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antiseptic1488 [7]
3 years ago
13

.What is the probability of two 6 sided diced rolled, adding up to lucky 13

Mathematics
1 answer:
BARSIC [14]3 years ago
5 0

Answer:

0

Step-by-step explanation:

Assuming the six sided dice are numbered 1 through six

The max each could roll is 6

6+6 = 12

They cannot reach a sum of 13

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Let h(x)=<img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%2B3%7D" id="TexFormula1" title="\sqrt{x+3}" alt="\sqrt{x+3}" align="absmi
PolarNik [594]
So if we have h(x) if multiplication isolate k-3andx-6 2x-3 = 6+7=13 value=62
7 0
3 years ago
Qs. If three masons construct a wall in.
nydimaria [60]

Answer:

2.5 days

Step-by-step explanation:

5 0
3 years ago
The client drank 6 ounces of orange juice and 4 ounces of coffee for breakfast, 8 ounces of tea and 6 ounces of ginger ale for l
algol13

Answer:

946.242 milliliters

Step-by-step explanation:

The client drank 6 ounces of orange juice and 4 ounces of coffee for breakfast. We need to convert ounces to milliliters

1 ounce = 29.574 milliliters

6 ounces of orange juice

= 6 × 29.574 = 177.444 milliliters

4 ounces of coffee = 4 × 29.574 = 118.296 milliliters

For lunch, he consumed 8 ounces of tea and 6 ounces of ginger ale.

8 ounces of tea = 8 × 29.574 =236.529 milliliters

6 ounces of ginger ale = 6 × 29.574 =177.444 milliliters

4 ounces of water at 10:00 a.m. and at 2:00 p.m. this means 8 ounces of water.

8 ounces of water = 8 × 29.574 =236.529 milliliters

Total consumption by the client =

177.444 + 118.296 + 236.529 + 236.529= 946.242 milliliters

5 0
3 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
A circle has a diameter with endpoints of (-3, 8) and (7, 4). What is the center of the circle?
Alexeev081 [22]
The center of the circle is expressed as (h,k) in the standard form  of equation. In this case, the diameter endpoints are <span>(-3, 8) and (7, 4). Hence the circle center has coordinates which are the midpoints of the two points. 

x = (-3 + 7) /2  = 2
y = (8 + 4) /2 = 6 

hence the center of the circle is (2,6)</span>
5 0
3 years ago
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