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Andreas93 [3]
3 years ago
10

Samantha leaves her house and drives 18 kilometers east to her friend's house. She then drives north to school. Finally, she dri

ves 30 kilometers directly home, in a straight line, from school. A map of Samantha's route is provided below.
How far does Samantha drive traveling from her friend's house to the school?

A.
42 kilometers

B.
12 kilometers

C.
35 kilometers

D.
24 kilometers

Mathematics
2 answers:
quester [9]3 years ago
8 0

Answer:

the answer is D) 35

Step-by-step explanation:

use the formula b²×h²

Kryger [21]3 years ago
8 0
I think it’s 12 because 30-18=12
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Kazeer [188]

$16

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4 0
3 years ago
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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

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3 0
2 years ago
7.07x10exponet2+3.51x10exponet3
Lynna [10]

Answer:

4217

Step-by-step explanation:

7.07(10^2)+3.51(10^3)

(7.07)(100)+3.51(10^3)

707+3.51(10^3)

707+(3.51)(1000)

707+3510

4217

4 0
3 years ago
Create three different expressions that are each equal to 20. Each expression should include only these three numbers: 4, -2, an
Archy [21]

Answer:

Step-by-step explanation:

you could do;

10 x -2 = 20

-2 x (-2) x 10 divided by (-2) + 10 + 10 + 10 +10 =20

4 + 4 - 10 x -2 - 4 - 4 = 20

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pochemuha
The answer is 1/4
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