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Marianna [84]
2 years ago
8

The tens digit is missing from the three digit number 8_9. If the tens digit is to be randomly selected from the ten different d

igits from 0 to 9, what is the probability that the resulting three-digit number will be a multiple of 9?
Mathematics
2 answers:
Nastasia [14]2 years ago
8 0

Answer:

9 times 91 = 819

Step-by-step explanation:

Ad libitum [116K]2 years ago
8 0

Answer:

3/9

Step-by-step explanation:

So,if they tell you to find a probability between 0-9 you should write down all the numbers from 0-9 and circle the multiples of what they give you.

eg. 0-9 = 0-1-2-3-4-5-6-7-8-9

      multiples of 3 = 3,6,9

then you STOP because they said from 0-9

so your answer is going to be 3 out of (the total numbers included) 9

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Yakvenalex [24]

Answer:

Friday, Feb. 8, at 3:16 p.m.

Step-by-step explanation:Just took it :3

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Eduardwww [97]

Answer:

E (3,-1)   D (-2, -2)   G (-1, 3)    F (3, 2)

Step-by-step explanation:

90 degrees = (y, -x)

that means you take the ordered pair you want to rotate and plug the x and y into the new spots.  

for example, point E is at (1,3) so the point rotated 90 degrees is (3,-1)

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VARVARA [1.3K]

Answer:

The four small cones combined will hold more ice cream than the big one.

Step-by-step explanation:

6 0
2 years ago
Pls help I will not let me add the screenshot so I'm going to try and explain. so there is a rectangle the top measures 10m and
-BARSIC- [3]
Answer:

£23

Explanation:

6x10=60
2x1.5=3
3.14x2^2=12.566
60-3-12.566=44.433
44.43/10=4.43
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3 0
2 years ago
Solve the given initial-value problem. x^2y'' + xy' + y = 0, y(1) = 1, y'(1) = 8
Kitty [74]
Substitute z=\ln x, so that

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dz}\cdot\dfrac{\mathrm dz}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)

Then the ODE becomes


x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}+x\dfrac{\mathrm dy}{\mathrm dx}+y=0\implies\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)+\dfrac{\mathrm dy}{\mathrm dz}+y=0
\implies\dfrac{\mathrm d^2y}{\mathrm dz^2}+y=0

which has the characteristic equation r^2+1=0 with roots at r=\pm i. This means the characteristic solution for y(z) is

y_C(z)=C_1\cos z+C_2\sin z

and in terms of y(x), this is

y_C(x)=C_1\cos(\ln x)+C_2\sin(\ln x)

From the given initial conditions, we find

y(1)=1\implies 1=C_1\cos0+C_2\sin0\implies C_1=1
y'(1)=8\implies 8=-C_1\dfrac{\sin0}1+C_2\dfrac{\cos0}1\implies C_2=8

so the particular solution to the IVP is

y(x)=\cos(\ln x)+8\sin(\ln x)
4 0
3 years ago
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