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3 years ago

Moe swam 2/3 of a mile in 12 minutes. If he swims at the same constant

Mathematics

2 answers:

AfilCa [17]3 years ago

6 0

Answer:

I do not understand the question?

tangare [24]3 years ago

4 0

Answer:

C. 18

$\frac{2}{3} \: miles = 12 \: minutes \\ 1 \: miles = (\frac{3}{2} \times 12) \: minutes \\ = 18 \: minutes$

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Two consecutive integers have a sum of 53. Find the integers.

Flura [38]

Answer:

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Step-by-step explanation:

Let one of the numbers be x,

Then the two numbers are x and x +1

And,

x+x+1 = 53

2x = 53 - 1 (Transposing 1 from LHS to RHS)

2x = 52

x = 52/2

x = 26

x + 1 = 27

So, The two numbers are 26 and 27.

7 0

2 years ago

Solve. 5 < x – 2 < 11

lions [1.4K]

The answer is 7 < x < 13.

5 < x - 2 < 11
Add 2 in all sections:
5 + 2 < x - 2 + 2 < 11 + 2
7 < x < 13

3 0

3 years ago

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