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Kay [80]
2 years ago
6

What is the expanded 97,367

Mathematics
2 answers:
brilliants [131]2 years ago
4 0

Answer:

Step-by-step explanation:

Kkkjjkjjkjkkmjjmjjjjjjjjjjjmjjjj

Ivahew [28]2 years ago
3 0

Answer:

Expanded Notation Form:

 97,367 =

 90,000  

+ 7,000  

+ 300  

+ 60  

+ 7  

Expanded Factors Form:

   97,367 =

 9 × 10,000  

+ 7 × 1,000  

+ 3 × 100  

+ 6 × 10  

+ 7 × 1  

Expanded Exponential Form:

 97,367 =

9 × 104

+ 7 × 103

+ 3 × 102

+ 6 × 101

+ 7 × 100

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Answer:

answer below

Step-by-step explanation:

∠3 = 76°

∠1 = ∠3 = 76°  (vertical angle)

∠2 = 180 - ∠1 = 180° - 76° = 104°  (supplementary adjacent angle)

∠4 = ∠2 = 104°   (vertical angle)

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3 years ago
Hola, estoy empezando a ver trigonometria queria ver si me pueden ayudar con este problema: Un extraterrestre diminuto pero horr
Kryger [21]

Answer:

El agente debe disparar a un ángulo aproximado de 87.7° con respecto a la horizontal.

Step-by-step explanation:

Sea E la localización del extraterrestre, O, el del hombre de negro, y B, el de la base del edificio. A partir de la información disponible, construimos el diagrama geométrico que indica la presencia de un triángulo rectángulo, el cual puede consultarse en la imagen adjunta.

Podemos determinar el ángulo de disparo mediante la siguiente relación trigonométrica inversa:

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2 years ago
If electricity power failures occur according to a Poisson distribution with an average of 3 failures every twenty weeks, calcul
Wittaler [7]

Answer:

0.9898 = 98.98% probability that there will not be more than one failure during a particular week.

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:

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

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

3 failures every twenty weeks

This means that for 1 week, \mu = \frac{3}{20} = 0.15

Calculate the probability that there will not be more than one failure during a particular week.

Probability of at most one failure, so:

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

Then

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

P(X = 0) = \frac{e^{-0.15}*0.15^{0}}{(0)!} = 0.8607

P(X = 1) = \frac{e^{-0.15}*0.15^{1}}{(1)!} = 0.1291

Then

P(X \leq 1) = P(X = 0) + P(X = 1) = 0.8607 + 0.1291 = 0.9898

0.9898 = 98.98% probability that there will not be more than one failure during a particular week.

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2 years ago
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