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shusha [124]
3 years ago
13

Your lunch account has $22 and you spend $2.20 on it each day write an equation for the line

Mathematics
1 answer:
Nikitich [7]3 years ago
8 0

Answer:

you would write 2.20x = 22

Step-by-step explanation:


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Suppose that E and F are two events and that P(E)=0.3 and P(F|E)=0.5. What is P(E and F)?
Sergio [31]

E and F are two events and that P(E)=0.3 and P(F|E)=0.5. Thus, P(E and F)=0.15


Bayes' theorem is transforming preceding probabilities into succeeding probabilities. It is based on the principle of conditional probability. Conditional probability is the possibility that an event will occur because it is dependent on another event.

P(F|E)=P(E and F)÷P(E)

It is given that P(E)=0.3,P(F|E)=0.5

Using Bayes' formula,

P(F|E)=P(E and F)÷P(E)

Rearranging the formula,

⇒P(E and F)=P(F|E)×P(E)

Substituting the given values in the formula, we get

⇒P(E and F)=0.5×0.3

⇒P(E and F)=0.15

∴The correct answer is 0.15.

If, E and F are two events and that P(E)=0.3 and P(F|E)=0.5. Thus, P(E and F)=0.15.

Learn more about Bayes' theorem on

brainly.com/question/17010130

#SPJ1

4 0
2 years ago
To go to dance class at 6:45p.m. bus,Kelly needs 35 minutes to walk home from a friend's house,30 minutes for dinner,and 20 minu
GREYUIT [131]
20+35+30= 1 hr&25 min so she should leave at 5:20
5 0
3 years ago
Mila Hawkins purchased a bookshelf for $210.00.
Alenkinab [10]

Answer:

$12.60

Step-by-step explanation:

6% of 210 is .06 x 210 which equals 12.60

6 0
2 years ago
Sharon travels a total of 60 mi to school each week. Aarushi travels 75% of the miles Sharon travels to
user100 [1]

..................'.60

6 0
2 years ago
Find an equation for the plane that is tangent to the surface z equals ln (x plus y )at the point Upper P (1 comma 0 comma 0 ).
alexira [117]

Let f(x,y,z)=z-\ln(x+y). The gradient of f at the point (1, 0, 0) is the normal vector to the surface, which is also orthogonal to the tangent plane at this point.

So the tangent plane has equation

\nabla f(1,0,0)\cdot(x-1,y,z)=0

Compute the gradient:

\nabla f(x,y,z)=\left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},\dfrac{\partial f}{\partial z}\right)=\left(-\dfrac1{x+y},-\dfrac1{x+y},1\right)

Evaluate the gradient at the given point:

\nabla f(1,0,0)=(-1,-1,1)

Then the equation of the tangent plane is

(-1,-1,1)\cdot(x-1,y,z)=0\implies-(x-1)-y+z=0\implies\boxed{z=x+y-1}

7 0
3 years ago
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