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Brut [27]
4 years ago
8

Suppose segment XY has one endpoint at X(0,0)

Mathematics
1 answer:
Kisachek [45]4 years ago
6 0

Answer:

The coordinates of point Y are (9 , 12)

Step-by-step explanation:

* Lets explain how to solve the problem

- If point (a , b) divides a line whose end points are (c , d) and (e , f) at

 ratio p : q from the point (c , d), then

 a=\frac{cq+ep}{p+q} and b=\frac{dq+fp}{p+q}

∵ Segment XY has one endpoint at X (0 , 0)

∵ Point (3 , 4) is 1/3 of the way from X to Y

∴ Point (3 , 4) divides the segment XY,  where the distance from X

  to the point (3 , 4) is 1 part and from point (3 , 4) to Y is 2 parts

∴ Point (3 , 4) divides segment XY to the ratio 1 : 2 from X

- Let point (a , b) is (3 , 4)

- Let point (c , d) is X (0 , 0)

- Let point (e , f) is Y

- Let p : q = 1 : 2

* <em>Lets find e and f</em>

∵ 3=\frac{(0)(2)+e(1)}{1+2}

∴ 3=\frac{0+e}{3}

∴ 3=\frac{e}{3}

- Multiply both sides by 3

∴ 9 = e

∴ <u><em>The x-coordinate of point Y is 9</em></u>

∵ 4=\frac{(0)(2)+f(1)}{1+2}

∴ 4=\frac{0+f}{3}

∴ 4=\frac{f}{3}

- Multiply both sides by 3

∴ 12 = f

∴ <u><em>The y-coordinate of point Y is 12</em></u>

* The coordinates of point Y are (9 , 12)

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Answer:

2.49

Step-by-step explanation:

Given  X= 58

N = 175.

P = 25% = 0.25.

P' = \frac{X}{N} = \frac{58}{175} = 0.3314\\Z = \frac{P'-P}{\sqrt{\frac{pq}{N} } }

Now plugging the values we get

= \frac{0.3314-0.25}{\sqrt{\frac{0.25\times0.75}{175} } }

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A nationwide survey of college seniors by the University of Michigan revealed that almost 70% disapprove of daily pot smoking. I
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Answer:

a) P(X≥6) = 0.9614

b) P(7≤ X ≤9)  = 0.628

c) P(X≤5) = 0.0386

Step-by-step explanation:

This can be solved using the binomial distribution formula:

P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ

Where p = probability of success

           q = probability of failure = 1-p

           n = number of trials

           x = number of successful trials

We have p = 70% = 0.7

               n = 12

a) Find the probability that the number who disapprove of smoking pot daily is no less than 6 i.e. P(X≥6). This can be calculated as:

P(X≥6) = 1 - P(X<6)

        = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)]

        = 1 - [¹²C₀ (0.7)⁰(0.3)¹² + ¹²C₁ (0.7)¹(0.3)¹¹ + ¹²C₂ (0.7)²(0.3)¹⁰ + ¹²C₃ (0.7)³(0.3)⁹ + ¹²C₄ (0.7)⁴(0.3)⁸ + ¹²C₅ (0.7)⁵(0.3)⁷]

         = 1 - (0.000000531 + 0.0000148 + 0.0001909 + 0.00148 + 0.00779 + 0.02911)

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b) P(7≤ X ≤9) = P(X=7) + P(X=8) + P(X=9)

                     = ¹²C₇ (0.7)⁷(0.3)¹²⁻⁷ + ¹²C₈ (0.7)⁸(0.3)¹²⁻⁸ + ¹²C₉ (0.7)⁹(0.3)¹²⁻⁹

                     = 0.158 + 0.231 + 0.239

   P(7≤ X ≤9)  = 0.628

c) P(X≤5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)

               = ¹²C₀ (0.7)⁰(0.3)¹² + ¹²C₁ (0.7)¹(0.3)¹¹ + ¹²C₂ (0.7)²(0.3)¹⁰ + ¹²C₃ (0.7)³(0.3)⁹ + ¹²C₄ (0.7)⁴(0.3)⁸ + ¹²C₅ (0.7)⁵(0.3)⁷

   P(X≤5) = 0.0386

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