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2 years ago

A swimming pool at a park measures 9.75 meters by 7.2 meters.

Mathematics

1 answer:

neonofarm [45]2 years ago

6 0

Answer:

Part a) The area of the swimming pool is $70.2\ m^2$

Part b) The total area of the swimming pool and the playground is $105.3\ m^2$

Step-by-step explanation:

Part a) Find the area of the swimming pool

we know that

The area of the swimming pool is

$A=LW$

where

L is the length side

W is the width side

we have

$L=9.75\ m\\W=7.2\ m$

substitute the values

$A=(9.75)(7.2)$

$A=70.2\ m^2$

therefore

The area of the swimming pool is $70.2\ m^2$

Part b) The area of the playground is one and a half times that of the swimming pool. Find the total area of the swimming pool and the playground

we know that

To obtain the area of the playground multiply the area of the swimming pool by one and a half

$\frac{1}{2}(70.2)=35.1\ m^2$

To obtain the total area of the swimming pool and the playground, adds the area of the swimming pool and the area of the playground

$70.2\ m^2+35.1\ m^2=105.3\ m^2$

therefore

The total area of the swimming pool and the playground is $105.3\ m^2$

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The number of flaws in a fiber optic cable follows a Poisson distribution. It is known that the mean number of flaws in 50m of c

boyakko [2]

Answer:

(a) The probability of exactly three flaws in 150 m of cable is 0.21246

(b) The probability of at least two flaws in 100m of cable is 0.69155

(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is 0.13063

Step-by-step explanation:

A random variable X has a Poisson distribution and it is referred to as Poisson random variable if and only if its probability distribution is given by

$p(x;\lambda)=\frac{\lambda e^{-\lambda}}{x!}$ for x = 0, 1, 2, ...

where $\lambda$ , the mean number of successes.

(a) To find the probability of exactly three flaws in 150 m of cable, we first need to find the mean number of flaws in 150 m, we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 150 m of cable is $1.2 \cdot 3 =3.6$

The probability of exactly three flaws in 150 m of cable is

$P(X=3)=p(3;3.6)=\frac{3.6^3e^{-3.6}}{3!} \approx 0.21246$

(b) The probability of at least two flaws in 100m of cable is,

we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 100 m of cable is $1.2 \cdot 2 =2.4$

$P(X\geq 2)=1-P(X$

$P(X\geq 2)=1-p(0;2.4)-p(1;2.4)\\\\P(X\geq 2)=1-\frac{2.4^0e^{-2.4}}{0!}-\frac{2.4^1e^{-2.4}}{1!}\\\\P(X\geq 2)\approx 0.69155$

(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is

$P(X=1)=p(1;1.2)=\frac{1.2^1e^{-1.2}}{1!}\\P(X=1)\approx 0.36143$

The occurrence of flaws in the first and second 50 m of cable are independent events. Therefore the probability of exactly one flaw in the first 50 m and exactly one flaw in the second 50 m is

$(0.36143)(0.36143) = 0.13063$

4 0

2 years ago

(a) Suppose one house from the city will be selected at random. Use the histogram to estimate the probability that the selected

vodka [1.7K]

Answer:

a. 0.71

b. 0.9863

Step-by-step explanation:

a. From the histogram, the relative frequency of houses with a value less than 500,000 is 0.34 and 0.37

-#The probability can therefore be calculated as:

$P(500000)=P(x=0)+P(x=500)\\\\\\\\=0.34+0.37\\\\\\\\=0.71$

Hence, the probability of the house value being less than 500,000 is o.71

-From the info provided, we calculate the mean=403 and the standard deviation is 278 The probability that the mean value of a sample of n=40 is less than 500000 can be calculated as below:

$P(\bar X$

Hence, the probability that the mean value of 40 randomly selected houses is less than 500,000 is 0.9863

8 0

2 years ago

-6y + 3(12y) = 20(y - 1) + 15 is it true?

xxMikexx [17]

Answer:

y= - 1/2 (negative half) = -0.5

Step-by-step explanation:

−6y+3(12y)=20(y−1)+15

Multiply 3 and 12 to get 36.

−6y+36y=20(y−1)+15

Combine −6y and 36y to get 30y.

30y=20(y−1)+15

Use the distributive property to multiply 20 by y−1.

30y=20y−20+15

Add −20 and 15 to get −5.

30y=20y−5

Subtract 20y from both sides.

30y−20y=−5

Combine 30y and −20y to get 10y.

10y=−5

Divide both sides by 10

y= -5/10

Reduce the fraction -5/10 = -0.5 to lowest terms by extracting and cancelling out 5 .

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2 years ago

Calculate the standard deviation. Variance = 107.76. Mean = 34.2

Llana [10]

Answer:

The standard deviation is 10.38

Step-by-step explanation:

Given;

Variance, v = 107.76

mean, x = 34.2

The standard deviation is given by;

standard deviation = √variance

standard deviation = √107.76

standard deviation = 10.381

standard deviation = 10.38 (two decimal places)

Therefore, the standard deviation is 10.38

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2 years ago

The service time distribution describes the probability P that the service time of the customer will be no more than T hours. If

Molodets [167]

Answer: option d.

Step-by-step explanation:

You have the following formul given in the problem:

$p=1-e^{-mt}$

You know that:

The number of customers serviced in an hour by the technical support representative is 6 costumbers, therefore:

$m=6$

As the problem asked for the probability that a costumber will be on hold less than 30 minutes, we know that:

$t=0.5$

Substitute the values above into the formula.

Then, you obtain:

$p=1-e^{-(6)(0.5)}=0.95$ or 95%

7 0

2 years ago