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OverLord2011 [107]
2 years ago
10

I need help with this question

Mathematics
1 answer:
V125BC [204]2 years ago
3 0

Answer:

Hi there, I like your pfp!

The correct answer should be 11 1/2!

Step-by-step explanation:

-48/-8 = 6 4/8 = 6 1/2

6 1/2 + 5 =

11 1/2

Anyways, hope this helped!

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D for the same reason percents

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Between x=2 and x=3, which function has a greater average rate of change than f(x)=2^x has?
maxonik [38]

Answer: f(3)

Step-by-step explanation:

First find the formula for the rate of change by taking the derivative of 2^x. Let f(x) equal some hypothetical y-value, then take the natural log of both sides.

y=2^x\\\ln(y)=x \ln(2)

Implicitly differentiate the left side and take the derivative of the right side

\frac{y'}{y} =\ln(2)

Multiply both sides by 'y' which was defined as 2^x

y'=\ln(2)*2^x

Plug in x = 2 and x = 3 to see which slope is larger

y'=\ln(2)*2^2=4\ln(2)\\y'=\ln(2)*2^3=8\ln(2)

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castortr0y [4]

Answer:

4 x 4 is 12

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2 years ago
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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 A 5 mL volume of the suspension is withdrawn.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Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows: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Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows: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

                             =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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