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luda_lava [24]
3 years ago
10

A 37 bag sample had a mean of 421 grams. Assume the population standard deviation is known to be 29. A level of significance of

0.05 will be used. State the null and alternative hypothesis.
Mathematics
1 answer:
Ierofanga [76]3 years ago
5 0

Answer:  H_0:\mu=421

H_a : \mu\neq421

Step-by-step explanation:

  • A null hypothesis is a type of hypothesis that is used in statistics that assumes there is no difference between particular characteristics of a population wheres the alternative hypothesis shows that there is a difference.

Given: A 37 bag sample had a mean of 421 grams.

Let \mu be the population mean.

Then, the null hypothesis would be:

H_0:\mu=421

whereas the alternative hypothesis would be:

H_a : \mu\neq421

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sleet_krkn [62]

Answer:

-6re−r [sin(6θ) - cos(6θ)]

Step-by-step explanation:

the Jacobian is ∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ

x = e−r sin(6θ), y = er cos(6θ)

δx/δθ = -6rcos(6θ)e−r sin(6θ), δx/δr = -sin(6θ)e−r sin(6θ)

δy/δθ = -6rsin(6θ)er cos(6θ), δy/δr = cos(6θ)er cos(6θ)

∂(x, y) /∂(r, θ) =  δx/δθ × δy/δr - δx/δr × δy/δθ

= -6rcos(6θ)e−r sin(6θ) × cos(6θ)er cos(6θ) - [-sin(6θ)e−r sin(6θ) × -6rsin(6θ)er cos(6θ)]

= -6rcos²(6θ)e−r (sin(6θ) - cos(6θ)) - 6rsin²(6θ)e−r (sin(6θ) - cos(6θ))

= -6re−r (sin(6θ) - cos(6θ)) [cos²(6θ) + sin²(6θ)]

= -6re−r [sin(6θ) - cos(6θ)]     since  [cos²(6θ) + sin²(6θ)] = 1

6 0
4 years ago
How do u find the percentage of a two way table
CaHeK987 [17]
By cuting it in time
8 0
3 years ago
Cesars net worth is 8,250.63 his liabilities are 3,222.35 what is cesars assets
SashulF [63]

Given parameters:

Net worth of Cesar = 8250.63

Liabilities = 3222.35

Unknown:

Cesar's asset = ?

Solution:

A person's net worth is their Assets minus liability;

             Net worth  = Assets - Liabilities

  Assets  = Net worth + Liabilities

              For Cesar;

 Assets  = 8250.63 + 3222.35 = 11472.98

Cesar's assets is  11472.98

8 0
3 years ago
Question 3 of 25
lys-0071 [83]
An ordered pair of the inverse of f(x)D, (1.2)
3 0
3 years ago
Construct a 99% confidthence interval for the population mean .Assume the population has a normal distribution. A group of 19 ra
Kobotan [32]
<h2>Answer with explanation:</h2>

Confidence interval for mean, when population standard deviation is unknown:

\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}

, where \overline{x} = sample mean

n= sample size

s= sample standard deviation

t_{\alpha/2} = Critical t-value for n-1 degrees of freedom

We assume the population has a normal distribution.

Given, n= 19 , s= 3.8 , \overline{x}=22.4

\alpha=1-0.99=0.01

A) Critical t value for \alpha/2=0.005 and degree of 18 freedom

t_{\alpha/2} = 2.8784

B) Required confidence interval:

22.4\pm ( 2.8744)\dfrac{3.8}{\sqrt{19}}\\\\=22.4\pm2.5058\\\\=(22.4-2.5058,\ 22.4+2.5058)=(19.8942,\ 24.9058)\approx(19.9,\ 24.9)

Lower bound = 19.9 years

Uppen bound = 24.9 years

C) Interpretation: We are 99% confident that the true population mean of lies in (19.9, 24.9) .

5 0
3 years ago
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