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amm1812
2 years ago
8

Pls help I really need it ASAP

Mathematics
2 answers:
gtnhenbr [62]2 years ago
8 0

Answer:

0, -1 for part A -1,-9 for part B

Step-by-step explanation:

gogolik [260]2 years ago
3 0
I would help but i’m not rlly a written response type of guy. hang in the tight the kid. someone will help soon
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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:

<h3>(a)</h3>

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.

<h3>(b)</h3>

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}.

<h3>(c)</h3>

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}.

<h3>(d)</h3>

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
3 years ago
A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y = 5 − x 2 . What are the dimensions
kherson [118]

Answer:

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola

y=5−x^2. What are the dimensions of such a rectangle with the greatest possible area?

Width =

Height =

Width =√10 and Height = \frac{10}{4}

Step-by-step explanation:

Let the coordinates of the vertices of the rectangle which lie on the given parabola y = 5 - x² ........ (1)

are (h,k) and (-h,k).

Hence, the area of the rectangle will be (h + h) × k

Therefore, A = h²k ..... (2).

Now, from equation (1) we can write k = 5 - h² ....... (3)

So, from equation (2), we can write

A =h^{2} [5-h^{2} ]=5h^{2} -h^{4}

For, A to be greatest ,

\frac{dA}{dh} =0 = 10h-4h^{3}

⇒ h[10-4h^{2} ]=0

⇒ h^{2} =\frac{10}{4} {Since, h≠ 0}

⇒ h = ±\frac{\sqrt{10} }{2}

Therefore, from equation (3), k = 5 - h²

⇒ k=5-\frac{10}{4} =\frac{10}{4}

Hence,

Width = 2h =√10 and

Height = k =\frac{10}{4}.

8 0
3 years ago
5.8 times 10 and show work (optional)
Step2247 [10]

Answer:

58

Step-by-step explanation:

A easy way to get this is to remove the decimal in 5.8 which gives you ----> 58

NOTE: This works for any problem multiplying 10 x a decimal.

7 0
3 years ago
A regular pentagon has a perimeter of 15x+35.
belka [17]

Answer:

-3.5 or -7/2

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
9. An Integer is 10 times the difference of a rational number and
amm1812

Answer:

yes

Step-by-step explanation:

The answer is yes because I need points to ask a question

5 0
3 years ago
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