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3 years ago

HELP WITH 4,5 and 6 PLEASE !!!!

Mathematics

1 answer:

WARRIOR [948]3 years ago

8 0

All you do for #4 and # 5 is plug the x values FROM THE TABLE into the equation, and see if you get the y values listed in the table. You don't, so you can see that the equations DO NOT REPRESENT THE DATA IN THE TABLE. For #6 you can see from the graph that one point on the line is (1,-2) which does represent the table of values. Looking further, as each x value decreases by 1, the y value decreases by 3, so (0,-5) makes sense too, again supporting the idea that the graph DOES represent the table of values

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Suppose you have an experiment where you flip a coin three times. You then count the number of heads. a.)State the random variab

WITCHER [35]

Answer:

a) X=number of heads observed when flipped the coin 3 times

b) the probability distribution is

P(X=x) = 3/4 * (1/[(3-x)!*x!)])

P(X)=1/8 for x=0 and x=3 and P(X)=3/8 for x=1 and x=2

Step-by-step explanation:

the random variable will be X=number of heads observed when flipped the coin 3 times . Since the result from each flip is independent of the others , then X has a binomial probability distribution , such that

P(X=x)= n!/[(n-x)!*x!)*p^x * (1-p)^(n-x)

where

n= number of times the coin is flipped = 3

p= probability of getting heads in a flip of the coin = 1/2 (assuming that the coin is fair)

therefore

P(X=x)= 3!/[(3-x)!*x!)*(1/2)^(3-x) * (1/2)^x = 3!/[(3-x)!*x!) * (1/2)³ = 3/4 * (1/[(3-x)!*x!)])

P(X=x)= 3/4 * (1/[(3-x)!*x!)]) , for x=[0,1,2,3]

for x=0 and x=3 → P(X)=3/4 * (1/[3!*0!)]) = 1/8

for x=1 and x=2 → P(X)=3/4 * (1/[2!*1!)]) = 3/8

we can verify that is correct since the sum of all the probabilities from x=0 to x=3 is 1/8 + 3/8+ 3/8+ 1/8 = 1

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3 years ago

Read 2 more answers

(x + y + 2)(y + 1)<br> Solve

RideAnS [48]

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$(x+y+2)(y+1)$

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$(a+b)(c)=ac+bc$

For this step, we are going to let:

$a=y$

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After applying this version of the Distributive Property, we are given the equation:

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We're going to use the distributive party again, but a slightly different version that's almost the same as first:

$(a)(b+c+d)=ab+ac+ad$

Using this on both terms of the equation changes the equation to:

$(xy+y^2+2y)+(x+y+2)$

Now, we just remove the parenthesis and combine like terms

$y^2+3y+xy+x+2$

This should be your answer. Let me know if you need any clarifications, thanks!

~ Padoru

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