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iris [78.8K]
3 years ago
6

Experience raising New Jersey Red chickens revealed the mean weight of the chickens at

Mathematics
1 answer:
Nikolay [14]3 years ago
6 0

Answer:

p-value = 0.1277

Step-by-step explanation:

p-value is the probability value tell us how likely it is to get a result like this if the Null Hypothesis is true.

Firstly we find the mean and standard deviation of the given data set.

⇒ Mean = \frac{4.41 +4.37+ 4.33+ 4.35 +4.30 +4.39 +4.36+ 4.38+ 4.40+ 4.39}{10}

⇒ Mean = 4.368

Standard deviation(\sigma) = \sqrt{\frac{1}{n}\sum_{i=1}^{n}{(x_{i}-\bar{x})^{2}} }

where, \bar{x} is mean of the distribution.

⇒ Standard Deviation = 0.034

Applying t- test:

Let out hypothesis is:

H₀: μ = 4.35

H₁: μ ≠ 4.35

Now,

Here, μ = Population Mean = 4.35

\bar{x}= Sample Mean = 4.368

σ = Standard Deviation = 0.034

n = 10

t=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}} }

Putting all values we get, t = 1.6777 with (10 -1) = 9 degree of freedom.

Then the p-value at 99% level of significance.

⇒ p-value = 0.1277

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The question is worded a bit strangely (in my opinion anyway), but I think your teacher wants you to describe how exponents work.

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3 years ago
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Answer:

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Step-by-step explanation:

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3 years ago
Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area: where V = volume (mm3
Alex

Answer:

V = 20.2969 mm^3 @ t = 10

r = 1.692 mm @ t = 10

Step-by-step explanation:

The solution to the first order ordinary differential equation:

\frac{dV}{dt} = -kA

Using Euler's method

\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i}    \\

Where initial droplet volume is:

V(0) = \frac{4pi}{3} * r(0)^3 =  \frac{4pi}{3} * 2.5^3 = 65.45 mm^3

Hence, the iterative solution will be as next:

  • i = 1, ti = 0, Vi = 65.45

V'_{i}  = -k *4pi*(\frac{3*65.45}{4pi})^(2/3)  = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88

  • i = 2, ti = 0.5, Vi = 63.88

V'_{i}  = -k *4pi*(\frac{3*63.88}{4pi})^(2/3)  = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33

  • i = 3, ti = 1, Vi = 62.33

V'_{i}  = -k *4pi*(\frac{3*62.33}{4pi})^(2/3)  = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813

We compute the next iterations in MATLAB (see attachment)

Volume @ t = 10 is = 20.2969

The droplet radius at t=10 mins

r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\

The average change of droplet radius with time is:

Δr/Δt = \frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min

The value of the evaporation rate is close the value of k = 0.08 mm/min

Hence, the results are accurate and consistent!

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Answer:

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Step-by-step explanation:

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