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navik [9.2K]
3 years ago
10

⚠️⚠️ DONT IGNOR ⚠️⚠️ WHAT IS THE ANSWER AND WHICH IS PROPORTIONAL ¿

Mathematics
1 answer:
Nadya [2.5K]3 years ago
5 0
The answer is c) y=2/3x
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1 year ago
Find the distance between the two points in simplest radical form.<br> (-5,8) and (-3, 1)
elixir [45]

Given:

The two points are (-5,8) and (-3,1).

To find:

The distance between the given two points in simplest radical form.

Solution:

Distance formula: The distance between two points is

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Using distance formula, the distance between (-5,8) and (-3,1) is

d=\sqrt{(-3-(-5))^2+(1-8)^2}

d=\sqrt{(-3+5)^2+(-7)^2}

d=\sqrt{(2)^2+(-7)^2}

d=\sqrt{4+49}

d=\sqrt{53}

Therefore, the distance between two points (-5,8) and (-3,1) is \sqrt{53} units.

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2 years ago
A sample of 260 students enrolled at a university were asked what they intended to do after graduation. The results are shown be
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Hi my name is Fabian and 69% because 69 is a man which is multi colour hair

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3 years ago
Given circle U with diameter ST and radius UQ. QR is tangent to U at Q. If
vfiekz [6]

Answer:

QT = 18

Step-by-step explanation:

4 0
2 years ago
Let X1 and X2 be two random variables following Binomial distribution Bin(n1,p) and Bin(n2,p), respectively. Assume that X1 and
ryzh [129]

Answer:

a) X1+X2 have distribution Bi(n1+n2, p)

b)

P(X1+X2 = 1 | X2 = 0) =  np(1-p)ⁿ¹⁻¹

P(X1+X2 = 1| X2 = 1) = (1-p)ⁿ¹

P(X1 + X2 = 1) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹

Step-by-step explanation:

Since both variables are independent but they have the same probability parameter, you can interpret that like if the experiment that models each try in both variables is the same. When you sum both random variables toguether, what you obtain as a result is the total amount of success in n1+n2 tries of the same experiment, thus X1+X2 have distribution Bi(n1+n2, p).

b)

Note that, if X2 = k, then X1+X2 = 1 is equivalent to X1 = 1-k. Since X1 and X2 are independent, then P(X1+X2 = 1| X2 = K) = P(X1=1-k|X2=k) = P(X1 = 1-k).

If k = 0, then this probability is equal to P(X1 = 1) = np(1-p)ⁿ¹⁻¹

If k = 1, then it is equal to P(X1 = 0) = (1-p)ⁿ¹

Thus,

P(X1+X2 = 1) = P(X1+X2 = 1| X2 = 1) * P(X2=1) + P(X1+X2 = 1| X2 = 0) * P(X2 = 0) = (1-p)ⁿ¹ * np(1-p)ⁿ²⁻¹+ (1-p)ⁿ²*np(1-p)ⁿ¹-¹

4 0
3 years ago
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