Each video game at an arcade requires 55 tokens.Tom has 100 t video game okens. He said he can play 2 games. Is he correct?

It takes Manisha 38 seconds to climb up the stairs of her house. If she spends 504 seconds doing this, how many minutes has she

Mrac [35]

Answer:

Manisha spent 8 minutes and 24 seconds in a week climibing the stairs of her house.

Step-by-step explanation: (totally my work, not copied)

Convert seconds to minutes.

60x8=480

60x9=540

so, 8 minutes

Find the remaining seconds.

504-480=24

Add remaining seconds to minutes

24sec+8min=8min24sec

So hence, your answer is 8 Minutes and 24 Seconds

Manisha spent 8 minutes and 24 seconds in a week climbing the stair of her house.

Thanks!

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5 0

1 year 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

3 0

1 year ago

Using the Exxon data as an example pretend Pennys Pickles divide their company into 10,000 shares of stock. If you buy 200 of th

nasty-shy [4]

You would own 2% of penny’s Picklesdo i hope this helps

4 0

2 years ago

Len tosses a coin three times. The coin shows heads every time. What are the chances the coin shows tails on the next toss? Expl

Kazeer [188]

Every coin flip has a 1/2 probability (50% chance) of landing on tails no matter what it landed on before.

so the probability of landing on tails id 1/2 or 50% ( however you need to answer)

5 0

2 years ago

Use the graph at the right. Find the vertices of the image of QRTW for a dilation with center (0,0) and a scale factor of 5.

Zanzabum

Answer:

What image? (9/ -0) Is 9 - 8 =1

Step-by-step explanation:

5 0

2 years ago