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zhannawk [14.2K]
3 years ago
14

When you are determining The probability of an event why must the probability be between 0 and 1

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

The probability of an event cannot be less than 0 because 0 means it's impossible. The probability of an event cannot be more than 1 because 1 means that it's certain that it will happen. That's why the probability must be between 0 and 1.

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Find the indicated function values for the function below<br> g(x)=9x + 8<br> g(0) =
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Answer:

g(0) = 8

Step-by-step explanation:

The problem gives us the following function:

g(x) = 9x + 8

g(0) =

This is the value of g when x = 0. So

g(x) = 9x + 8

g(0) = 9*0 + 8

g(0) = 8

So the answer to this question is:

g(0) = 8

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If 32^2c= 8^c+7, what is the value of c
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3 years ago
Use the equation a = IaIâ
german

Answer:

a) \:\:=\sqrt{14}\cdot \frac{\:\:}{\sqrt{14} }

b)\:\:=\sqrt{29} \cdot \frac{\:\:}{\sqrt{29} }

c) \:\:=7\cdot \frac{\:\:}{7}

Step-by-step explanation:

a) Let <u>a</u>=<2,1,-3>

The magnitude of <u>a</u> is |a|=\sqrt{2^2+1^2+(-3)^2}

|a|=\sqrt{4+1+9}=\sqrt{14}

The unit vector in the direction of a is

\hat{a}=\frac{\:\:}{\sqrt{14} }

Using the relation a=|a|\hat{a}, we have

\:\:=\sqrt{14}\cdot \frac{\:\:}{\sqrt{14} }

b) Let a=2i - 3j + 4k

|a|=\sqrt{2^2+(-3)^2+4^2}

|a|=\sqrt{4+9+16}=\sqrt{29}

\hat{a}=\frac{\:\:}{\sqrt{29} }

Using the relation a=|a|\hat{a}, we have

\:\:=\sqrt{29} \cdot \frac{\:\:}{\sqrt{29} }

c) Let us first find the sum of <1, 2, -3> and <2, 4, 1> to get:

<1+2, 2+4, -3+1>=<3, 6, -2>

Let a=<3, 6, -2>

The magnitude is

|a|=\sqrt{3^2+6^2+(-2)^2}

|a|=\sqrt{9+36+4}=\sqrt{49}=7

The unit vector in the direction of <u>a</u> is

\hat{a}=\frac{\:\:}{7}

Using the relation a=|a|\hat{a}, we have

\:\:=7\cdot \frac{\:\:}{7}

5 0
3 years ago
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