1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Sedaia [141]
4 years ago
12

A rectangular pool is surrounded by a walk 4 meters wide. The pool is 5 meters longer than its width. If the total area of the p

ool and walk is 488 square meters more than the area of the​ pool, find the dimensions of the pool. The width of the pool is ____m and the length of the pool is____ m.
Mathematics
2 answers:
larisa [96]4 years ago
6 0

Answer:

45 and 55

Step-by-step explanation:

emmasim [6.3K]4 years ago
4 0

Answer: The width of the pool is 24m and the length of the pool is 29m

Step-by-step explanation: The dimensions of the rectangular pool is not given, but the clues we have are that the length of the pool is 5 meters longer than it’s width. With that information, we can say that the width is W, while the length is W + 5.

Also the pool is surrounded by a walk 4m wide. That means the total length including the two walkways is W + 5 + 4 + 4 (that equals W + 13), and the total width including the two walkways is W + 4 + 4 (that equals W + 8). We have yet another clue, which states that the total area of the pool and walk is 488m more than the area of the pool.

We shall label the area of the pool as AP. If Area = L x W, then area of pool is

AP = (W + 5) x W

AP = W^2 + 5W

Also the total area of the pool and walk shall be labeled as APW. Therefore,

APW = (W + 13) x (W + 8)

APW = W^2 + 8W + 13W + 104

APW = W^2 + 21W + 104

Having calculated the area of the pool as W^2 + 5W and the total area of the pool and walk as W^2 + 21W + 104, and remembering that the area of pool and walk is 488 meters more than the area of the pool, we can now write the following equation;

W^2 + 21W + 104 = W^2 + 5W + 488

Collecting like terms we now have

W^2 - W^2 + 21W - 5W = 488 - 104

16W = 384

Divide both sides of the equation by 16

W = 24.

If the length of the pool has been given as 5 meters more than its width, then length of the pool equals 24 + 5 and that equals 29 meters. Therefore

Length = 29m and Width = 24m

You might be interested in
The service life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of 10 batteries
Nat2105 [25]

Answer:

(1) We are conducting the one-sample t-test and will be testing for the hypothesis of mean battery greater than 25 h. So, we will reject the null hypothesis for mean battery life equal to 25 hr against the alternative hypothesis for mean battery life greater than 25 hr.

(2) A 95% lower confidence interval on mean battery life is 25.06 hr.

Step-by-step explanation:

We are given that a random sample of 10 batteries is subjected to an accelerated life test by running them continuously at an elevated temperature until failure, and the following lifetimes (in hours) are obtained: 25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                              P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~  t_n_-_1

where, \bar X = sample mean BAC = \frac{\sum X}{n} = 26 hr

             s = sample standard deviation = \sqrt{\frac{\sum(X - \bar X)^{2} }{n-1} } = 1.625 hr

             n = sample of batteries = 10

             \mu = population mean battery life

<em> Here for constructing a 95% lower confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

(1) It is stated that the manufacturer wants to be certain that the mean battery life exceeds 25 h.

Since we are conducting the one-sample t-test and will be testing for the hypothesis of mean battery greater than 25 h. So, we will reject the null hypothesis for mean battery life equal to 25 hr against the alternative hypothesis for mean battery life greater than 25 hr.

(2) So, a 95% lower confidence interval for the population mean, \mu is;

P(-1.833 < t_9) = 0.95  {As the lower critical value of t at 9

                                             degrees of  freedom is -1.833 with P = 5%}    

P(-1.833 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }) = 0.95

P( -1.833 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} ) = 0.95

P( \bar X-1.833 \times {\frac{s}{\sqrt{n} } } < \mu) = 0.95

<u>95% lower confidence interval for</u> \mu = [ \bar X-1.833 \times {\frac{s}{\sqrt{n} } } ]

                                                             = [ 26-1.833 \times {\frac{1.625}{\sqrt{10} } } ]

                                                            = [25.06 hr]

Therefore, a 95% lower confidence interval on mean battery life is 25.06 hr.

5 0
3 years ago
I will give brainliest!!!
Cloud [144]

Answer:

1) 36 square units.

2) 190 square units.

Step-by-step explanation:

As you can see, the figures are trapezoids, therefore, you can calculate the area of each one of them with the following formula:

A=(\frac{B+b}{2})h

Where:

<em>B</em> is the longer base,<em> b</em> is the shorter base and <em>h</em> is the height.

Then:

1) B=4units+(8units-2units)=4units+6units=10units\\b=8units\\h=4units

A=(\frac{10units+8units}{2})(4units)=36units^{2}

2) B=3units+15units+5units=23units\\b=15units\\h=10units

A=(\frac{23units+15units}{2})(10units)=190units^{2}


3 0
4 years ago
!! HELP ASAP !!
kherson [118]
Can you provide a picture?
6 0
3 years ago
For y = 4x + 2x2 – x3, show how to find the value for y, when x = 2. Show your work
shusha [124]
To find the value of y in the given equation, substitute the given value of x to the equation.
   
                               y = 4x + 2x² -x³

Substitute 2 for all x's.
  
                               y = 4(2) + 2(2²) - 2³ = 8

Therefore, the value of y is 8. 


6 0
3 years ago
How are terminating decimals different from repeating decimals
Nady [450]
Hi. The difference between terminating decimals and different from repeating decimals is:

Terminating decimals is a decimal that ends. It is a decimal with an infinite number of digits.

Repeating decimals have a digit or a block of digits that repeat over and over again.

Hope this helps.

Take care,
Diana
8 0
4 years ago
Other questions:
  • How many times greater is 1,000,000,000 than 1,000,000?that is how many groups of 1 million are there in 1 billion? Please help.
    8·2 answers
  • 1
    14·1 answer
  • The Giant Machinery has the current capital structure of 65% equity and 35% debt. Its net income in the current year is $250,000
    9·1 answer
  • The rule which allows us to change the sign of an exponent is called the ____ rule.
    5·1 answer
  • Solve this system of linear equations. Separate
    8·1 answer
  • What is the volume of the rectangular prism? A prism has a length of 8 inches, width of 2 inches, and height of 12 and one-half
    15·2 answers
  • What property is used to show that 6(x + 7) = 6x + 42?
    6·1 answer
  • Mrs. Bush uses 9 gallons of gas per day. If she drove for 16 days and went 2,592 miles, how many milos is she able to go on 1 ga
    14·1 answer
  • Prove this identify : 2 cos^2 (45°-A ) = 1 +sin 2 A​
    5·1 answer
  • Write an equation of the line that passes through the point (2,3) and is perpendicular to the line x+4y-8
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!