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Firlakuza [10]
3 years ago
11

The owner of an office building is expanding the length and width of a parking lot by the same amount. The lot currently measure

s 120 ft by 80 ft, and the expansion will increase its area by 4,400 ft2. By how many feet should the length of the parking lot be increased? A = lw 1.2 f

Mathematics
2 answers:
Hatshy [7]3 years ago
6 0

Answer: The length of the parking lot should be increased 20 ft.

Please, see the attached files.

Thanks.

Arlecino [84]3 years ago
6 0

Answer:

b on edge

Step-by-step explanation:

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How long will it take you to double an amount of $500 if you invest it at a rate of 7% compounded monthly?   Round to 3 decimal
Lorico [155]

Answer:

9.931 years

Step-by-step explanation:

6 0
3 years ago
Let X and Y be two independent random variables following beta distributions Beta(120, 2020).
Alla [95]

With X,Y\sim\mathrm{Beta}(120,2020), we have identical PDFs

P(X=x)=\dfrac{x^{119}(1-x)^{2019}}{B(120,2020)}

for 0, and 0 otherwise, where

B(a,b)=\dfrac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}

Since X,Y are independent, the joint PDF is

P(X=x,Y=y)=P(X=x)P(Y=y)=\dfrac{(xy)^{119}((1-x)(1-y))^{2019}}{B(120,2020)^2}

for points (x,y) in the unit square, and 0 otherwise.

1. The distribution is continuous, so P(X=0.5)=\boxed0.

2. X+3 is the region in the x,y plane contained within the unit square and above the line y=\frac{x+3}2. This region is empty, because this line lies above the square altogether, so P(X+3.

3. X>Y is the region in the same square below the line y=x. So we have

P(X>Y)=\displaystyle\int_0^1\int_0^xP(X=x,Y=y)\,\mathrm dy\,\mathrm dx=\boxed{\frac12}

4 0
3 years ago
A water trough is 7 feet long, and its cross section is an equilateral triangle with sides 4 feet long. Water is pumped into the
Tju [1.3M]

Answer:

a. An equilateral triangle main characteristic is that all the sides lenght are the same (s). To find the height (h), we could divide the triangle (as seen in the picture) and apply Pytagorean theorem.

(\frac{s}{2}) ^{2} +h^{2}=s^{2}

Clearing the expression, we obtain: h= \frac{\sqrt{3}}{2}s

b. Knowing the rate at which the volume is changing 4 ft^{3}, we can find the relation between the change in the volume and the height.

V=A*h

As we want to express the volume in terms of the height, we have to find the area in terms of height

A=\frac{base*height}{2}

Base=s=\frac{2h}{\sqrt{3}}

Therefore, A=\frac{\frac{2h}{\sqrt{3}} *h}{2} =\frac{h^{2} }{\sqrt{3}}

V=7 ft*\frac{h^{2}}{\sqrt{3}}

Therefore the change in the volume with the height, will be the derivate of this expression

\frac{dV}{dt} =2*7*\frac{h}{\sqrt{3}} (\frac{dh}{dt}  )

Knowing dV/dt=4 cubic feet per second, and h=1/2 foot, we can know dh/dt

\frac{dh}{dt}=\frac{\frac{dV}{dt}*\sqrt{3}}{7*2*h} =\frac{4*\sqrt{3} }{14*\frac{1}{2} }=0.98 ft/sec

Step-by-step explanation:

7 0
4 years ago
Subtract and simplify.
yan [13]
2y^2+4y-9 should be the answer
4 0
4 years ago
Read 2 more answers
Helpppppppppppppppppp
Anuta_ua [19.1K]

Option A:

\left(x^{4}\right)\left(3 x^{3}-2\right)\left(4 x^{2}+5 x\right)=12 x^{9} +15 x^{8} -  8 x^{6}-10 x^{5}

Solution:

Given expression \left(x^{4}\right)\left(3 x^{3}-2\right)\left(4 x^{2}+5 x\right).

To find the product of the above expression:

\left(x^{4}\right)\left(3 x^{3}-2\right)\left(4 x^{2}+5 x\right)

First multiply first two factors with each term.

           =(x^{4} \times 3 x^{3}- x^{4} \times 2 ) \left(4 x^{2}+5 x\right)

Using exponent rule: a^m \cdot a^n=a^{m+n}

            =(3 x^{7}- 2x^{4} ) \left(4 x^{2}+5 x\right)

Now multiply these two factors with each term.

            =3 x^{7} (4 x^{2}+5 x)- 2x^{4} \left(4 x^{2}+5 x\right)

            =(4 x^{2} \times 3 x^{7} +5 x \times 3 x^{7} )-  \left(4 x^{2} \times 2x^{4}+5 x \times 2x^{4}\right)

Using exponent rule: a^m \cdot a^n=a^{m+n}

            =(12 x^{9} +15 x^{8} )-  (8 x^{6}+10 x^{5})

           =12 x^{9} +15 x^{8} -  8 x^{6}-10 x^{5}

\left(x^{4}\right)\left(3 x^{3}-2\right)\left(4 x^{2}+5 x\right)=12 x^{9} +15 x^{8} -  8 x^{6}-10 x^{5}

Hence option A is the correct answer.

4 0
4 years ago
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