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3 years ago

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What is the range of the function shown in the graph below?

1 answer:

Alex_Xolod [135]3 years ago

3 0

We need to see the graph. There is no pic attached.

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Math question down below

Bas_tet [7]

I would say 25 is the answer to it

4 0

3 years ago

Please help ASAP first right answer gets brainliest

ICE Princess25 [194]

Answer:

1. 31/99 2. 215/999 3. 6704/9999

3 0

3 years ago

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The equation tan(x- pi/3) is equal to _____.

kherson [118]

Answer:

B

Step-by-step explanation:

Using the expansion

tan(x - y) = $\frac{tanx-tany}{1+tanxtany}$ , then

tan(x - $\frac{\pi }{3}$ )

= $\frac{tanx-tan(-\frac{\pi }{3}) }{1+tanxtan(-\frac{\pi }{3}) }$

note that tan( - $\frac{\pi }{3}$ ) = - tan( $\frac{\pi }{3}$ ) = - $\sqrt{3}$

= $\frac{tanx-(-\sqrt{3}) }{1+tanx(-\sqrt{3}) }$

= $\frac{tanx+\sqrt{3} }{1-\sqrt{3}tanx }$ → B

6 0

3 years ago

Read 2 more answers

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per

ICE Princess25 [194]

Answer:

1.76% probability that in one hour more than 5 clients arrive

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 time interval.

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour.

This means that $\mu = 2$

What is the probability that in one hour more than 5 clients arrive

Either 5 or less clients arrive, or more than 5 do. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 5) + P(X > 5) = 1$

We want P(X > 5). So

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353$

$P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707$

$P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707$

$P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1804$

$P(X = 4) = \frac{e^{-2}*2^{4}}{(4)!} = 0.0902$

$P(X = 5) = \frac{e^{-2}*2^{5}}{(5)!} = 0.0361$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1353 + 0.2702 + 0.2702 + 0.1804 + 0.0902 + 0.0361 = 0.9824$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9824 = 0.0176$

1.76% probability that in one hour more than 5 clients arrive

8 0

3 years ago

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Solve –4(6x + 3) = –12(x + 10). A. x = 2 B. x = 9 C. x = 5 D. x = –5

sergey [27]

–4(6x + 3) = -12<span>(x + 10)
</span><span> -24x - 12 = -12x - 120
12x = 108
x = 9

answer
</span><span>B. x = 9</span><span>

</span>

6 0

3 years ago