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2 years ago

Given the following sequence:3,6,12,24

Mathematics

1 answer:

Mars2501 [29]2 years ago

8 0

Answer:

It is a GP.

Step-by-step explanation:

a = 3

r = 2

a(n) term = ar^n

As it increases as multiples, it is a geometric progression or a GP.

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Giving right answer a branleist

Marta_Voda [28]

Answer:

yes, (8×10) - (8×3)

hope this helps :)

6 0

2 years ago

Read 2 more answers

Yahoo creates a test to classify emails as spam or not spam based on the contained words. This test accurately identifies spam (

Amiraneli [1.4K]

Answer and Step-by-step explanation:

The computation is shown below:

Let us assume that

Spam Email be S

And, test spam positive be T

Given that

P(S) = 0.3

$P(\frac{T}{S}) = 0.95$

$P(\frac{T}{S^c}) = 0.05$

Now based on the above information, the probabilities are as follows

i. P(Spam Email) is

= P(S)

= 0.3

$P(S^c) = 1 - P(S)$

= 1 - 0.3

= 0.7

ii. $P(\frac{S}{T}) = \frac{P(S\cap\ T}{P(T)}$

$= \frac{P(\frac{T}{S}) . P(S) }{P(\frac{T}{S}) . P(S) + P(\frac{T}{S^c}) . P(S^c) }$

$= \frac{0.95 \times 0.3}{0.95 \times 0.3 + 0.05 \times 0.7}$

= 0.8906

iii. $P(\frac{S}{T^c}) = \frac{P(S\cap\ T^c}{P(T^c)}$

$= \frac{P(\frac{T^c}{S}) . P(S) }{P(\frac{T^c}{S}) . P(S) + P(\frac{T^c}{S^c}) . P(S^c) }$

$= \frac{(1 - 0.95)\times 0.3}{ (1 -0.95)0.95 \times 0.3 + (1 - 0.05) \times 0.7}$

= 0.0221

We simply applied the above formulas so that the each part could come

8 0

2 years ago

A spacecraft is traveling with a velocity of v0x = 5320 m/s along the +x direction. Two engines are turned on for a time of 739

OlgaM077 [116]

Answer:

The velocities after 739 s of firing of each engine would be 6642.81 m/s in the x direction and 5306.02 in the y direction

Step-by-step explanation:

For a constant acceleration: $v_{f}=v_{0}+at$ , where $v_{f}$ is the final velocity in a direction after the acceleration is applied, $v_{0}$ is the initial velocity in that direction before the acceleration is applied, a is the acceleration applied in such direction, and t is the amount of time during where that acceleration was applied.
<em>Then for the x direction</em> it is known that the initial velocity is $v_{0x} =$ 5320 m/s, the acceleration (the applied by the engine) in x direction is $a_{x}$ 1.79 m/s2 and, the time during the acceleration was applied (the time during the engines were fired) of the is 739 s. Then: $v_{fx}=v_{0x}+a_{x}t=5320\frac{m}{s} +1.79\frac{m}{s^{2} }*739s=6642.81\frac{m}{s}$
In the same fashion, <em>for the y direction</em>, the initial velocity is $v_{0y} =$ 0 m/s, the acceleration in y direction is $a_{y}$ 7.18 m/s2, and the time is the same that in the x direction, 739 s, then for the final velocity in the y direction: $v_{fy}=v_{0y}+a_{y}t=0\frac{m}{s} +7.18\frac{m}{s^{2} }*739s=5306.02\frac{m}{s}$