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natka813 [3]
3 years ago
7

The first step in determining the solution to the system of equations, y = -x2 - 4x – 3 and y = 2x + 5, akyettically is to set t

he
two equations equal as -* - 4x - 3 = 2x + 5. What is the next step?

A.Set y = 0 in y = -x2 - 4x - 3.
B.Factor each side of the equation
C.Use substitution to create a one variable equation
D.Combine like terms onto one side of the equation.
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Mathematics
1 answer:
riadik2000 [5.3K]3 years ago
6 0

Answer:

D

Step-by-step explanation:

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Find k so that the following function is continuous:<br> f(x)={kx8x2if0≤x&lt;5if5≤x.
tankabanditka [31]

Check the one-sided limits:

\displaystyle \lim_{x\to5^-}f(x) = \lim_{x\to5}kx = 5k

\displaystyle \lim_{x\to5^+}f(x) = \lim_{x\to5}8x^2 = 200

If <em>f(x)</em> is to be continuous at <em>x</em> = 5, then these two limits should have the same value, which means

5<em>k</em> = 200

<em>k</em> = 200/5

<em>k</em> = 40

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Cheryl has $100 on her subway gift card. Each week she buys the same lunch for $10. She wants to continue using the card until t
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Answer:

100-60=40

Step-by-step explanation:

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3 years ago
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Mini muffins cost $3 per dozen
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Answer:

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3 years ago
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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