Sajia can sell 21 books she can sell if she comes back on sunday and the required inequality is 
<em><u>Solution:</u></em>
Given that Sajia has 30 Books in her library
She sold 9 books at the thrift store on Saturday
To find: We have to write and solve an inequality to determine number of more books she can sell if she comes back on sunday
Let "x" be the number of more books she can sell if she comes back on sunday
<em><u>We can write a inequality as:</u></em>


Now moving 9 from L.H.S to R.H.S we get,

On solving 30 - 9 = 21,

So Sajia can sell 21 books she can sell if she comes back on sunday
Answer:
only 1 and 2
(mean and median)
the mean absolute deviation is the measure of variability
Step-by-step explanation:
Answer:
108 degrees
Step-by-step explanation:
Since angles A and B are supplementary, they both add up to 180. Angle A = 7x+10, and angle B = 7x-26. You can set up the equation A + B = 180, and then plug in angles A and B, which gives you 7x+10 + 7x-26 = 180. Then you just need to isolate x:
Combine like terms: 14x - 16 = 180
Add 16 to both sides: 14x = 196
Divide both sides by 14: x = 14
Now that you know x, you can plug it into the expression for angle A (7x+10):
A = 7(14) + 10 = 98+10 = 108
Answer:
a) 
b) 
c) 
With a frequency of 4
d) 
<u>e)</u>
And we can find the limits without any outliers using two deviations from the mean and we got:

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case
Step-by-step explanation:
We have the following data set given:
49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000
Part a
The mean can be calculated with this formula:

Replacing we got:

Part b
Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

Part c
The mode is the most repeated value in the sample and for this case is:

With a frequency of 4
Part d
The midrange for this case is defined as:

Part e
For this case we can calculate the deviation given by:

And replacing we got:

And we can find the limits without any outliers using two deviations from the mean and we got:

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case