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Arte-miy333 [17]
3 years ago
12

A spinner with 10 equally sized slices has 10 yellow slices. The dial is spun and stops on a slice at random. What is the probab

ility that the dial stops on a yellow slice?
Write your answer as a fraction in simplest form.

Mathematics
1 answer:
djverab [1.8K]3 years ago
5 0

Answer:

1/10

Step-by-step explanation:

there are 10 slices, so if the probability of it landing on one of them is 1/10

You might be interested in
Given the function f(x) = x2 and k = -3, which of the following represents a vertical shift?
MAVERICK [17]
I think is 4 the correct answer
5 0
2 years ago
Write (-3, -5) (-1, -19) in point slope form
stiv31 [10]

Answer:

(y + 5) = -7(x + 3)

Step-by-step explanation:

Point Slope Form:  (y - y1) = m(x - x1)

<u>Step 1:  Find Slope</u>

m = \frac{y2 - y1}{x2 - x1}

m = \frac{-19 - (-5)}{-1 - (-3)}

m = \frac{-19 + 5}{-1 + 3}

m = \frac{-14}{2}

m = -7

<u>Step 2:  Put into point slope form</u>

Point Slope Form:  (y - y1) = m(x - x1)

(y - (-5)) = -7(x - (-3))

(y + 5) = -7(x + 3)

8 0
3 years ago
Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.) 5, 1,
Dahasolnce [82]

Answer:

The direction cosines are:

\frac{5}{\sqrt{42} }, \frac{1}{\sqrt{42} }  and  \frac{4}{\sqrt{42} }  with respect to the x, y and z axes respectively.

The direction angles are:

40°,  81° and  52° with respect to the x, y and z axes respectively.

Step-by-step explanation:

For a given vector a = ai + aj + ak, its direction cosines are the cosines of the angles which it makes with the x, y and z axes.

If a makes angles α, β, and γ (which are the direction angles) with the x, y and z axes respectively, then its direction cosines are: cos α, cos β and cos γ in the x, y and z axes respectively.

Where;

cos α = \frac{a . i}{|a| . |i|}               ---------------------(i)

cos β = \frac{a.j}{|a||j|}               ---------------------(ii)

cos γ = \frac{a.k}{|a|.|k|}             ----------------------(iii)

<em>And from these we can get the direction angles as follows;</em>

α =  cos⁻¹ ( \frac{a . i}{|a| . |i|} )

β = cos⁻¹ ( \frac{a.j}{|a||j|} )

γ = cos⁻¹ ( \frac{a.k}{|a|.|k|} )

Now to the question:

Let the given vector be

a = 5i + j + 4k

a . i =  (5i + j + 4k) . (i)

a . i = 5         [a.i <em>is just the x component of the vector</em>]

a . j = 1            [<em>the y component of the vector</em>]

a . k = 4          [<em>the z component of the vector</em>]

<em>Also</em>

|a|. |i| = |a|. |j| = |a|. |k| = |a|           [since |i| = |j| = |k| = 1]

|a| = \sqrt{5^2 + 1^2 + 4^2}

|a| = \sqrt{25 + 1 + 16}

|a| = \sqrt{42}

Now substitute these values into equations (i) - (iii) to get the direction cosines. i.e

cos α = \frac{5}{\sqrt{42} }

cos β =  \frac{1}{\sqrt{42} }              

cos γ =  \frac{4}{\sqrt{42} }

From the value, now find the direction angles as follows;

α =  cos⁻¹ ( \frac{a . i}{|a| . |i|} )

α =  cos⁻¹ ( \frac{5}{\sqrt{42} } )

α =  cos⁻¹ (\frac{5}{6.481} )

α =  cos⁻¹ (0.7715)

α = 39.51

α = 40°

β = cos⁻¹ ( \frac{a.j}{|a||j|} )

β = cos⁻¹ ( \frac{1}{\sqrt{42} } )

β = cos⁻¹ ( \frac{1}{6.481 } )

β = cos⁻¹ ( 0.1543 )

β = 81.12

β = 81°

γ = cos⁻¹ ( \frac{a.k}{|a|.|k|} )

γ = cos⁻¹ (\frac{4}{\sqrt{42} })

γ = cos⁻¹ (\frac{4}{6.481})

γ = cos⁻¹ (0.6172)

γ = 51.89

γ = 52°

<u>Conclusion:</u>

The direction cosines are:

\frac{5}{\sqrt{42} }, \frac{1}{\sqrt{42} }  and  \frac{4}{\sqrt{42} }  with respect to the x, y and z axes respectively.

The direction angles are:

40°,  81° and  52° with respect to the x, y and z axes respectively.

3 0
3 years ago
2.the solution set of x +y =5 and x-y =3
Nataliya [291]

Answer:

Check the solution below

Step-by-step explanation:

2) Given the equation

x +y =5... 1 and

x-y =3 ... 2

Add both equations

x+x = 5+3

2x = 8

x = 8/2

x = 4

Substitute x = 4 into 1:

From 1: x+y = 5

4+y= 5

y = 5-4

y = 1

3) Given

x+3y =15 ... 1

2x+7y=19 .... 2

From 2: x = 15-3y

Substitute into 2

2(15-3y)+7y = 19

30-6y+7y = 19

30+y = 19

y = 19-30

y = -11

Substitute y=-11 into x = 15-3y

x =15-3(-11)

x = 15+33

x = 48

The solution set is (48, -11)

4) given

x/2 +y/3 =0 and x+2y=1

From 1

(3x+2y)/6 = 0

3x+2y = 0.. 3

x+2y= 1... 4

From 4: x = 1-2y

Substutute

3(1-2y) +2y = 0

3-6y+2y = 0

3 -4y = 0

4y = 3

y = 3/4

Since x = 1-2y

x = 1-2(3/4)

x = 1-3/2

x= -1/2

The solution set is (-1/2, 3/4)

5) Given

5.x=1/2 and y =x +1 then solution is

We already know the vkue of x

Get y

y= x+1

y = 1/2 + 1

y = 3/2

Hence the solution set is (1/2, 3/2)

6) Given

3x +y =5 and x -3y =5

From 3; x = 5+3y

Substitute into 1;

3(5+3y)+y = 5

15+9y+y = 5

10y = 5-15

10y =-10

y = -1

Get x;

x = 5+3y

x = 5+3(-1)

x = 5-3

x = 2

Hence two solution set is (2,-1)

6 0
3 years ago
Solve for f. show your work! 12=f-13-2
tankabanditka [31]
12=f-13-2
12=f-15
f=27
3 0
3 years ago
Read 2 more answers
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