is 2, and the mean is 4. So, this is:

P(4 < X \leq 6) = P(X = 5) + P(X = 6) =  0.1563 + 0.1042 = 0.2605

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