How many times does the minute hand move around the clock face in one day?

2 answers:

8 0

To find that, we will start from one minute. How many seconds are there in one minute? There are 60 seconds in one minute.
Now, how many minutes are there in an hour? 60 minutes. So we have to do
60 × 60 = seconds in an hour = 3600 seconds
How many hours are in a day? Well, there are 24 hours in a day.
So 3600 × 24 = seconds in a day = 86400 seconds in a day

There are 86400 seconds in a day.

Alexus [3.1K]2 years ago

4 0

The hour hand will go around the clock twice during a day, every 12 hours. During a single rotation, the minute hand will go around 12 times. The minute hand will pass the hour hand 11 times in the first 12 hours, and 11 times in the second 12 hours. This means that they will meet 22 times.

You might be interested in

Five cards are drawn from a standard 52-card playing deck. A gambler has been dealt five cards—two aces, one king, one 3, and on

Nookie1986 [14]

Answer:

The probability that he ends up with a full house is 0.0083.

Step-by-step explanation:

We are given that a gambler has been dealt five cards—two aces, one king, one 3, and one 6. He discards the 3 and the 6 and is dealt two more cards.

We have to find the probability that he ends up with a full house (3 cards of one kind, 2 cards of another kind).

We know that gambler will end up with a full house in two different ways (knowing that he has given two more cards);

If he is given with two kings.
If he is given one king and one ace.

Only in these two situations, he will end up with a full house.

Now, there are three kings and two aces left which means at the time of drawing cards from the deck, the available cards will be 47.

So, the ways in which we can draw two kings from available three kings is given by = $\frac{^{3}C_2 }{^{47}C_2}$ {∵ one king is already there}

= $\frac{3!}{2! \times 1!}\times \frac{2! \times 45!}{47!}$ {∵ $^{n}C_r = \frac{n!}{r! \times (n-r)!}$ }

= $\frac{3}{1081}$ = 0.0028

Similarly, the ways in which one king and one ace can be drawn from available 3 kings and 2 aces is given by = $\frac{^{3}C_1 \times ^{2}C_1 }{^{47}C_2}$