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4 years ago

Given a directed line segment with endpoints A(3, 2) and B(6, 11), what is the point that

Mathematics

2 answers:

mrs_skeptik [129]4 years ago

7 0

Answer:

$(\frac{21}{5} ,\frac{28}{5} )$ is the point

Step-by-step explanation:

A(3, 2) and B(6, 11)

The ratio that divided AB is 2:3

we use section formula

2:3 is m:n

formula is $(\frac{mx_2+nx_1}{m+n} ,\frac{my_2+ny_1}{m+n} )$

A(3, 2) is (x1,y1) and B(6, 11) is (x2,y2)

Plug in the value in the formula

$(\frac{mx_2+nx_1}{m+n} ,\frac{my_2+ny_1}{m+n} )$

$(\frac{(2)(6)+3(3)}{2+3} ,\frac{2(11)+3(2)}{2+3} )$

$(\frac{21}{5} ,\frac{28}{5} )$

bixtya [17]4 years ago

5 0

The formula for a directed line is:

X = K1(X2) + (K2(X1) / K3

Y = K1(Y2) + (K2(Y1) / K3

K1 = numerator = 2

K2 = denominator - numerator = 3-2 = 1

K3 = denominator = 3

X = (2(6) + 1(3)) /3

X = (12+3) / 3

X = 12/3

X = 5

Y = (2(11) + 1(2)) /3

Y = (22 + 2) /3

Y = 24/3

Y = 8

The point is (5,8)

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Suppose you have two urns with poker chips in them. Urn I contains two red chips and four white chips. Urn II contains three red

Neporo4naja [7]

Answer:

Multiple answers

Step-by-step explanation:

The original urns have:

Urn 1 = 2 red + 4 white = 6 chips
Urn 2 = 3 red + 1 white = 4 chips

We take one chip from the first urn, so we have:

The probability of take a red one is : $\frac{1}{3}$ (2 red from 6 chips(2/6=1/2))

For a white one is: $\frac{2}{3}$ (4 white from 6 chips(4/6=(2/3))

Then we put this chip into the second urn:

We have two possible cases:

First if the chip we got from the first urn was white. The urn 2 now has 3 red + 2 whites = 5 chips
Second if the chip we got from the first urn was red. The urn two now has 4 red + 1 white = 5 chips

If we select a chip from the urn two:

In the first case the probability of taking a white one is of: $\frac{2}{5}$ = 40% ( 2 whites of 5 chips)
In the second case the probability of taking a white one is of: $\frac{1}{5}$ = 20% ( 1 whites of 5 chips)

This problem is a dependent event because the final result depends of the first chip we got from the urn 1.

For the fist case we multiply :

$\frac{4}{6}$ x $\frac{2}{5}$ = $\frac{4}{15}$ = 26.66% ( $\frac{4}{6}$ the probability of taking a white chip from the urn 1, $\frac{2}{5}$ the probability of taking a white chip from urn two)

For the second case we multiply:

$\frac{1}{3}$ x $\frac{1}{5}$ = $\frac{1}{30}$ = .06% ( $\frac{1}{3}$ the probability of taking a red chip from the urn 1, $\frac{1}{5}$ the probability of taking a white chip from the urn two)