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3 years ago

8912 butterflies on one branch they estimate 1000 times as large. How many butterflies were at the site in all?

2 answers:

6 0

<span>There are 8 912 butterflies on one branch.
And they estimated to have 1000 times as large.
now, find the total number of the butterflies found in their site.
=> 8 912 x 1000 (because it is stated that the number of butterflies in all is 1 000 times as large as the given number of butterflies in a one branch.)
=> Multiply 8 912 by 1 000
=> 8 912 000
Thus, the total of butterflies in their site is approximately 8 912 000</span><span>

</span>

Alex17521 [72]3 years ago

4 0

Answer:

There are 8 912 butterflies on one branch.

And they estimated to have 1000 times as large.

now, find the total number of the butterflies found in their site.

=> 8 912 x 1000 (because it is stated that the number of butterflies in all is 1 000 times as large as the given number of butterflies in a one branch.)

=> Multiply 8 912 by 1 000

=> 8 912 000

Thus, the total of butterflies in their site is approximately 8 912 000

Read more on Brainly.com - brainly.com/question/620276#readmore

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A new shopping mall is considering setting up an information desk manned by one employee. Based upon information obtained from s

quester [9]

Answer:

a) $P=1-\frac{\lambda}{\mu}=1-\frac{20}{30}=0.33$ and that represent the 33%

b) $p_x =\frac{\lambda}{\mu}=\frac{20}{30}=0.66$

c) $L_s =\frac{20}{30-20}=\frac{20}{10}=2 people$

d) $L_q =\frac{20^2}{30(30-20)}=1.333 people$

e) $W_s =\frac{1}{\lambda -\mu}=\frac{1}{30-20}=0.1hours$

f) $W_q =\frac{\lambda}{\mu(\mu -\lambda)}=\frac{20}{30(30-20)}=0.0667 hours$

Step-by-step explanation:

Notation

P represent the probability that the employee is idle

$p_x$ represent the probability that the employee is busy

$L_s$ represent the average number of people receiving and waiting to receive some information

$L_q$ represent the average number of people waiting in line to get some information

$W_s$ represent the average time a person seeking information spends in the system

$W_q$ represent the expected time a person spends just waiting in line to have a question answered

This an special case of Single channel model

Single Channel Queuing Model. "That division of service channels happen in regards to number of servers that are present at each of the queues that are formed. Poisson distribution determines the number of arrivals on a per unit time basis, where mean arrival rate is denoted by λ".

Part a

Find the probability that the employee is idle

The probability on this case is given by:

In order to find the mean we can do this:

$\mu = \frac{1question}{2minutes}\frac{60minutes}{1hr}=\frac{30 question}{hr}$

And in order to find the probability we can do this:

$P=1-\frac{\lambda}{\mu}=1-\frac{20}{30}=0.33$ and that represent the 33%

Part b

Find the proportion of the time that the employee is busy

This proportion is given by:

$p_x =\frac{\lambda}{\mu}=\frac{20}{30}=0.66$

Part c

Find the average number of people receiving and waiting to receive some information

In order to find this average we can use this formula:

$L_s= \frac{\lambda}{\lambda -\mu}$

And replacing we got:

$L_s =\frac{20}{30-20}=\frac{20}{10}=2 people$

Part d

Find the average number of people waiting in line to get some information.

For the number of people wiating we can us ethe following formula"

$L_q =\frac{\lambda^2}{\mu(\mu-\lambda)}$

And replacing we got this:

$L_q =\frac{20^2}{30(30-20)}=1.333 people$

Part e

Find the average time a person seeking information spends in the system

For this average we can use the following formula:

$W_s =\frac{1}{\lambda -\mu}=\frac{1}{30-20}=0.1hours$

Part f

Find the expected time a person spends just waiting in line to have a question answered (time in the queue).

For this case the waiting time to answer a question we can use this formula:

$W_q =\frac{\lambda}{\mu(\mu -\lambda)}=\frac{20}{30(30-20)}=0.0667 hours$

6 0

3 years ago

Read 2 more answers

Find three consecutive even integers. Such that 7 times of first integer is 4 more than the sum of second and third integers

attashe74 [19]

Answer:

1st even integer = 2

2nd even integer = 4

3rd even integer = 6

Step-by-step explanation:

Let the consecutive even integers be:

1st = 2(x)

2nd= 2(x +1) = 2x + 2

3rd = 2(x + 2) = 2x + 4

According to Given conditions:

7(2x) = 4 + 2x +2 + 2x + 4

By Simplifying:

14x = 10 + 4x

Subtracting 4x from both sides

14x - 4x = 10 + 4x -4x

10x = 10

Dividing both sides by 10 we get:

x = 1

Now putting value of x in supposed integers:

1st even integer = 2(1) = 2

2nd even integer = 2(1)+2 = 4

3rd even integer = 2(1) + 4 = 6

I hope it will help you!

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3 years ago

Lucy rolls 2 fair dice and adds the results from each. Work out the probability of getting a total that is a multiple of 5

Sergio039 [100]

Answer:

(1,4) ( 4,1) (2,3) (3,2)

Step-by-step explanation:

Multiples of 5 = 5 and 10

Ways to roll a 5 = (1,4) ( 4,1) (2,3) (3,2)

Ways to roll a 10 = (5,5) (4,6) (6,4)

We have 7 possible outcomes for a multiple of 5...and there are 36 possible outcomes in total

So P (multiple of 5) = 7 / 36

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3 years ago

If the terminal side of angle θ , in standard position, passes through the point (16,63)

Fantom [35]

Answer:

Step-by-step explanation:

sinθ = 63/√(16² + 63²) = 63/65

cosθ = 16/65

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2 years ago

4. At $2.329 per gallon for regular

rosijanka [135]

Answer:

$20.47

hope that's right.

7 0

3 years ago

Read 2 more answers