Distance of the person from the building (x)
tan 24 = 20/x => x = 20/tan 24 = 44.92 ft
The height of the building above the height of the person and the distance of the person from the building forms two legs of right angled triangle.
Therefore, if h is the height of the building, then;
tan 31 = (h-20)/44.92
h-20 = 44.92 tan 31
h = (44.92 tan 31) + 20 = 26.99 + 20 = 46.99 ft
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:
$135
Step-by-step explanation:
id k how to explain but since he saves 5 dollars he basically has 15 dollars in his bank account u just need to multiply 15 with however many weeks he saves up for
I don’t think you can lol
V=(1/3)(area of the base)(<span>altitude)
V=</span><span>(1/3)(</span>6²)(6)= 72 m³
