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LenaWriter [7]
3 years ago
5

ASAP!!! PLEASE help me solve this question!!!! I really need help...

Mathematics
2 answers:
GenaCL600 [577]3 years ago
6 0

Answer:

304(pi) g

Step-by-step explanation:

First we find the volume of the hollow ball. Then we find the mass using the volume and density.

Let R = exterior radius = 3 cm

Let r = interior radius = 2 cm

volume = exterior volume - interior volume

volume = (4/3)(pi)R^3 - (4/3)(pi)r^3

volume = (4/3)(pi)(R^3 - r^3)

volume = (4/3)(pi)(3^3 - 2^3) cm^3

volume = (4/3)(pi)(27 - 8) cm^3

volume = (76/3)pi cm^3

Now we use the density and the volume to find the mass.

density = mass/volume

mass = density * volume

mass = 12 g/cm^3 * (76/3)pi cm^3

mass = 304(pi) g

Answer: 304(pi) g

bija089 [108]3 years ago
4 0

Answer:

m=(76π/3 )(12)=304πg

Step-by-step explanation:

volume of sphere=(4πr³)/3

V=4π(2)³/3=32π/3 of the interior

V of exterior=4π(3)³/3=36π

Volume of the whole metal =108π/332π/3=76π/3=25 1/3 π

m=V*D ( volume * density)=

m=(76π/3 )(12)=304πg

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Answer:

  (x, y) = (77/240, -3/10)

Step-by-step explanation:

It is convenient to write the equations in standard form.

Multiplying the first equation by 21 gives ...

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Multiplying the second equation by 8 gives ...

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Then the system of equations in standard form is ...

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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 A 5 mL volume of the suspension is withdrawn.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Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows: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

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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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(d)

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

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

<em>z</em> = 1.96

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

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Thus, the sample selected must be of size 240.

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