I NEED HELP WITH THIS, SHOW YOUR WORK!!! BRAINLY TO WHOEVER CAN GET IT RIGHT

Dvinal [7]

Well its
900+0+0
basically
like if it was
9,234
then it would be
9000+200+30+4

7 0

3 years ago

The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(

Umnica [9.8K]

Answer:

(a) $A_1$ and $A_2$ are indeed mutually-exclusive.

(b) $\displaystyle P(A_1\; \cap \; B) = \frac{1}{20}$ , whereas $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(c) $\displaystyle P(B) = \frac{9}{100}$ .

(d) $\displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}$ , whereas $P(A_1 \; |\; B) = \displaystyle \frac{4}{9}$

Step-by-step explanation:

$P(A_1 \; \cap \; A_2) = 0$ means that it is impossible for events $A_1$ and $A_2$ to happen at the same time. Therefore, event $A_1$ and $A_2$ are mutually-exclusive.

By the definition of conditional probability:

$\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}$ .

Rearrange to obtain:

$\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot P(A_1) = 0.25 \times 0.20 = \frac{1}{20}$ .

Similarly:

$\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot P(A_2) = 0.80 \times 0.05 = \frac{1}{25}$ .

Note that:

$\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}$ .

In other words, $A_1$ and $A_2$ are collectively-exhaustive. Since $A_1$ and $A_2$ are collectively-exhaustive and mutually-exclusive at the same time:

$\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ .

By Bayes' Theorem:

$\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}$ .

Similarly:

$\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}$ .

6 0

2 years ago

A bouncy ball is dropped such that the height of its first bounce is 5.5 feet and each successive bounce is 64% of the previous

ahrayia [7]

Answer:

0.4 ft

Step-by-step explanation:

It can be convenient to let a spreadsheet compute the values for you. Here, we have written an explicit formula for the height of the n-th bounce:

h = 5.5×0.64^(n-1)

We have written the formula this way because we are given the height of the first bounce, not the starting height. Each bounce multiplies the height by a factor of 0.64. Then the 7th bounce will have a height of ...

h = 5.5×0.64^6 ≈ 0.378 ≈ 0.4 . . . . feet

6 0

3 years ago

If f(x)= x-3/x , g(x)=x+3 , and h(x)=2x+1 , what is (g*h*f)(x) ?

alexgriva [62]

Answer:

Option B is correct

Step-by-step explanation:

We need to find g(h(f(x))

We first find h(f(x))

f(x) = x-3/x

Putting value of f(x) in place of x in h(x)

h(x) = 2x+1

h(f(x)) = 2(x-3/x) + 1

= 2(x-3/x) + 1

= 2x -6 /x +1

= 2x-6 +x/x

= 3x-6/x

Now, putting value of h(f(x)) in g(x)

g(x) = x+3

g(h(f(x)))= (3x-6/x) + 3

= 3x-6/x + 3

= 3x-6+3x/x

= 6x-6/x

So, Option B is correct.

6 0

3 years ago

Solving quadratic equation<br>7r^2-14=-7

Gala2k [10]

I hope this helps you

7 0

3 years ago