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3 years ago

HELPPPPP ASAPPPPPPPPPPPPPPPPP

Mathematics

1 answer:

Oksanka [162]3 years ago

3 0

If n=21 just plug it into the second equation
N+10=_
21+10=_
31=_
31 is the answer

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If y varies inversely as x and y 27 when x 40 find y when x 10

Iteru [2.4K]

When y varies inversely as x ⇒ y*x=k

y=27 and x=40 ⇒27*40=1080=k

y*10=1080
y=1080:10
y=108

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3 years ago

Which is the correct answer?

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2 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.

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