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3 years ago

After the strand of measurement why do you think that graphing and data analysis are easier? Please help ASAP!!! :(

Mathematics

1 answer:

Vikki [24]3 years ago

8 0

It’s organized and you can see it more visually , gives you a way to double check and write down the data properly.

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3 years ago

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James's garden has a length of 12 1/4 feet and a width of 9 2/3 feet. What length of fencing will he need to surround his garden

Lina20 [59]

43 and 5/6 because 12 and 1/4*2 is 24 and 1/2 then 9 and 2/3*2 is 19 and 1/3 and when you add that you get 43 and 5/6

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4 years ago

A rectangle box with a square base and no top needs to be made using 300ft^2 of material. Find the dimensions of the box with th

Yuri [45]

The dimensions of the box are 10 ft and 5 ft

The maximum volume is 500 ft³

Step-by-step explanation:

A rectangle box with

A square base and no top
It needs to be made using 300 ft² of material
It has greatest volume

Surface area of a box without top (SA) = perimeter of base × height + area of the base

Volume of a box (V) = base area × height

Assume that the length of the side of the square base is x and the height of the box is y

∵ It needs to be made using 300 ft² of material

∴ The surface area of the box is 300 ft²

∵ Its base is a square of side length x ft

∴ Perimeter of the base = 4 × x = 4 x

∴ Area of the base = x²

∵ The height of the box = y ft

∵ SA = perimeter of base × height + area of the base

∵ SA = (4x)(y) + x²

∴ SA = 4xy + x²

∵ SA of the box = 300 ft²

- Equate the two expressions of SA

∴ 4xy + x² = 300

Now let us find y in terms of x

- Subtract x² from both sides

∴ 4xy = 300 - x²

- Divide each term by 4x to find y

∴ $y=\frac{75}{x}-\frac{1}{4}x$

∵ V = area of the base × height

∴ V = x² × y = x²y

- Substitute y by the equation of it above

∴ $V=x^{2}(\frac{75}{x}-\frac{1}{4}x)$

∴ $V=75x-\frac{1}{4}x^{3}$

∵ The volume of the box is greatest

- That means differentiate V and equate it by 0

∵ $\frac{dV}{dx}=75-\frac{3}{4}x^{2}$

∵ $\frac{dV}{dx}=0$ ⇒ greatest volume

∴ $75-\frac{3}{4}x^{2}=0$

- Subtract 75 from both sides

∴ $-\frac{3}{4}x^{2}=-75$

- Divide both sides by $-\frac{3}{4}$

∴ x² = 100

- Take √ for both sides

∴ x = 10

Substitute the value of x in the equation of y

∵ $y=\frac{75}{10}-\frac{1}{4}(10)$

∴ y = 5

The dimensions of the box are 10 ft and 5 ft

∵ $V=75x-\frac{1}{4}x^{3}$

∵ x = 10

∴ $V=75(10)-\frac{1}{4}(10)^{3}$

∴ $V=750-\frac{1}{4}(1000)$

∴ V = 750 - 250

∴ V = 500 ft³

The maximum volume is 500 ft³

Learn more:

You can learn more about the volume in brainly.com/question/6443737

#LearnwithBrainly

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3 years ago

Is 0.9727 a positive correlation, negative correlation, or no correlation?

lapo4ka [179]

I think it’s positive correlation

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3 years ago

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