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solong [7]
3 years ago
9

The table shows the height of a plant as it grows. Which equation in point ­slope form gives the plant’s height at any time

Mathematics
2 answers:
vodka [1.7K]3 years ago
5 0

Answer:

Option A is correct.

y-16=8(x-2) is the equation represent the point slope form gives the plant's height at any time.

Step-by-step explanation:

Point slope intercept form: For any two points (x_1, y_1) and  (x_2, y_2) then,

the general form

y-y_1=m(x-x_1) for linear equations;  where m is the slope given by:

m =\frac{y_2-y_1}{x_2-x_1}

Consider any two points from the table;

let A= (2 , 16) and B =(4, 32)

First calculate the slope of the line AB:

m =\frac{y_2-y_1}{x_2-x_1}=\frac{32-16}{4-2}=\frac{16}{2} = 8

Therefore, slope of the line m = 8

Then,

the equation of line is:

y-y_1=m(x-x_1)

Substitute the value of m=8 and (2, 16) above we get;

y-16=8(x-2)

Therefore, the equation in point slope form which gives the plant's height at any time is; y-16=8(x-2) , where x is the time(months) and y is the plant height (cm)


lozanna [386]3 years ago
3 0
A. is the answer it should be<span />
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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.
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  • 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:

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#SPJ1

3 0
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35b. If the order is wrong the image would be shrunk instead of enlarged, with a ratio of 3.5\div 6.5

36. We know that

\dfrac{p}{q}=\dfrac{2}{3}

Observe that

\dfrac{6p}{2q}=\dfrac{6}{2}\cdot\dfrac{p}{q}=3\cdot\dfrac{p}{q}=3\cdot\dfrac{3}{2}=\dfrac{9}{2}

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