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Suppose a rectangular plot is to be fenced in, with the fencing on each of the four sides (labeled east, west, north, south) cos

Damm [24]

Answer:

A = 1701,38 ft²

Dimensions :

x (north and south sides ) = 38.89 ft

y ( east and west sides ) = 43,75 ft

Step-by-step explanation:

North and south (sides of same length) equal "y" cost (4 + 5 ) = 4,5 $/ft²

East and west (sides of same length) equal "x" cost ( 3 + 5 ) = 4 $ /ft²

Equation of cost is

C = Cost of (north + south ) + Cost (east + west)

C = 2 * 4,5 * x + 4*2* y

C = 9x + 8y

700 = 9x + 8y ⇒ y = ( 700- 9x)/ 8

A = x*y

A(x) = x * ( 700 - 9x ) /8

A(x) = ( 700 x -9x²) / 8 A´(x) = ( 700 - 18 x )/ 8 A´(x) = 0

( 700 - 18 x )/ 8 = 0 ⇒ 700 - 18 x = 0 ⇒ x = 700/18

x = 38.89 ft

y = ( 700 - 9x )/8 ⇒ y = 349.99 / 8 ⇒ y = 43.75

And maximum ara is

A = x*y A = 38.89 * 43.75 = 1701,38 ft²

5 0

4 years ago

What Is m∠KNL?????????????????????

Sloan [31]

U have some vertical angles....vertical angles are equal

5x + 2 = 3(x + 14)
5x + 2 = 3x + 42
5x - 3x = 42 - 2
2x = 40
x = 40/2
x = 20

JNM = 5x + 2.....5(20) + 2 = 102
KNL = 3(x + 14).....3(20 + 14)....3(34) = 102 <===

7 0

3 years ago

Math question geometry please hlep

Mashcka [7]

Answer:

A) True

Step-by-step explanation:

A seems like the right answer because the line is equal all the way up.

Hope this helps, have a great day/night!

4 0

4 years ago

Read 2 more answers

Examine the required sample size needed to be able to

Xelga [282]

Answer:

$n=(\frac{2.33(500)}{10})^2 =13572.25 \approx 13573$

And if we use a sample level of n =2000 the margin of error would be higher as we can see here:

$ME=2.33\frac{500}{\sqrt{2000}}=26.05$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=500$ represent the population standard deviation assumed

n represent the sample size (variable of interest)

Confidence =98% or 0.98

ME = 10 represent the margin of error desired

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =10 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 98% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.01;0;1)", and we got $z_{\alpha/2}=2.33$ , replacing into formula (b) we got:

$n=(\frac{2.33(500)}{10})^2 =13572.25 \approx 13573$

So the answer for this case would be n=13573 rounded up to the nearest integer

And if we use a sample level of n =2000 the margin of error would be higher as we can see here:

$ME=2.33\frac{500}{\sqrt{2000}}=26.05$

3 0

4 years ago

What is the rotational symmetry of a rectangle

Fudgin [204]

Order 2 rectangle has only 2

6 0

4 years ago