Answer:
the answer to your question is yes, but the answer to #21 is ;
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Step-by-step explanation:
Answer:
ur answer
hope u will finish it fast
Answer:
The interquartile range is 5.
Step-by-step explanation:
Ah, a throwback to interquartile range... let me help :)
4,5,6,8,9,10,11,12
First, you need to know how to use the IQR. The interquartile range is basically known as the process of subtracting the upper quartile and the lower quartile of a set of data. The lower quartile should be written as Q1, and the upper quartile would be labeled as Q3. This would make the midpoint (median) data set Q2, and the highest possible point would be labeled Q4. Next, you have to always understand what you are looking at. For example, let's split the set 5,6,7,8,9,10,11,12 into groups. 5 and 6 would be Q1, 7 and 8 would be Q2, 9 and 10 would be Q3, and last but not least, 11 and 12 would be labeled as Q4. Now take Q1 and subtract it from Q3 and that is how you get your IQR.
Subtract 32 to both sides to the equation becomes -5x^2 + 7x + 9 = 0.
To solve this equation, we can use the quadratic formula. The quadratic formula solves equations of the form ax^2 + bx + c = 0
x = [ -b ± √(b^2 - 4ac) ] / (2a)
x = [ -7 ± √(7^2 - 4(-5)(9)) ] / ( 2(-5) )
x = [ -7 ± √(49 - (-180) ) ] / ( -10 )
x = [ -7 ± √(229) ] / ( -10)
x = [ -7 ± sqrt(229) ] / ( -10 )
x = 7/10 ± -sqrt(229)/10
The answers are 7/10 + sqrt(229)/10 and 7/10 - sqrt(229)/10.
Answer:
No
Step-by-step explanation:
