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AURORKA [14]
3 years ago
10

Almeida collects data on the ages of the students in her aerobics class and creates a dot plot to represent the ages.

Mathematics
2 answers:
vazorg [7]3 years ago
6 0
I believe your question is incomplete.
ella [17]3 years ago
3 0

Answer:

21,21,22,23,25,25,25,27

Step-by-step explanation:

2nd option in the drop-down box

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1 4/11 divided by 1/11
s344n2d4d5 [400]

Answer:

15

Step-by-step explanation:

hope this helps

4 0
3 years ago
Cos ( α ) = √ 6/ 6 and sin ( β ) = √ 2/4 . Find tan ( α − β )
Zina [86]

Answer:

\purple{ \bold{ \tan( \alpha  -  \beta ) = 1.00701798}}

Step-by-step explanation:

\cos( \alpha ) =  \frac{ \sqrt{6} }{6}  =  \frac{1}{ \sqrt{6} }  \\  \\  \therefore \:  \sin( \alpha )  =  \sqrt{1 -  { \cos}^{2} ( \alpha ) }  \\  \\  =  \sqrt{1 -  \bigg( {\frac{1}{ \sqrt{6} } \bigg )}^{2} }  \\  \\ =  \sqrt{1 -  {\frac{1}{ {6} }}}  \\  \\ =  \sqrt{ {\frac{6 - 1}{ {6} }}}   \\  \\  \red{\sin( \alpha ) =  \sqrt{ { \frac{5}{ {6} }}} } \\  \\  \tan( \alpha ) =  \frac{\sin( \alpha ) }{\cos( \alpha ) }  =  \sqrt{5}  \\  \\ \sin( \beta )  =  \frac{ \sqrt{2} }{4}  \\  \\  \implies \: \cos( \beta )  =   \sqrt{ \frac{7}{8} }  \\  \\ \tan( \beta )  =  \frac{\sin( \beta ) }{\cos( \beta ) } =  \frac{1}{ \sqrt{7} }   \\  \\  \tan( \alpha  -  \beta ) =  \frac{ \tan \alpha  -  \tan \beta }{1 +  \tan \alpha .  \tan \beta}  \\  \\  =  \frac{ \sqrt{5} -  \frac{1}{ \sqrt{7} }  }{1 +  \sqrt{5} . \frac{1}{ \sqrt{7} } }  \\  \\  =  \frac{ \sqrt{35} - 1 }{ \sqrt{7}  +  \sqrt{5} }  \\  \\  \purple{ \bold{ \tan( \alpha  -  \beta ) = 1.00701798}}

8 0
3 years ago
What is Jackie's monthly salary if her annual gross pay is $65,400
frez [133]

Answer:

The answer is $5450 per month

8 0
3 years ago
Find equations of the tangent plane and the normal line to the given surface at the specified point. x + y + z = 8exyz, (0, 0, 8
Dima020 [189]

Let f(x,y,z)=x+y+z-8e^{xyz}. The tangent plane to the surface at (0, 0, 8) is

\nabla f(0,0,8)\cdot(x,y,z-8)=0

The gradient is

\nabla f(x,y,z)=\left(1-8yze^{xyz},1-8xze^{xyz},1-8xye^{xyz}\right)

so the tangent plane's equation is

(1,1,1)\cdot(x,y,z-8)=0\implies x+y+(z-8)=0\implies x+y+z=8

The normal vector to the plane at (0, 0, 8) is the same as the gradient of the surface at this point, (1, 1, 1). We can get all points along the line containing this vector by scaling the vector by t, then ensure it passes through (0, 0, 8) by translating the line so that it does. Then the line has parametric equation

(1,1,1)t+(0,0,8)=(t,t,t+8)

or x(t)=t, y(t)=t, and z(t)=t+8.

(See the attached plot; the given surface is orange, (0, 0, 8) is the black point, the tangent plane is blue, and the red line is the normal at this point)

4 0
3 years ago
Barry buys a package of pasta for $2.39 and a jar of tomato sauce for $3.09. He uses a $0.75 coupon and a $0.50 coupon.
seraphim [82]

Answer:$4.23

Step-by-step explanation:

6 0
3 years ago
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