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8090 [49]
2 years ago
10

If events X and Y are independent, what must be true? Check all that apply.

Mathematics
2 answers:
Luda [366]2 years ago
7 0

3. P(Y | X) = P(Y) and 6. P(X | Y) = P(X)

larisa86 [58]2 years ago
6 0
All these are true when X and Y are independent events:

1) The occurrence or lack of event X does not influence the occurrence of event Y, and the occurrence  or lack of event Y does not influence the occurrence of event X.

2) P(X∩Y) = P(X) * P(Y)

3) P(X | Y) = P(X).... this is the probability of X given Y is equal to the probability of X.

4) P (Y | X) = P(Y)

5) P(A∪B) = P(A) + P(B) - P(A∩B)
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A car company sells a scale model to of the size of one of its cars.
mixer [17]

Answer:

w = 13500 --- real windscreen

v = 408 --- scale volume

Step-by-step explanation:

Given

k = \frac{1}{10} --- scale factor

See attachment

Required

Complete the table

To get the windscreen (w) of the real car, we do the following computation

w = \frac{w_{scale}}{k^2}

This gives:

w = \frac{135}{(1/10)^2}

w = \frac{135}{1/100}

w = 13500

Next, the volume (v) of the scale mode

We make use of the following computation

v = v_{real} * k^3

So, we have:

v = 408000 * (1/10)^3

v = 408000 * 1/1000

v = 408

7 0
2 years ago
100/225 square feet of your bakery will be used as your storefront, where customers come visit. The rest kill be used for storag
Lelu [443]
The answer is 5/9. I showed a bit more work on the other post
7 0
2 years ago
Explain how to use a graph of the function f(x) to find f(3).
liraira [26]
Usually there will be a line through the graph, labeling f(x) = ..... so plug in f(3) and follow the line on the x-axis to look for 3 then find y
7 0
2 years ago
Read 2 more answers
Prove tan3A= (cos(2A)-cos(4A))/ (sin(4A)- Sin(2A)) ...?
aivan3 [116]
The solution to the problem is as follows:

<span>cos2A - cos4A = -2 sin(6A/2).sin(-2A/2) = +2 sin(3A).sinA
 
and
 
sin4A - sin2A = 2 cos(6A/2).sin(2A/2) = 2 cos(3A).sinA

Hence RHS = (cos(2A)-cos(4A))/ (sin(4A)- Sin(2A)) = sin 3A / cos 3A = tan 3A = LHS
</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
5 0
2 years ago
A particle moves on the circle x2+y2=25 in the xy-plane for time t≥0. At the time when the particle is at the point (3,4), dx/dt
yaroslaw [1]

Answer:

\displaystyle y'=-\frac{9}{2}

Step-by-step explanation:

<u>Differentiation</u>

We have a relationship between x and y as follows:

x^2+y^2=25

Both variables depend on time t for t≥0.

Differentiating with respect to time:

(x^2)'+(y^2)'=(25)'

Applying the derivative of a power function:

2xx'+2yy'=0

Recall the derivative of a constant is 0.

Dividing by 2:

xx'+yy'=0

Solving for y':

\displaystyle y'=-\frac{xx'}{y}

At some specific time, we have:

x=3, y=4, dx/dt = x' = 6. Substituting:

\displaystyle y'=-\frac{3\cdot 6}{4}

Operating:

\displaystyle y'=-\frac{18}{4}

Simplifying:

\mathbf{\displaystyle y'=-\frac{9}{2}}

8 0
2 years ago
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