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cupoosta [38]
3 years ago
8

Compute the function table Draw the graph of each function f(x) = 2x+1

Mathematics
1 answer:
Wittaler [7]3 years ago
8 0

Answer:

<u>Function Table</u>

<u>x     |   f(x)</u>

0     |   1

1      |   3

2     |   5

View Image For Graph

Step-by-step explanation:

This function follows the general equation for the slope-intercept formula:

y = mx + b

m is the slope, which is how much the graph move up whenever to move to the right by 1. In our case, our slope is m=2, so every time we move right by 1, we have to move up by 2.

b is the y-intercept. it's basically the starting point when x is 0. In our case, b=1 so we start at (0, 1)

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Solve for y 2x+y-3z=5
Elza [17]

Answer:

<h3>A. y=-2x+3z+25</h3>

Step-by-step explanation:

Isolate the term of x and y from one side of the equation.

<u>To solve:</u>

  • The value of y.

\Longrightarrow: \sf{\sqrt{2x+y-3z}=5}

<h3>2x+y-3z=25</h3>

<u>First, you have to subtract by 2x-3z from both sides.</u>

\Longrightarrow: \sf{2x+y-3z-\left(2x-3z\right)=25-\left(2x-3z\right)}}

<u>Solve.</u>

\Longrightarrow: \boxed{\sf{y=25-2x+3z}}}

  • <u>Therefore, the correct answer is "A. y=-2x+3z+25".</u>

I hope this helps, let me know if you have any questions.

4 0
2 years ago
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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

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