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daser333 [38]
3 years ago
13

The width of a rectangular pool is 5 meters longer than the length. The perimeter of the pool is 58 meters. Use and equation to

find the pools length.
Mathematics
1 answer:
garri49 [273]3 years ago
8 0
The perimeter of a rectangular pool = 2l + 2w

If the width is 5m longer than the length, 

Then Perimeter = 2(l + 5) + 2l
                          = 4l + 10  {THE EQUATION}

Since Perimeter = 58 m
Then 4l + 10 = 58 m

⇒  l  =  12 m

Thus the Length of the Pool = 12 m
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You invest $2,500 in an account that grows 5% each year. what will be your investment amount after 6 years?
lesantik [10]

Answer:  b. $3,374.65

Step-by-step explanation:

The exponential equation of growth (continuously) is given by :-

y=Ae^{rx}, where A is the initial amount, r is the rate of growth ( in decimal) and x is the time period.

Given : You invest $2,500 in an account that grows 5% each year.

i.e.  A= $2,500  and r= 5%=0.05

Then, the equation model this situation will be :-

y=2500e^{0.05x}=2500(1.05)^x

Now, At x= 6

y=2500e^{0.05\times6}\\\\\Rightarrow\ y=2500e^{0.3}=3374.64701894\approx3374.65

Hence, the investment amount after 6 years will be $3,374.65.

4 0
3 years ago
If 6 cost $60, how much do 8 cost​
madreJ [45]

Answer:

80

Step-by-step explanation:

8 0
3 years ago
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4 0
3 years ago
A particular sale involves four items randomly selected from a large lot that is known to contain 9% defectives. Let X denote th
umka2103 [35]

Answer:

The expected repair cost is $3.73.

Step-by-step explanation:

The random variable <em>X</em> is defined as the number of defectives among the 4 items sold.

The probability of a large lot of items containing defectives is, <em>p</em> = 0.09.

An item is defective irrespective of the others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 9 and <em>p</em> = 0.09.

The repair cost of the item is given by:

C=3X^{2}+X+2

Compute the expected cost of repair as follows:

E(C)=E(3X^{2}+X+2)

        =3E(X^{2})+E(X)+2

Compute the expected value of <em>X</em> as follows:

E(X)=np

         =4\times 0.09\\=0.36

The expected value of <em>X</em> is 0.36.

Compute the variance of <em>X</em> as follows:

V(X)=np(1-p)

         =4\times 0.09\times 0.91\\=0.3276\\

The variance of <em>X</em> is 0.3276.

The variance can also be computed using the formula:

V(X)=E(Y^{2})-(E(Y))^{2}

Then the formula of E(Y^{2}) is:

E(Y^{2})=V(X)+(E(Y))^{2}

Compute the value of E(Y^{2}) as follows:

E(Y^{2})=V(X)+(E(Y))^{2}

          =0.3276+(0.36)^{2}\\=0.4572

The expected repair cost is:

E(C)=3E(X^{2})+E(X)+2

         =(3\times 0.4572)+0.36+2\\=3.7316\\\approx 3.73

Thus, the expected repair cost is $3.73.

4 0
3 years ago
3 less than the quotient of a number y and 4
mr Goodwill [35]

3 less than the quotient of a number y and 4

3       <        y / 4

3    <     y/4

x 4          x 4

12 < y

6 0
3 years ago
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