Answer:
15
Step-by-step explanation:
Let us first write an equation where the unknown number is represented with variable x. <em>The sum of a fifth of x and two is five:</em>

Great! Let's solve for x by isolating the variable.

<u>Therefore, the unknown number is 15.</u>
<u />
<em>I hope this helps! Let me know if you have any questions :)</em>
The correct statement about the data collected by Ms. Pearson is that there is no association between a student's absences and the final average grades.
<h3>When do variables have a linear relationship?</h3>
The equation that represents a linear relationship is: a + bx
Where x represents the rate of increase. Thus, for linear equations, the functiion increases by a constant term.
Looking at the table, the average final grade does not increase by a constant term.
To learn more about linear functions, please check: brainly.com/question/26434260
How to find the area of a parallelogram is length times height. 12 times what gets you to 60? 5 does. 12 times 5 is 60.
HEIGHT: 5
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: