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Romashka-Z-Leto [24]

3 years ago

8

The graph of y=sin(x-3x/2) is the graph of the y=sin(x) shifted in which direction?

2 answers:

Tems11 [23]3 years ago

6 0

3x/2 units to the right, so b).
It's not c) or d), because the -3x/2 is within the sin, rather than outside like y=sin(x)-3x/2
If it were shifted to the left, the equation would be y=sin(x+3x/2)

Paul [167]3 years ago

5 0

Answer:

b. $\frac{3\pi}{2}$ to the right.

Step-by-step explanation:

We are asked to find the translation applied on graph of $y=\text{sin}(x)$ to get $y=\text{sin}(x-\frac{3\pi}{2})$

Let us recall the translation rules.

$f(x+a)=\text{Graph shifted to left by a units}$

$f(x-a)=\text{Graph shifted to right by a units}$

Upon looking at our given transformation we can see that graph of $y=\text{sin}(x)$ is shifted to right by $\frac{3\pi}{2}$ units to get $y=\text{sin}(x-\frac{3\pi}{2})$ .

Therefore, option 'b' is the correct choice.

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So we want to solve the following:

Conditioned on this event, show that the probability that her paper is in drawer $j$ , is given by:

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this gives us:

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Solution to Part (1):

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this means that

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Solution to Part(2):

so we have now that if $j$ = $i$ , we get that:

$P(C_{j}|A ) = \frac{P(A \cap C_{j})}{P(A)}$

remember that:

$P(A|C_{j} ) = \frac{P(A \cap C_{j})}{P(C_{j})}$

this implies that:

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so we just need to combine the above relations to get:

$P(C_{j}|A) = \frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i} }$

as needed so part two is solved.

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