How do you divide (n^2-n-29)/(n-6) ??

Mathematics

1 answer:

Orlov [11]2 years ago

5 0

9514 1404 393

Answer:

see attached

Step-by-step explanation:

Polynomial long division is done the way any long division is done. Find a "partial quotient", subtract from the dividend the product of that partial quotient and the divisor. The result is a new dividend. Repeat until the degree of the dividend is less than that of the divisor.

In the attached, the "Hints" show you how the partial quotient is found, and they show you how the product of the partial quotient and divisor is found.

The partial quotient term is simply the ratio of the highest degree terms of dividend and divisor. (Unlike numerical long division, there is no guessing.)

The remainder is the dividend of lower degree than the divisor. As in numerical long division, the full quotient expresses the remainder over the divisor.

For example, 5 ÷ 3 = 1 r 2 = 1 + 2/3.

Your full quotient is (n+5) +1/(n-6).

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. A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W bulbs. Suppose that three bulbs ar

yaroslaw [1]

Answer:

a) 59.34%

b) 44.82%

c) 26.37%

d) 4.19%

Step-by-step explanation:

(a)

There are in total 4+5+6 = 15 bulbs. If we want to select 3 randomly there are K ways of doing this, where K is the combination of 15 elements taken 3 at a time

$K=\binom{15}{3}=\frac{15!}{3!(15-3)!}=\frac{15!}{3!12!}=\frac{15.14.13}{6}=455$

As there are 9 non 75-W bulbs, by the fundamental rule of counting, there are 6*5*9 = 270 ways of selecting 3 bulbs with exactly two 75-W bulbs.

So, the probability of selecting exactly 2 bulbs of 75 W is

$\frac{270}{455}=0.5934=59.34\%$

(b)

The probability of selecting three 40-W bulbs is

$\frac{4*3*2}{455}=0.0527=5.27\%$

The probability of selecting three 60-W bulbs is

$\frac{5*4*3}{455}=0.1318=13.18\%$

The probability of selecting three 75-W bulbs is

$\frac{6*5*4}{455}=0.2637=26.37\%$

Since the events are disjoint, the probability of taking 3 bulbs of the same kind is the sum 0.0527+0.1318+0.2637 = 0.4482 = 44.82%

(c)

There are 6*5*4 ways of selecting one bulb of each type, so the probability of selecting 3 bulbs of each type is

$\frac{6*5*4}{455}=0.2637=26.37\%$

(d)

The probability that it is necessary to examine at least six bulbs until a 75-W bulb is found, supposing there is no replacement, is the same as the probability of taking 5 bulbs one after another without replacement and none of them is 75-W.

As there are 15 bulbs and 9 of them are not 75-W, the probability a non 75-W bulb is $\frac{9}{15}=0.6$