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Soloha48 [4]
3 years ago
11

What is the correct expansion of (3x-2)(2x^2+5)

Mathematics
2 answers:
Murljashka [212]3 years ago
7 0
Foil or whatever
just multiply

remember
(a-b)(c+d)=ac-bc+ad-bd
and
(ax^m)(bx^n)=(a)(x^m)(b)(x^n)=abx^{m+n}

so

(3x-2)(2x²+5)=
(3x)(2x²)+(-2)(2x²)+(3x)(5)+(-2)(5)=
6x³-4x²+15x-10
wolverine [178]3 years ago
6 0
3x*2x^2=6x^3
3x*5=15x
-2*2x^2=-4x^2
-2*5=-10
6x^3+15x-4x^2-10=
6x^3-4x^2+15x-10
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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 A 5 mL volume of the suspension is withdrawn.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
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Answer:

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(b) 0.2961

(c) 0.3108

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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,

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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

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Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows: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Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows: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

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(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A 10 mL sample is withdrawn.

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows: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

                             =P(-0.40

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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