Simply the expression.

True or False. The Electromagnetic Spectrum can be described in terms of wave energy, frequency, and wavelength.

NeTakaya

True, because energy, frequency, and wavelength are all properties of sound, and sound is an electromagnetic wave. The electromagnetic spectrum encompasses a continuous range of frequencies or wavelengths of electromagnetic radiation, ranging from long wavelength, low energy radio waves to short wavelength, high frequency, high energy gamma rays.

5 0

2 years ago

2. Which equation is equivalent to 2 + 4x = 5 - 3(x + 1)? A. 2 + 4x = 2x - 3 B. 2 + 4% = 5 - 3 -3. C. 6x = 5-(+ 1). D. 2 + 4 = 2

Andre45 [30]

Answer:

B) 2 + 4x = 5 - 3 - 3x

Step-by-step explanation:

By simplifying JUST the right side and re-arranging them

2 + 4x = 5 - 3(x + 1),

2 + 4x = 5 - 3x -3,

2 + 4x = 5 -3 - 3x

3 0

3 years ago

Find the nth term of this sequence 2 5 8 11 14 ...

mariarad [96]

Each number is added by 3 so it's 17 20 23 26 29 32 35 38 41 and so on.

8 0

3 years ago

Read 2 more answers

Please help me with Q 7 c

frez [133]

Answer:

11.25 km h^-2

Step-by-step explanation:

Acceleration = increase in velocity / time

= 90 - 0 / 8

= 11.25 km h^-2.

8 0

2 years ago

Read 2 more answers

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