<h2>
Answer with explanation:</h2>
Let
be the average weight of chocolate chips in each bag.
As per given , we have
![H_0: \mu =439.0\\\\ H_a: \mu](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu%20%3D439.0%5C%5C%5C%5C%20H_a%3A%20%5Cmu%3C439.0)
Since
is left-tailed , so we perform a left-tailed test.
Also, the population standard deviation is known
.
So we use z-test.
Test statistic : ![z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqt{n}}}](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B%5Coverline%7Bx%7D-%5Cmu%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqt%7Bn%7D%7D%7D)
Substitute given value ,![\overline{x}=433.0\ , n=47,\ \sigma=21 \ , \mu=439](https://tex.z-dn.net/?f=%5Coverline%7Bx%7D%3D433.0%5C%20%2C%20n%3D47%2C%5C%20%5Csigma%3D21%20%5C%20%2C%20%5Cmu%3D439)
![z=\dfrac{433-439}{\dfrac{21}{\sqrt{47}}}\approx-1.96](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B433-439%7D%7B%5Cdfrac%7B21%7D%7B%5Csqrt%7B47%7D%7D%7D%5Capprox-1.96)
Significance level = 0.05
Decision rule : Reject
when p-value < 0.05.
By z-table , P- value for left-tailed test = P(z<-1.96)= 1- P(z<1.96)
[∵ P(Z<-z)=1-P(Z<z)]
= 1- 0.9750
=0.025
Since , P-value(0.025) < 0.05 , so we reject the null hypothesis.
We support the claim that the machine is underfilling the bags at 0.05 significance level .
(1800 g)/(9 g/cm^3) = 1800/9 cm^3 = 200 cm^3
<span><span>I have <span>already answered...
</span></span>It's very easy, to resolve this exercise is necessary to use
the cartesian coordinate system. However the answear is: right triangle. ;)</span>
Wouldn't you just multiply this, which is 3x9 = 27