Question

work out the probability that they own a dog​

Answer 1

Step-by-step explanation:

$\blue{\mathfrak{\underline{\large{Given}}}:} \\ C = student \: who \: own \: a \: cat \\ D = student \: who \: own \: a \: dog \\ U = 40 = number \: of \: students \\\blue{\mathfrak{\underline{\large{To \: find}}}:} \\ probability \: of \: owning \: a \: dog \\ \blue{\mathfrak{\underline{\large{Finding}}}:} \\ n(C) = x(x + 3) + 7 \\ = {x }^{2} + 3x + 7\\ n(D) = 4x + 7\\ n(C \cap \: D) = 7 \\ n(C \: U \: D)' =3x - 6 \\ \\ n(C \: U \: D)' =U - n(C \: U \: D) \\ n(C \: U \: D)' =U - (n(C) + n(D) - n(C \cap \: D)) \\ 3x - 6 = 40 - ( {x }^{2} + 3x + 7 + 4x + 7 - 7) \\ 3x - 6 = 40 - {x}^{2} - 7x - 7 \\ {x}^{2} + 10x - 39 = 0 \\ {x}^{2} + 13x - 3x - 39 = 0 \\ x(x + 13) - 3(x + 13) \\( x - 3)(x + 13) = 0 \\ x = 3, \: \: x = - 13 \\ \blue{\mathfrak{\large{ \rightarrow \: we \: have \: to \: take \: positive \: value}}} \\ \blue{\mathfrak{\large{because \: number \: of \: animals \: can't \: be \: 0}}} \\ Number \: of students \: owns\: dogs \: and \: cats \: ( n(C \: U \: D) ) \\ = {x}^{2} + 7x + 7 \\ = {3}^{2} + 7 \times 3 + 7 \\ = 9 + 21 + 7 = > 37 \\ Number \: of \: students \: who \: owns\: dogs \: only\\ = 4x \\ = 4 \times 3 \\ = 12 + 7 \\ = 12 \\\blue{\mathfrak{\underline{\large{probability \: of \: owning \: a \: dog}}}:} \\ = \frac{Number \: of \: students \: who \: owns\: dogs}{Total \: number \: student} \\ = \frac{12}{40} = \frac{3}{10} \\ \blue{\mathfrak{\boxed{If \: it \: specify \: that \: student \: chosen }}} \\ \blue{\mathfrak{ \boxed{from \: owners \: of \: cats \:and \: dogs \: }}} \\ \blue{\mathfrak{\boxed{then \: you \: can \: use \: value \: 37 \: in \: replace \: of \: total \: students}}} \\ \\ \red{\mathfrak{\underline{\large{Hope \: It \: Helps \: You }}}} \\ \green{\mathfrak{\underline{\large{Mark \: Me \: Brainliest}}}}$