A cylindrical paint can has the capcity of one gallon. for another size can, the height is doubled what is the capacity of the larger size?

Answers

Answer 1

Answer: To understand the answer you will need to know that when you change any number of dimensions, the outcome will be affect in the same way based on the number of dimensions that were changed and what measurement you need. A three dimensional measurement such as volume will only be affect by how many of those 3 dimensions are actually changing. In this example, there is only 1 dimension changing (the height), so the total volume will be affected in the same way, only one time doubled.

1 gallons x 2 = 2 gallons.

The capacity will be 2 gallons.

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How many times does 4 go into 48?

Answers

Answer:

Number of times 4 is gone through to get 48 = 12

Step-by-step explanation:

The number which is formed after the product is given to be 48

Now, we need to find how many times 4 is gone through to get the final result as 48

4 × 1 = 4

4 × 2 = 8

4 × 3 = 12

4 × 4 = 16

4 × 5 = 20

4 × 6 = 24

4 × 7 = 28

4 × 8 = 32

4 × 9 = 36

4 × 10 = 40

4 × 11 = 44

4 × 12 = 48

Hence, Number of times 4 is gone through to get 48 = 12

12, because 48 divided by 4 is 12

A swimming pool can be filled using either a pipe, a hose, or both. Using the pipe alone takes 12 hours. Using both it takes 8 $(4)/(7)$ hours. How long does it take using the hose alone?

Answers

$(1)/(x)+(1)/(12)=(1)/(8(4)/(7))\n(1)/(x)=(1)/((60)/(7))-(1)/(12)\n(1)/(x)=(7)/(60)-(1)/(12)\n(1)/(x)=(7)/(60)-(5)/(60)\n(1)/(x)=(2)/(60)\n(1)/(x)=(1)/(30)\nx=30$

30 hrs

Final answer:

The time it would take to fill the pool using the hose alone is 15 hours. This was calculated using the information given on the time it takes to fill the pool using the hose and the pipe together, and the time it takes to fill the pool using just the pipe, to find the individual rates of filling and then solve for the time for the hose alone.

Explanation:

The subject of this problem is rates, specifically the rate at which the pool is being filled with water. Let's denote the rate of the pool filling with the pipe as P, and the rate of the pool filling with the hose as H. We know that the pool fills in 12 hours with only P, so P = 1/12 pools/hour.

Additionally, we know that it takes 8 4/7 hours to fill the pool with both P and H. So, the rate of P + H = 1/(8 4/7) pools/hour, which simplifies to 7/60.

Given that we know P, we can solve for H by subtracting P from the combined rate. H = P + H - P = 7/60 - 1/12 = 1/15 pools/hour. Therefore, to fill the pool using the hose alone, it will take 1/H = 1/(1/15) = 15 hours.

Learn more about Rates here:

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Can someone help me with this

Answers

treat each side as if it's their own equation.
14-2*2
First multiply
14-4
Now subtract
10/5
Now divide
2
2 is the answer

calculate the volume of the cylinder . Round your answer to the nearest hundredth . How do I get the answer ?

Answers

Multiply the are of the circle face by the height. To get the are of the face it is πr squared. Round this to 2 decimal places

πr²
3 place decimal
this is my answer

#5.Chose one of two tables below to create your own Question (with solution).

My Conditional Frequency Question is:

My solution (work and answer):

Please explain how/why you chose this question

Answers

Therefore , the solution of the given problem of unitary method comes out to be the likelihood that a particular child has a curfew given that they have tasks is 3/7, or roughly 0.43.

What is an unitary method?

The job can be completed by bringing together what was learned and applying this variable technique, that also includes all supplementary data from two people that utilized a specific tactic. To put it another way, if the desired outcome materialises, either the entity stated in the calculation will be recognised, or both expression essential processes will truly skip the colour. For forty pencils, a refundable charge of Rupees ($1.01) might be required.

Here,

Using the following formula, one can determine the conditional chance that a child will have a curfew if they have chores:

Curfew and duties are equal, so

=>P(curfew | chores) = P (chores)

The odds are listed in the table below:

=> Curfew and errands P = 3/10

=> P(tasks) = 7/9

Therefore,

=> P(curfew | chores)=3/10/ (7/10)=3/7

Therefore, the likelihood that a particular child has a curfew given that they have tasks is 3/7, or roughly 0.43.

To know more about unitary method visit:

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