Mathematics, 31.01.2020 14:52 lolz55678

Geometry
15 pts
answer is in fractions

line segment jm has endpoints with coordinates 0 and 25 on a number line. points k and l are on segment jm. k has a coordinate of 5 and point l has a coordinate of 12. find the probability that a point on jm is placed first on jl and a second point is not placed on kl.

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Answer from: zack66828

On a Number Line, if only whole numbers are marked

Points J, M, K, and L are marked, having coordinates 0, 25, 5, and 12.

Two points are again marked on the number line.

Probability,that a point on J M is placed first on J L

= There are 10 natural numbers in between J L and 12 natural numbers between L M.

So, Required Probability

$=\frac{_{1}^{C}\textrm{10}\times_{1}^{C}\textrm{12}}{_{1}^{C}\textrm{25}\times_{1}^{C}\textrm{25}}\\\\=\frac{10*12}{25*25}\\\\=\frac{120}{625}\\\\=\frac{24}{125}$

Now, Probability that second point is not placed on KL, it means that point is either is on J K or L M.

There are 4 natural number between 0 and 5 and 12 natural number between L and M.

Probability of marking second point on J M is

$=\frac{_{1}^{C}\textrm{4}\times_{1}^{C}\textrm{12}}{_{1}^{C}\textrm{25}\times_{1}^{C}\textrm{25}}\\\\=\frac{4*12}{25*25}\\\\=\frac{48}{625}$

Probability of marking two points on the number line, with given condition is

$=\frac{120}{625}+\frac{48}{625}\\\\=\frac{168}{625}$

If you consider ,points on the number line which are real numbers, then we can't find the required Probability that is marking two points on the line segment.

$Geometry 15 pts answer is in fractions line segment jm has endpoints with coordinates 0 and 25 on$

Answer from: Quest

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf =

pf = 6,25 u

ver más en -

step-by-step explanation:

Answer from: Quest

lo98765432wqasdfghjkloi87654567890

step-by-step explanation:

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Geometry
15 pts
answer is in fractions

line segment jm has endpoints with...

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