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san4es73 [151]
4 years ago
12

You just discovered that you have 100 feet of fencing and you have decided to make a rectangular garden. What is the largest are

a you can enclose using the materials you have? You must set up an equation and solve.
Mathematics
1 answer:
Rudik [331]4 years ago
8 0
The area is:
 A = x * y
 The perimeter is:
 P = 2x + 2y = 100
 We clear y:
 2y = 100-2x
 y = 50-x
 We write the area in terms of x:
 A (x) = x * (50-x)
 Rewriting:
 A (x) = 50x-x ^ 2
 Deriving:
 A '(x) = 50-2x
 We equal zero and clear x:
 50-2x = 0
 x = 50/2
 x = 25
 Then, the other dimension is given by:
 y = 50-x
 y = 50-25
 y = 25
 Therefore, the largest area is:
 A = (25) * (25)
 A = 625 feet ^ 2
 Answer:
 
the largest area you can enclose using the materials you have is:
 
A = 625 feet ^ 2
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mario62 [17]

x=2/9

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

−4−3x=6x−6

−4+−3x=6x+−6

−3x−4=6x−6

Step 2: Subtract 6x from both sides.

−3x−4−6x=6x−6−6x

−9x−4=−6

Step 3: Add 4 to both sides.

−9x−4+4=−6+4

−9x=−2

Step 4: Divide both sides by -9.

−9x

−9

=

−2

−9

6 0
3 years ago
Read 2 more answers
The person is a member for super for a female boss over in the and loss increased approximately linearly from 5% in 1974 to 9% i
natali 33 [55]

Answer:

1998

y=6x+1944

Step-by-step explanation:

The percentage of Male workers who prefer a female boss over a male boss increased approximately linearly from 5% in 1974 to 9% in 1998. Predict when 9% of male workers will prefer a female boss

it is explicit from the question that 9% of male workers prefer female boss in 1998. but we can predict a model for this by getting the slope of the graph

y=the year

x=the percentage of men who prefer a female boss

s=y2-y1/(x2-x1)

s=1998-1974/(9-5)

s=24/4

s=6

therefore we have

y=mx+c

y=6x+c........1

when y=1998,x=9

1998=6(9)+c

c=1944

from equation 1

y=6x+1944

8 0
3 years ago
The gypsy moth is a serious threat to oak and aspen trees. A state agriculture department places traps throughout the state to d
k0ka [10]
Part A:

From the central limit theorem, since the number of samples is large enough (up to 30), the mean of the the mean of the average number of moths in 30 traps is 0.6.



Part B:

The standard deviation is given by the population deviation divided by the square root of the sample size.

standard \ deviation= \frac{\sigma}{\sqrt{n}} = \frac{0.4}{\sqrt{30}} = \frac{0.4}{5.477} =0.073



Part C:

The probability that an approximately normally distributed data with a mean, μ, and the standard deviation, σ, with a sample size of n is greater than a number, x, given by

P(X\ \textgreater \ x)=1-P(X\ \textless \ x) \\  \\ =1-P\left(z\ \textless \  \frac{x-\mu}{\sigma/\sqrt{n}} \right)

Thus, given that the mean is 0.6 and the standard deviation is 0.4, the probability that <span>the average number of moths in 30 traps is greater than 0.7</span> given by:

P(X\ \textgreater \ 0.7)=1-P(X\ \textless \ 0.7) \\ \\ =1-P\left(z\ \textless \ \frac{0.7-0.6}{0.073} \right)=1-P(z\ \textless \ 1.369) \\  \\ =1-0.91455=\bold{0.0855}
7 0
4 years ago
A hypothesis test is conducted at the 5 percent level of significance to test whether the population correlation is zero. If the
Nonamiya [84]

Answer:

Which is the output of the formula =AND(12>6;6>3;3>9)?

A.

TRUE

B.

FALSE

C.

12

D.

9

Step-by-step explanation:

5 0
3 years ago
a shopkeeper sold goods for rs 2400 and made a profit of 25% in the process. find his profit per cent if he had sold his goods f
miv72 [106K]

case 1,

Let the CP be ₹x,

SP = ₹2400

Profit = SP – CP

= 2400 – x

Profit % = {(2400–x)/ x} × 100%

According to the question,

{(2400–x)/ x} × 100 = 25

=> (2400–x)/ x= 25 /100

=> 100(2400–x) = 25x [ cross multiplication]

=> 240000 – 100x = 25x

=> 240000 = 25x + 100x

=> 240000 = 125x

=> 240000/125 = x

=> x = 1920

So, CP = ₹1920

case 2,

SP = ₹2040

Profit = SP – CP

= 2040 – 1920

= ₹120

profit % = 120/1920 × 100%

= 16%

<h3>Thus, his profit would be 16% if he had sold his goods for ₹2040.</h3>
6 0
3 years ago
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