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Zielflug [23.3K]
3 years ago
8

Suppose that a point (X,Y) us chosen at random from the rectangle S defined as follows: S={(x,y) : 0<= x <=2 and 1<=y&l

t;=4.} a. Determine the joint p.d.f. of X and Y, the marginal p.d.f. of X, and the marginal p.d.f. of Y. b. Are X and Y independent?
Mathematics
1 answer:
mrs_skeptik [129]3 years ago
7 0
For a 19 foot scale find the frequency of a note to the frequency of the preceding note?
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10 points! Solve the equation. If necessary, round your answer to the nearest tenth.
Julli [10]
If x² = 196, then we must do the inverse (square root) to find x. However, the square root of any positive number will always be ± (plus or minus).

That means you have to do:

√196 = x
x = <span>±14. 

We can check this by doing:

14 </span>× 14 = 196
-14 × -14 = 196

Therefore the answer is B) -14, 14. 
7 0
3 years ago
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log8x=2 solve for x by converting the logarithm to an exponential equation. (i already know the answer is 12.5 but i need the wr
swat32

Answer:

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Step-by-step explanation:

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3 litters were born at broadway kennel last year and each litter had 7 puppies how many puppies were born at broadway kennel las
frez [133]
3*7=21
21 puppies were born last year.
6 0
3 years ago
Can someone help me please?
alekssr [168]

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Step-by-step explanation:

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3 0
3 years ago
A couple intends to have two children, and suppose that approximately 52% of births are male and 48% are female.
Pachacha [2.7K]

a) Probability of both being males is 27%

b) Probability of both being females is 23%

c) Probability of having exactly one male and one female is 50%

Step-by-step explanation:

a)

The probability that the birth is a male can be written as

p(m) = 0.52 (which corresponds to 52%)

While the probability that the birth is a female can be written as

p(f) = 0.48 (which corresponds to 48%)

Here we want to calculate the probability that over  2 births, both are male. Since the two births are two independent events (the probability of the 2nd to be a male  does not depend on the fact that the 1st one is a male), then the probability of both being males is given by the product of the individual probabilities:

p(mm)=p(m)\cdot p(m)

And substituting, we find

p(mm)=0.52\cdot 0.52 = 0.27

So, 27%.

b)

In this case, we want to find the probability that both children are female, so the probability

p(ff)

As in the previous case, the probability of the 2nd child to be a female is independent from whether the 1st one is a male or a female: therefore, we can apply the rule for independent events, and this means that the probability that both children are females is the product of the individual probability of a child being a female:

p(ff)=p(f)\cdot p(f)

And substituting

p(f)=0.48

We find:

p(ff)=0.48\cdot 0.48=0.23

Which means 23%.

c)

In this case, we want to find the probability they have exactly one male and exactly one female child. This is given by the sum of two probabilities:

- The probability that 1st child is a male and 2nd child is a female, namely p(mf)

- The probability that 1st child is a female and 2nd child is a male, namely p(fm)

So, this probability is

p(mf Ufm)=p(mf)+p(fm)

We have:

p(mf)=p(m)\cdot p(f)=0.52\cdot 0.48=0.25

p(fm)=p(f)\cdot p(m)=0.48\cdot 0.52=0.25

Therefore, this probability is

p(mfUfm)=0.25+0.25=0.50

So, 50%.

Learn more about probabilities:

brainly.com/question/5751004

brainly.com/question/6649771

brainly.com/question/8799684

brainly.com/question/7888686

#LearnwithBrainly

5 0
3 years ago
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