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emmasim [6.3K]
2 years ago
5

10. Expand and Simplify (r+3) (r+11)=

Mathematics
1 answer:
mihalych1998 [28]2 years ago
6 0

Answer:

r^2 + 14r + 33

Step-by-step explanation:

Apply the "FOIL" method for multiplying two binomials:

(r+3) (r+11) = r^2 (first terms) +

                    + 33 (last terms) +

                     + 11r (outer terms) +

                      + 3r (inner terms)

Combining ilke terms, we get:

r^2 + 14r + 33

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Answer:

(a) 7.69%

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Step-by-step explanation:

(a)

In a deck of 52 cards there are 4 aces.

The odds in favor or the probability of selecting an ace is:

P(Ace) = \frac{Number\ of\ aces}{Number\ of\ cards\ in\ total}\\ =\frac{4}{52}\\ =0.076923\\\approx7.69\%

Thus, the probability of selecting an ace from a random deck of 562 cards is 7.69%.

(b)

The outcomes of each toss of a coin is independent of the other, since the result of the previous toss does not affect the result of the current toss.

The probability that both the tosses will end up in heads is:

P(2\ Heads)=P(1^{st}\ Head)\times P(2^{nd}\ Head)\\=\frac{1}{2}\times \frac{1}{2}\\ =\frac{1}{4}\\ =0.25\ or\ 25\%\\

Thus, the probability that both the tosses will end up in heads is 25%.

(c)

The sample space of two dice consists of 36 outcomes in total.

Out of these 36 outcomes there is only 1 Boxcar, i.e. two sixes.

The probability of a boxcar when two dice are rolled is:

P(Boxcar)=\frac{Favorable\ outcomes}{Total\ no.\ of\ outcomes}\\= \frac{1}{36}\\ =0.027777\\\approx2.78\%

Thus, the probability of a boxcar when two dice are rolled is 2.78%.

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