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2 years ago

If the maximum of a data set is 50 and the minimum of the set is 10, what is the range?

Mathematics

1 answer:

Nonamiya [84]2 years ago

5 0

Answer:

explanation:

The range is the difference between the smallest and highest numbers in a list or set. To find the range, first put all the numbers in order. Then subtract the lowest number from the highest. The answer gives you the range of the list.

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Can you please help

Dmitry [639]

Answer is in the file below

tinyurl.com/wpazsebu

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2 years ago

He nine ring wraiths want to fly from barad-dur to rivendell. rivendell is directly north of barad-dur. the dark tower reports t

kiruha [24]

This problem is better understood with a given figure. Assuming that the flight is in a perfect northwest direction such that the angle is 45°, therefore I believe I have the correct figure to simulate the situation (see attached).

Now we are asked to find for the value of the hypotenuse (flight speed) given the angle and the side opposite to the angle. In this case, we use the sin function:

sin θ = opposite side / hypotenuse

sin 45 = 68 miles per hr / flight

flight = 68 miles per hr / sin 45

<span>flight = 96.17 miles per hr</span>

6 0

3 years ago

Helppppppppppppppppppppp!

In-s [12.5K]

Answer:

1:3

Step-by-step explanation:

Since your scaling down you divide the smaller number by the larger number

7/21=1/3

=1:3

Hope this helps :)

5 0

2 years ago

General solutions of sin(x-90)+cos(x+270)=-1<br> {both 90 and 270 are in degrees}

mixer [17]

Answer:

$\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

Step-by-step explanation:

Given:

$\sin (x-90^{\circ})+\cos(x+270^{\circ})=-1$

First, note that

$\sin (x-90^{\circ})=-\cos x\\ \\\cos(x+270^{\circ})=\sin x$

So, the equation is

$-\cos x+\sin x= -1$

Multiply this equation by $\frac{\sqrt{2}}{2}:$

$-\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x= -\dfrac{\sqrt{2}}{2}\\ \\\dfrac{\sqrt{2}}{2}\cos x-\dfrac{\sqrt{2}}{2}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos (x+45^{\circ})=\dfrac{\sqrt{2}}{2}$

The general solution is

$x+45^{\circ}=\pm \arccos \left(\dfrac{\sqrt{2}}{2}\right)+2\pi k,\ \ k\in Z\\ \\x+\dfrac{\pi }{4}=\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{4}\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

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3 years ago

Two more than five times a number is less than or equal to 47

Sliva [168]

5x + 2 < 47

Side note: put the line under the

7 0

2 years ago