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3 years ago

Volume cylinder pi 3.14 d=12 cm h = 6cm d=2r

Mathematics

1 answer:

alexira [117]3 years ago

8 0

Answer:

678.24cm³

Step-by-step explanation:

Volume of cylinder = # r² h

Where # = 3.14

r = radius = d/2 = 12/2 = 6cm

h = 6cm

Therefore

Volume = #r²h

= 3.14 × 6² × 6

= 3.14 × 36 × 6

= 678.24cm³

I hope this was helpful, please rate as brainliest

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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 $\frac{9}{15}=0.6$

Since there are no replacement, the probability of taking a second non 75-W bulb is now $\frac{8}{14}=0.5714$

Following this procedure 5 times, we find the probabilities

$\frac{9}{15},\frac{8}{14},\frac{7}{13},\frac{6}{12},\frac{5}{11}$

which are

0.6, 0.5714, 0.5384, 0.5, 0.4545

As the events are independent, the probability of choosing 5 non 75-W bulbs is the product

0.6*0.5714*0.5384*0.5*0.4545 = 0.0419 = 4.19%

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Answer:

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#54 Simplify and write the answers using positive exponents only.

kramer

Gg easy

remember
(x^m)/(x^n)=x^(m-n)
and
(x^m)^n=x^(mn)
and
(xyz)^m=(x^m)(y^m)(z^m)
and
(m/n)^a=(m^a)/(n^a)
and
x^-n=1/(x^n)

so
$( \frac{6mn^{-2}}{3m^{-1}n^2} )^{-3}$ =
$( \frac{6}{3} )^{-3}( \frac{m}{m^{-1}} )^{-3}( \frac{n^{-2}}{n^2} )^{-3}$ =
$(2^{-3})((m^2)^{-3})((n^{-4})^{-3})$ =
$( \frac{1}{2^3})(m^{-6})(n^{12})$ =
$( \frac{1}{8} )( \frac{1}{m^6})(n^{12})$ =
$\frac{n^{12}}{8m^6}$

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3 years ago