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3 years ago

Simplify the expression: 8f–10f+5f+6f

Mathematics

1 answer:

ANEK [815]3 years ago

4 0

You began to do the equation from left to right. 8-10 = -2, so -2+5= 3, 3+6= 9f

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3 years ago

Suppose you have a bag containing 15 red beads, 12 white beads and 8 black beads. You are going to draw one bead out of the bag.

iren [92.7K]

Answer:

probability that you draw a black bead or a white bead that is P(E1 U E2)

= 20/35 = 0.571

Step-by-step explanation:

Given a bag containing 15 red beads, 12 white beads and 8 black beads

n(S) = 15 + 12 +8 = 35

Let 'E1 be the event of selecting black beads and E2 be the event of selecting white beads

n(E1) = 8 and n(E2) = 12

Probability of draw a black bead P( E1 ) = $\frac{n(E_{1} )}{n(S)}$

$P(E_{1}) = \frac{8}{35}$

Probability of draw a white bead P( E2 ) = $\frac{n(E_{2} )}{n(S)}$

$P(E_{2}) = \frac{12}{35}$

probability that you draw a black bead or a white bead that is P(E1 U E2)

and E1 n E2 = ∅ (disjoint events)

Axiom of union

$P(E_{1}UE_{2} ) = P(E_{1})+P(E_{2}) - P(E_{1}nE_{2})\\$

E1 n E2 = ∅ ⇒ P(E1 n E2) = p(∅) = 0

$P(E_{1}UE_{2} ) = P(E_{1})+P(E_{2}) - 0$

$P(E_{1}UE_{2} ) = \frac{8}{35} +\frac{12}{35}$

$P(E_{1}UE_{2} ) = \frac{20}{35}$

$P(E_{1}UE_{2} ) = 0.571$

8 0

4 years ago