What is the period for tangent functions? (Answer in radians)

Find the midpoint of (-3,-5), (4,5) *<br> (3,2)<br> O (-6,4)<br> (0.5,0)<br> O (-1,-3)

Sergeu [11.5K]

Answer:

(0.5, 0) will be the midpoint by applying formula which is

add both "x" quadrant and divide by it by two and then add both "you" quadrant and then divide it also by 2 you will get answer

3 0

3 years ago

How do I find the equation of the axis of symmetry and the coordinates of the vertex?

Firlakuza [10]

<span>For a quadratic function in standard form, y=ax2+bx+c , the axis of symmetry is a vertical line x=−b2a . Example 1: Find the axis of symmetry of the parabola shown. The x -coordinate of the vertex is the equation of the axis of symmetry of the parabola.</span>

5 0

3 years ago

Put the numbers in order from least to greatest.

loris [4]

Answer:

B. -.9,-4, 1, .2, .8

Hope this helps!:)

7 0

3 years ago

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A helicopter takes off from the ground 800 feet from the base of a building. It flies in a straight line directly to the top of

Xelga [282]

The helicopter took off from the angle of 11°.

7 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