All categories

3 years ago

Find the mean, median, mode, and the range. The picture is for number 12 to 15

Mathematics

1 answer:

Neporo4naja [7]3 years ago

5 0

Answer:

mean - 5

median - 5

mode - 6

range - 7

mean - 444

median - 453

mode - 345

range - 198

mean - 6

median - 6

mode - 8

range - 7

mean - 10

median - 9

mode - 9

range - 17

mean - 5

median - 4

mode - 3

range - 7

mean - 143

median - 148

mode - 157

range - 33

12) 9

13) 10

14) 10, 5

15)

If the price of swimming lessons went up to mean and mode would change.

mean - 9.66666667 (9.6 repeating)

median - 10

mode - 10

range - 13

You might be interested in

Plsss solve it!!!! ASAP???!!

klio [65]

Answer: 3 $\sqrt{5}$

Step-by-step explanation:

as you see c is the hypotenuse of the triangle that has two of the side 6 and 3

c^2=a^+b^2

c^2= 3^2+6^2

c^2=9+36

c^2=45

c= $\sqrt{45}$ =3 $\sqrt{5}$

3 0

3 years ago

Find the volume v of the described solid s. the base of s is an elliptical region with boundary curve 4x2 + 9y2 = 36. cross-sect

Tasya [4]

$4x^2+9y^2=36\iff\dfrac{x^2}9+\dfrac{y^2}4=1$

defines an ellipse centered at $(0,0)$ with semi-major axis length 3 and semi-minor axis length 2. The semi-major axis lies on the $x$ -axis. So if cross sections are taken perpendicular to the $x$ -axis, any such triangular section will have a base that is determined by the vertical distance between the lower and upper halves of the ellipse. That is, any cross section taken at $x=x_0$ will have a base of length

$\dfrac{x^2}9+\dfrac{y^2}4=1\implies y=\pm\dfrac23\sqrt{9-x^2}$
$\implies \text{base}=\dfrac23\sqrt{9-{x_0}^2}-\left(-\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac43\sqrt{9-{x_0}^2}$

I've attached a graphic of what a sample section would look like.

Any such isosceles triangle will have a hypotenuse that occurs in a $\sqrt2:1$ ratio with either of the remaining legs. So if the hypotenuse is $\dfrac43\sqrt{9-{x_0}^2}$ , then either leg will have length $\dfrac4{3\sqrt2}\sqrt{9-{x_0}^2}$ .

Now the legs form a similar triangle with the height of the triangle, where the legs of the larger triangle section are the hypotenuses and the height is one of the legs. This means the height of the triangular section is $\dfrac4{3(\sqrt2)^2}\sqrt{9-{x_0}^2}=\dfrac23\sqrt{9-{x_0}^2}$ .

Finally, $x_0$ can be chosen from any value in $-3\le x_0\le3$ . We're now ready to set up the integral to find the volume of the solid. The volume is the sum of the infinitely many triangular sections' areas, which are

$\dfrac12\left(\dfrac43\sqrt{9-{x_0}^2}\right)\left(\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac49(9-{x_0}^2)$

and so the volume would be

$\displaystyle\int_{x=-3}^{x=3}\frac49(9-x^2)\,\mathrm dx$
$=\left(4x-\dfrac4{27}x^3\right)\bigg|_{x=-3}^{x=3}$
$=16$

6 0

3 years ago

Can someone please help me work out the missing angles please I’m struggling

Vladimir [108]

Step-by-step explanation:

a =45°

b=45°

c=30°

d=30°

e=30°

f=30°

total of the angle

90×4=360

6 0

3 years ago

Read 2 more answers

Colin wants to set up an aquarium.. He already has a tank, but needs to purchase fish, filters, and plants. If the cost of the f

ahrayia [7]

Answer:

340.65

Step-by-step explanation: Add 84.79+44.75+18.66=148.2

148.2-$488.85= 340.65

7 0

3 years ago

Read 2 more answers

Does anyone know how to do this <br><br>csc x ( sin x + cos x) = 1 + cot x

Dima020 [189]

9514 1404 393

Explanation:

Use the identities ...

csc(x) = 1/sin(x)

cot(x) = cos(x)/sin(x)

$\csc(x)(\sin(x)+\cos(x))=1+\cot(x)\\\\\dfrac{1}{\sin(x)}(\sin(x)+\cos(x))=1+\cot(x)\\\\\dfrac{\sin(x)}{\sin(x)}+\dfrac{\cos(x)}{\sin(x)}=1+\cot(x)\\\\1+\cot(x)=1+\cot(x)$

8 0

3 years ago