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4 years ago

CAN Someone help me out with Experimental probability.Ill give brainless

Mathematics

1 answer:

pentagon [3]4 years ago

4 0

Answer:

1/4

Step-by-step explanation:

It is split into 8 equal parts, if you want to see how likely you're going to get red, look at how many red parts there are. 2. Now make it into a fraction! 2/8, but we need it to be simplified so it is 1/4.

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Answers...............

umka21 [38]

Answer: Two acute angles

7 0

3 years ago

For integers a, b, and c, consider the linear Diophantine equation ax C by D c: Suppose integers x0 and y0 satisfy the equation;

Dmitrij [34]

Answer:

$x = x_1+r(\frac{b}{gcd(a, b)} )\\y=y_1-r(\frac{a}{gcd(a, b)} )$

b. x = -8 and y = 4

Step-by-step explanation:

This question is incomplete. I will type the complete question below before giving my solution.

For integers a, b, c, consider the linear Diophantine equation

$ax+by=c$

Suppose integers x0 and yo satisfy the equation; that is,

$ax_0+by_0 = c$

what other values

$x = x_0+h$ and $y=y_0+k$

also satisfy ax + by = c? Formulate a conjecture that answers this question.

Devise some numerical examples to ground your exploration. For example, 6(-3) + 15*2 = 12.

Can you find other integers x and y such that 6x + 15y = 12?

How many other pairs of integers x and y can you find ?

Can you find infinitely many other solutions?

From the Extended Euclidean Algorithm, given any integers a and b, integers s and t can be found such that

$as+bt=gcd(a,b)$

the numbers s and t are not unique, but you only need one pair. Once s and t are found, since we are assuming that gcd(a,b) divides c, there exists an integer k such that gcd(a,b)k = c.

Multiplying as + bt = gcd(a,b) through by k you get

$a(sk) + b(tk) = gcd(a,b)k = c$

So this gives one solution, with x = sk and y = tk.

Now assuming that ax1 + by1 = c is a solution, and ax + by = c is some other solution. Taking the difference between the two, we get

$a(x_1-x) + b(y_1-y)=0$

Therefore,

$a(x_1-x) = b(y-y_1)$

This means that a divides b(y−y1), and therefore a/gcd(a,b) divides y−y1. Hence,

$y = y_1+r(\frac{a}{gcd(a, b)})$ for some integer r. Substituting into the equation

$a(x_1-x)=rb(\frac{a}{gcd(a, b)} )\\gcd(a, b)*a(x_1-x)=rba$

$x = x_1-r(\frac{b}{gcd(a, b)} )$

Thus if ax1 + by1 = c is any solution, then all solutions are of the form

$x = x_1+r(\frac{b}{gcd(a, b)} )\\y=y_1-r(\frac{a}{gcd(a, b)} )$

In order to find all integer solutions to 6x + 15y = 12

we first use the Euclidean algorithm to find gcd(15,6); the parenthetical equation is how we will use this equality after we complete the computation.

$15 = 6*2+3\\6=3*2+0$

Therefore gcd(6,15) = 3. Since 3|12, the equation has integral solutions.

We then find a way of representing 3 as a linear combination of 6 and 15, using the Euclidean algorithm computation and the equalities, we have,

$3 = 15-6*2$

Because 4 multiplies 3 to give 12, we multiply by 4

$12 = 15*4-6*8$

So one solution is

$x=-8$ & $y = 4$

All other solutions will have the form

$x=-8+\frac{15r}{3} = -8+5r\\y=4-\frac{6r}{3} =4-2r$

where $r$ ∈ Ζ

Hence by putting r values, we get many (x, y)

3 0

3 years ago

How to write eight ten-thousandths of thirty-five million in scientific notation?

VashaNatasha [74]

Eight ten-thousandths = 8*(1/10,000) = 8/10⁴ = 8 x 10⁻⁴.
Thirty-five million = 35*(1,000,000) = 35 x 10⁶

Therefore eight ten-thousandths of thirty-five million is
(8 x 10⁻⁴) * (35 x 10⁶)
= 280 x 10⁶⁻⁴
= 280 x 10²
= (2.8 x 10²) * 10²
= 2.8 x 10⁴

Answer: 2.8 x 10⁴

5 0

4 years ago

Jack says 0.90 is between 0.9 and 0.91. Is he correct? Explain.

nevsk [136]

Yes he is right because 0.90 is before 0.91.

4 0

3 years ago

Read 2 more answers

Y = 5 - 4x<br>3x – 2y = 12

VMariaS [17]

Answer:

T3 qgt3 gg aez

Step-by-step explanation:

6 0

3 years ago

CAN Someone help me out with Experimental probability.Ill give brainless​

CAN Someone help me out with Experimental probability.Ill give brainless