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Evgen [1.6K]
3 years ago
15

Which steps will verify that a parrallelogram is a rectangle

Mathematics
1 answer:
Xelga [282]3 years ago
7 0

Step-by-step explanation:

"To verify that a given parallelogram is a rectangle, you can calculate the lengths of all sides, and show that both pairs of opposite sides are congruent and calculate the slopes of every side, and show that adjacent sides are perpendicular."

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5 1/2 + 2 2/3 + 3 2/5=
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3 years ago
Suppose you play a game in which 5 dice (6-sided) are rolled simultaneously.
goblinko [34]

The expected value of the game in which 5 dice (6-sided) are rolled simultaneously is -0.2094.

<h3>How to find the mean (expectation) and variance of a random variable?</h3>

Supposing that the considered random variable is discrete, we get:

Mean =  E(X) = \sum_{\forall x_i} f(x_i)x_i

Here,   x_i; \: \: i = 1,2, ... , n is its n data values and f(x_i)is the probability of  X = x_i

Suppose you play a game in which 5 dice (6-sided) are rolled simultaneously.

  • If a "4" is rolled, then you win $2 for each "4" showing.
  • If all the dice are showing "4", you win $1000.
  • If none of the dice are showing "4", then you lose $5.

Let Y is the amount of money player won. The value of X can be,

Y=2,4,6,8,1000,-5

<h3>How to find that a given condition can be modeled by binomial distribution?</h3>

Binomial distributions consists of n independent Bernoulli trials. Bernoulli trials are those trials which end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

Suppose we have random variable X pertaining binomial distribution with parameters n and p, then it is written as

X \sim B(n,p)

The probability that out of n trials, there'd be x successes is given by

P(X =x) = \: ^nC_xp^x(1-p)^{n-x}

The expected value and variance of X are:

E(X) = np\\ Var(X) = np(1-p)

Put the values as 5 trials for each time 4 appears.

P(X =0) = \: ^5C_1(\dfrac{1}{6})^0(1-\dfrac{1}{6})^{5-0}=0.4018 \:\\P(X =1) = \: ^5C_1(\dfrac{1}{6})^1(1-\dfrac{1}{6})^{5-1}=0.402\\P(X =2) = \: ^5C_2(\dfrac{1}{6})^2(1-\dfrac{1}{6})^{5-2}=0.162\\P(X =3) = \: ^5C_3(\dfrac{1}{6})^3(1-\dfrac{1}{6})^{5-3}=0.032\\P(X =4) = \: ^5C_4(\dfrac{1}{6})^4(1-\dfrac{1}{6})^{5-4}=0.0032\\P(X =5) = \: ^5C_5(\dfrac{1}{6})^5(1-\dfrac{1}{6})^{5-5}=0.00013\\

The probability of loosing $5 equal probability of 0 success.

P(Y=-5)=P(x=0)

Similarly, for probability of getting profit are,

P(Y=2)=P(x=1)\\P(Y=4)=P(x=2)\\P(Y=6)=P(x=3)\\P(Y=8)=P(x=4)\\P(Y=1000)=P(x=5)

Expected value of game,

E(Y)=\sum y .P(Y=y)\\E(Y)=-5.P(X=0)+2.P(X=1)+4.P(X=2)+6.P(X=3)+8.P(X=4)+1000.P(X=5)\\E(Y)=-0.2094

Thus, the expected value of the game in which 5 dice (6-sided) are rolled simultaneously is -0.2094.

Learn more about expectation of a random variable here:

brainly.com/question/4515179

Learn more about binomial distribution here:

brainly.com/question/13609688

#SPJ1

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