A point has the coordinates (m, 0) and m ≠ 0.

108 rounded to the nearest tenth

Artemon [7]

Answer:

110

Step-by-step explanation:

8 0

3 years ago

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A circle has radius 50 cm. Which of these is closest to its area?

juin [17]

Answer:

7,854cm^2

Step-by-step explanation:

π50²=2500π

2500π=7853.98

6 0

2 years ago

The histogram shows the number of minutes people were required to wait when they went to register for school. How many users are

UkoKoshka [18]

Answer:

4

Step-by-step explanation:

60 minutes = 1 hour → 120 minutes = 2 hours. The purple bar represents at least 121 minutes, which is greater than 2 hours.

7 0

3 years ago

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The system of equations below represents p, the number of photo files, and v, the number of video files Helena uploaded to an on

ss7ja [257]

P + v = 650....p = 650 - v
4p + 9v = 2925

4(650 - v) + 9v = 2925
2600 - 4v + 9v = 2925
-4v + 9v = 2925 - 2600
5v = 325
v = 325/5
v = 65.....video files

p + v = 650
p + 65 = 650
p = 650 - 65
p = 585....photo files

answer : Helena uploaded 9 times as many photo files as video files
65 * 9 = 585

5 0

3 years ago

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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:

$P(A_1 \; \cap \; A_2) = 0$ 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