Can someone please help me with this

John and Annie have two distinguishable ponds. Initially, each of the ponds contains four ducks and five geese. John first picks

Sladkaya [172]

Answer: 4/9

Step-by-step explanation:

The probability that annie picks a duck Pt= the probability of picking a duck without accounting for the added one (Po)+ probability of picking the added bird and it's a duck(Pi)

Pt = Po+Pi

Since the total number of birds in the right pond is 10 after the addition of one by john

Po= 4/10

Pi= the of john adding a duck × the probability of annie picking the added bird

Pi= 4/9 × 1/10

Pi = 4/90

Pt= 4/10 + 4/90

Pt = (36+4)/90

Pt=40/90

Pt= 4/9

(This implies that the probability of picking a duck remain the same even after the addition of one bird from the left ponds because they both have equal proportions of duck and geese i.e the initial number of duck and geese in both right and left ponds are 4 and 5 respectively)

Thanks.

4 0

4 years ago

Prove diagonals of a rhombus are perpendicular

Lisa [10]

Answer:

A rhombus is a parallelogram with four congruent sides.

So, all sides of rhombus ABCD are congruent.

i.e, $\overline{JM} \cong \overline{JK} \cong \overline{KL}\cong \overline{LM}$

Also, we know that the diagonals of a parallelogram bisect each other.

Since a rhombus is a parallelogram.

By property of rhombus , if point N is the intersection of the diagonals as shown in the figure, then

$\overline{MN} \cong \overline{NK}$ .....[1]

$\overline{JN} \cong \overline{NL}$

In ΔJNM and ΔJNK

$\overline{MN} \cong \overline{NK}$ [side] [by (1)]

$\overline{JM} \cong \overline{JK}$ [side] [Given]

By reflexive property states that a segment is congruent to itself:

$\overline{JN} \cong \overline{JN}$ [Side] [Reflexive Property]

SSS(Side-Side-Side) postulates states that if three sides of one triangle are congruent to three sides of another triangle, then the two triangles are congruent.

then by SSS congruence,

$\triangle JNM \cong \triangle JNK$