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Assoli18 [71]
3 years ago
12

Write an algebraic expression for the perimeter of a rectangular garden with a width of 3 feet and a length of 2y feet. Simplify

your expression and find the perimeter of the garden if y=6. Draw a model and show your work.
Mathematics
1 answer:
frutty [35]3 years ago
8 0

Answer:

i. P=4y+6

ii. P=30\:ft

Step-by-step explanation:

The given rectangular garden has width, w=3ft and length, l=2y\:ft.

The perimeter of a rectangle is calculated using the formula;

P=2l+2w

We substitute the given dimensions into the formula to get;

P=2(2y)+2(3)

\Rightarrow P=4y+6

We substitute;

y=6 to obtain;

\Rightarrow P=4(6)+6

Simplify;

\Rightarrow P=24+6

\Rightarrow P=30ft

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Step-by-step explanation:

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16 + 8 + 4= 28  

4 0
3 years ago
PLEASE HELP ME ASAP!!!!<br> The top says "Coffee cost $18.96 for 3 pounds."
mezya [45]
I think you divide 18.96 by 3 to get the amount for one pound cost then go pound for pound there
4 0
3 years ago
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Can someone please explain how to do this multi step problem? Annie and Dustin took a beginner's programming course over several
Anarel [89]
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3 0
2 years ago
Can someone please help me find the side length area, and please draw the shape on how it would look if it were finished
Mumz [18]

The side length is 5 units, and the area is 25 square units

<h3>What are areas?</h3>

The area of a shape is the amount of space it occupies

<h3>What is the area of a square</h3>

The area of a square is the product of its dimensions

From the graph, the coordinates of the square can be assumed to be:

A = (0,0)

B = (4,3)

The side length of the square is then calculated using the following distance formula

d = \sqrt{(x_2- x_1)^2 + (y_2 -y_1)^2}

So, we have:

AB = \sqrt{(0 - 4)^2 + (0-3)^2}

AB = \sqrt{25}

AB = 5

The area of the square is the calculated as:

Area = AB^2

This gives

Area = 5^2

Area = 25

Hence, the side length is 5 units, and the area is 25 square units

See attachment for the diagram of the shape

Read more about areas at:

brainly.com/question/24487155

6 0
2 years ago
onsider the following hypothesis test: H 0: 50 H a: &gt; 50 A sample of 50 is used and the population standard deviation is 6. U
kondaur [170]

Answer:

a) z(e)  >  z(c)   2.94 > 1.64  we are in the rejection zone for H₀  we can conclude sample mean is great than 50. We don´t know how big is the population .We can not conclude population mean is greater than 50

b) z(e) < z(c)  1.18 < 1.64  we are in the acceptance region for   H₀  we can conclude H₀ should be true. we can conclude population mean is 50

c) 2.12  > 1.64 and we can conclude the same as in case a

Step-by-step explanation:

The problem is concerning test hypothesis on one tail (the right one)

The critical point  z(c) ;  α = 0.05  fom z table w get   z(c) = 1.64 we need to compare values (between z(c)  and z(e) )

The test hypothesis is:  

a) H₀      ⇒      μ₀  = 50     a)  Hₐ    μ > 50   ;    for value 52.5

                                          b) Hₐ    μ > 50   ;     for value 51

                                          c) Hₐ    μ > 50   ;      for value 51.8

With value 52.5

The test statistic    z(e)  ??

a)  z(e) =  ( μ  -  μ₀ ) /( σ/√50)      z(e) = (2.5*√50 )/6   z(e) = 2.94

2.94 > 1.64  we are in the rejected zone for H₀  we can conclude sample mean is great than 50. We don´t know how big is the population .We can not conclude population mean is greater than 50

b) With value 51

z(e) =  ( μ  -  μ₀ ) /( σ/√50)    ⇒  z(e) =  √50/6    ⇒  z(e) = 1.18

z(e) < z(c)  we are in the acceptance region for   H₀  we can conclude H₀ should be true. we can conclude population mean is 50

c) the value 51.8

z(e)  =  ( μ  -  μ₀ ) /( σ/√50)    ⇒ z(e)  = (1.8*√50)/ 6   ⇒ z(e) = 2.12

2.12  > 1.64 and we can conclude the same as in case a

8 0
3 years ago
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