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3 years ago

Izumi is running on a quarter-mile oval track. After running 110 yards, his coach records his time as 16 seconds.

Mathematics

1 answer:

pickupchik [31]3 years ago

3 0

Answer:

14.04 miles per hour

Step-by-step explanation:

The problem is asking for Izumi's speed. The formula of speed is:

s = $\frac{d}{t}$ , where "s" means speed, "d" means distance and "t" means time.

The problem is also asking for the unit miles per hour ( $\frac{miles}{hour}$ ) so, this means that we have to know how many miles Izumi ran, given that the problem only mentioned yards (110 yards).

Let's convert 110 yards to miles, provided that he Izumi ran 1,760 yards in a mile.

110 yards ÷ 1760 $\frac{yards}{mile}$ = 0.0625 miles (this is the distance covered by Izumi in miles)

Let's go back again to the formula: s = $\frac{d}{t}$

s = $\frac{0.0625 miles}{16 seconds}$ = 0.0039 $\frac{miles}{sec}$

Since, the we arrived at a miles per second unit, we have to convert it to miles per hour.

So, if a minute has 60 seconds, then an hour has 3,600 seconds.

Thus, 0.0039 $\frac{miles}{sec}$ × 3,600 $\frac{seconds}{hour}$ = 14.04 miles per hour (the answer)

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