Answer: 
Step-by-step explanation:
The missing figure is attached.
The volume of an oblique cylinders and the volume of a right cylinder can be found with this formula:
Where "r" is the radius and "h" is the height.
The volume of an oblique cone and the volume of a right cone can be found with this formula:
Where "r" is the radius and "h" is the height.
According to the information given in the exercise, you know that the volume of the cylinder and also the radius of the cylinder and the cone ,are the following:

Therefore, in order to find the volume of the cone, you only need to multiply the volume of the cylinder by
.
Then, you get:

Answer:
4. x= 18
Step-by-step explanation:
We solve the problem like this:
Since 3 is proportional to 6 and 9 is proportional to x, we set up a proportion.
3/9 = 6/x
We then use cross-multiplication to solve for x.
See attached image
First, we must know this: Complementary angles are two angles whose sum is equal to 90°, while supplementary angles are two angles whose sum is equal to 180°. That been said, the only statement which is true is the second statement, <span>
MNL is complementary to KNL
Reasons why others are False</span>GNJ is supplementary to JNK, not complementary
MNG is supplementary to GNJ, not complementary
LNJ (not KNJ) <span>is supplementary to MNL
</span>GNM is equal to JNK, not supplementary
We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:

The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:
![CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack](https://tex.z-dn.net/?f=CI%28%5Cmu%29%3D%5Clbrack%20x-Z_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ccdot%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%5B%5D%7Bn%7D%7D%2Cx%2BZ_%7B1-%5Cfrac%7B%5Calpha%7D%7B2%7D%7D%5Ccdot%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%5B%5D%7Bn%7D%7D%5Crbrack)
Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:
![CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack](https://tex.z-dn.net/?f=CI%28%5Cmu%29%3D%5Clbrack30.0-Z_%7B0.99%7D%5Ccdot%5Cfrac%7B2.40%7D%7B%5Csqrt%5B%5D%7B28%7D%7D%2C30.0%2BZ_%7B0.99%7D%5Ccdot%5Cfrac%7B2.40%7D%7B%5Csqrt%5B%5D%7B28%7D%7D%5Crbrack)
Where (from tables):

Finally, the interval at 98% confidence level is: