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wariber [46]
2 years ago
14

Five bells begin to ring together and they ring at intervals of 3, 6, 10, 12 and 15 seconds, respectively. How many times will t

hey ring together at the same second in one hour excluding the one at the end?
Mathematics
1 answer:
Evgen [1.6K]2 years ago
6 0

Answer:

60 times will they ring together at the same second in one hour excluding the one at the end.

Step-by-step explanation:

Given : Five bells begin to ring together and they ring at intervals of 3, 6, 10, 12 and 15 seconds, respectively.

To find : How many times will they ring together at the same second in one hour excluding the one at the end?

Solution :

First we find the LCM of 3, 6, 10, 12 and 15.

2 | 3  6  10  12  15

2 | 3  3   5   6  15

3 | 3  3   5   3  15

5 | 1    1  5   1    5

  | 1    1   1   1     1

LCM(3, 6, 10, 12,15)=2\times 2\times 3\times 5

LCM(3, 6, 10, 12,15)=60

So, the bells will ring together after every 60 seconds i.e. 1 minutes.

i.e. in 1 minute they rand together 1 time.

We know, 1 hour = 60 minutes

So, in 60 minute they rang together 60 times.

Therefore, 60 times will they ring together at the same second in one hour excluding the one at the end.

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10. Makenzie is at the bakery. She sees a sign that says 8 pastries cost
kotegsom [21]

Answer:

$100

Step-by-step explanation:

8x= 160

160/8 = 20

x=20

5x = 100

3 0
2 years ago
Please Help Me!!!!!!!!<br><br> hg
Phantasy [73]

Answer:

YEAH

Step-by-step explanation:

8 0
3 years ago
Your cell phone company Offers you a choice between two promotional deals you can either get 500 free text messages with the cha
ollegr [7]

Answer:

The number of messages that will cost same both plans is <u>1000</u>.

Step-by-step explanation:

Given,

For 1st offer:

Total number of messages = 500

Charge of each message after 500 messages = $0.10

For 2nd offer:

Total number of messages = 400

Charge of each message after 400 messages = $0.20

We need to find out the number of messages can be sent for equal cost of both plans.

Solution,

Let the number of messages be 'm'.

So For 1st offer:

Total cost 'c' is equal to fixed cost for 500 messages plus charge of each message multiplied with number of messages.

framing in equation form, we get;

c=500+0.10m

Again for 2nd offer:

Total cost 'c' is equal to fixed cost for 400 messages plus charge of each message multiplied with number of messages.

framing in equation form, we get;

c=400+0.20m

Now the question said that the charge would be the same.

So we can say that;

500+0.10m=400+0.20m

On combining the like terms, we get;

500-400=0.20m-0.10m\\\\100=0.1m\\\\

Now using multiplication property, we will multiply both side by '10' and get;

100\times 10=0.1m\times10\\\\1000=m

Hence The number of messages that will cost same both plans is <u>1000</u>.

5 0
2 years ago
A motorboat travels due east 6 m/s across a 100m wide river. The current
sladkih [1.3K]

as far as I can make it, the vectors will be like the ones in the picture below, where the green one is the motorboat and the current is in red, and their sum will be the gray one.


since the boat is going 6 m/s in 1 second the motorboat has a magnitude of 6, and since it's due East, is just a horizontal line with an angle of 0°.


the current is due North, therefore has an angle of 90° and since the current is going 3 m/s, so in 1 second it has a magnitude of 3.


\bf \stackrel{\textit{motorboat}}{\begin{cases}x=rcos(\theta )\\\qquad 6cos(0^o)\\\qquad 6\\y=rsin(\theta )\\\qquad 6sin(0^o)\\\qquad 0\end{cases}}\implies \qquad \stackrel{\textit{current}}{\begin{cases}x=3cos(90^o)\\\qquad 0\\y=3sin(90^o)\\\qquad 3\end{cases}}\implies \\\\\\+\implies \implies \\\\\\\measuredangle \theta =tan^{-1}\left(\cfrac{3}{6}  \right)\implies \measuredangle \theta \approx 26.57^o


so, that will be the angle of the resultant vector, now, we know the river is 100 meters wide, so we can use cosine of θ to get the length and therefore the magnitude of the resultant vector. Keeping in mind that the angle of elevation θ is about 26.57° and that the adjacent side is 100.


\bf cos(26.57^o)=\cfrac{\stackrel{adjacent}{100}}{\stackrel{hypotenuse}{h}}\implies h=\cfrac{100}{cos(26.57^o)}\implies h\approx 111.8

7 0
3 years ago
A survey reported that 70% of those responding to a national survey of college freshmen were interested in taking summer courses
olya-2409 [2.1K]

Answer:

95% confidence interval: (0.5717,0.8283)          

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 49

p = 70% = 0.7

Alpha, α = 0.05

The confidence interval can be calculated as:

p \pm z_{critical}\sqrt{\dfrac{p(1-p)}{n}}

Putting the values, we get,

z_{critical}\text{ at}~\alpha_{0.05} = 1.96

0.7 \pm 1.96(\sqrt{\dfrac{0.7(1-0.7)}{49}} ) = 0.7 \pm 0.1283 =(0.5717,0.8283)

95% confidence interval: (0.5717,0.8283)      

5 0
3 years ago
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