What is the interquartile range of the sequence 5,5,8,8,13,14,16,16,19,22,23,27,31 ?
Romashka-Z-Leto [24]
Answer:
The Interquartile range is 10.
Step-by-step explanation:
First, we will need to find the mean, the mean of this sequence is 16, you will now need to find quartile 1 and quartile 3. Quartile 1 is 13, and quartile 3 is 23. Lastly, subtract Quartile 3 and Quartile 1 will be the answer.
So, 23-13=10
The Answer will be 10, the interquartile range is 10.
Hope this helps!
Answer:
- |t -(-5)| = 1.5
- [-6.5, -3.5] = [minimum, maximum]
Step-by-step explanation:
The magnitude of the difference between the temperature (t) and -5 will be 1.5 at the limits:
|t -(-5)| = 1.5
This is equivalent to two equations:
In each case, the equation is solved by subtracting 5 from both sides:
- t = -6.5 . . . . minimum allowed temperature
- t = -3.5 . . . . maximum allowed temperature
I graphed it out and it seems that both A and C are correct. I would check them to make sure.
Answer:
A.
Step-by-step explanation:
<span>4x+2y=-6
2y = -4x - 6
y = - 2x - 3
slope = -2
</span><span>y-intercept when x = 0
so y = -3
answer
slope = -2
</span><span>y-intercept (0,-3)</span>