)Information about the age of participants in a dance competition is below:

Mathematics

2 answers:

abruzzese [7]3 years ago

7 0

The middle 50% of the data is between 17 and 27. I have attached a picture of it.

Each section is representing 25% of the data.

The distance between the 17 and the 27 is called the interquartile range.

Inga [223]3 years ago

7 0

It would be in between 17 and 27 because if you look at the line it will be on top of 21 and its in between them because if you look on the outer lines of the box those lines point on 17 and 27 and they are 8 pieces in between so between 17 and 27 it would be 8 and 10 so the answer would be c 17 and 27. I hope I could help!

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Answer:

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A trough of water is 20 meters in length and its ends are in the shape of an isosceles triangle whose width is 7 meters and heig

Vaselesa [24]

Answer:

a) Depth changing rate of change is $0.24m/min$ , When the water is 6 meters deep

b) The width of the top of the water is changing at a rate of $0.17m/min$ , When the water is 6 meters deep

Step-by-step explanation:

As we can see in the attachment part II, there are similar triangles, so we have the following relation between them $\frac{3.5}{10} =\frac{a}{h}$ , then $a=0.35h$ .

a) As we have that volume is $V=\frac{1}{2} 2ahL=ahL$ , then $V=(0.35h^{2})L$ , so we can derivate it $\frac{dV}{dt}=2(0.35h)L\frac{dh}{dt}$ due to the chain rule, then we clean this expression for $\frac{dh}{dt}=\frac{1}{0.7hL}\frac{dV}{dt}$ and compute with the knowns $\frac{dh}{dt}=\frac{1}{0.7(6m)(20m)}2m^{3}/min=0.24m/min$ , is the depth changing rate of change when the water is 6 meters deep.

b) As the width of the top is $2a=0.7h$ , we can derivate it and obtain $\frac{da}{dt}=0.7\frac{dh}{dt} =0.7*0.24m/min=0.17m/min$ The width of the top of the water is changing, When the water is 6 meters deep at this rate