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Pavlova-9 [17]
3 years ago
10

Given :f(x) =1/x+5 and g(x) =x-2 What are the restrictions of the domain of f(g(x))?

Mathematics
2 answers:
LUCKY_DIMON [66]3 years ago
8 0

Answer:x≠-3

Step-by-step explanation:

emmainna [20.7K]3 years ago
5 0
I don't know if it's (1/x )+ 5 or 1/(x+5) anyways
g(x) = x-2 so in f(g(x)) you replace every x with x-2
f(g(x)) = (1/(x-2)) + 5 or 1/(x-2-5) = 1/(x-7)
if the function is like the first form so you avoid numbers which will make your denominator equals zero
x-2 = 0, x=2 or x-7 = 0, x=7
so if it's like first one answer is R - 2
if it's second answer is R - 7
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The concentration of particles in a suspension is 50 per mL. A 5 mL volume of the suspension is withdrawn. a. What is the probab
kolezko [41]

Answer:

(a) 0.6579

(b) 0.2961

(c) 0.3108

(d) 240

Step-by-step explanation:

The random variable <em>X</em> can be defined as the number of particles in a suspension.

The concentration of particles in a suspension is 50 per ml.

Then in 5 mL volume of the suspension the number of particles will be,

5 × 50 = 250.

The random variable <em>X</em> thus follows a Poisson distribution with parameter, <em>λ</em> = 250.

The Poisson distribution with parameter λ, can be approximated by the Normal distribution, when λ is large say λ > 10.  

The mean of the approximated distribution of X is:

μ = λ  = 250

The standard deviation of the approximated distribution of X is:

σ = √λ  = √250 = 15.8114

Thus, X\sim N(250, 250)

(a)

Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows:

P(235

                             =P(-0.95

Thus, the value of P (235 < <em>X</em> < 265) = 0.6579.

(b)

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

P(48

                             =P(-0.28

Thus, the value of P(48.

(c)

A 10 mL sample is withdrawn.

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

P(48

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Thus, the value of P(48.

(d)

Let the sample size be <em>n</em>.

P(48

                             0.95=P(-z

The value of <em>z</em> for this probability is,

<em>z</em> = 1.96

Compute the value of <em>n</em> as follows:

z=\frac{\bar X-\mu}{\sigma/\sqrt{n}}\\\\1.96=\frac{48-50}{15.8114/\sqrt{n}}\\\\n=[\frac{1.96\times 15.8114}{48-50}]^{2}\\\\n=240.1004\\\\n\approx 241

Thus, the sample selected must be of size 240.

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Answer:

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Step-by-step explanation:

If you want a line parallel to the equation, the line must have the same slope.

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You are looking for a line where the slope is 2/7.

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y = 3/4x + 1

Slope is 3/4.

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And there's your answer: e.

Hope this helps!

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Step-by-step explanation:

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