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3 years ago

Nathan has a rectangular plot of grass in his backyard that has dimensions of meters in width and meters in length.

Mathematics

1 answer:

NNADVOKAT [17]3 years ago

6 0

Answer:

use the area of a rectangle to calculate for the total area of the rectangular grass plot

area = length × width

Step-by-step explanation:

The grass plot in Nathan's backyard has a rectangular shape. The dimensions of the sides are in meters . The width and the length are in metres . Since the grass plot is rectangular , the area of the plot can be computed using the formula for area of a rectangle.

Area of a rectangle = L × W

where

L = length

W = width

The area of a rectangle is the width multiplied by the length. For example if the width of the grass plot is 100 meters and the length is 200 meters .The area of the grass rectangular plot will be

area = L × W

area = 200 × 100

area = 20000 m²

The unit of the area will be in square meters (m²)

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2 years ago

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Estimate the total cost that each friend will have to pay for gas and hotels. Explain how you got your answer. Here are some figures that may help you out:

•Tony's car can travel 28 miles for each gallon of gas.

•The average fuel cost at the time of their trip is $3 per gallon.

•They plan to drive about 650 miles each day.

•They estimate the average cost of a hotel each night is $85.

•They will drive approximately 2,240 miles to get from Albuquerque to Boston.

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3 years ago

The number of defective circuit boards coming off a soldering machine follows a Poisson distribution. During a specific ten-hour

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Answer:

a) the probability that the defective board was produced during the first hour of operation is $\frac{1}{10}$ or 0.1000

b) the probability that the defective board was produced during the last hour of operation is $\frac{1}{10}$ or 0.1000

c) the required probability is 0.2000

Step-by-step explanation:

Given the data in the question;

During a specific ten-hour period, one defective circuit board was found.

Lets X represent the number of defective circuit boards coming out of the machine , following Poisson distribution on a particular 10-hours workday which one defective board was found.

Also let Y represent the event of producing one defective circuit board, Y is uniformly distributed over ( 0, 10 ) intervals.

f(y) = $\left \{ {{\frac{1}{b-a} }\\\ }} \right _0$ ; ( a ≤ y ≤ b ) $_{elsewhere$

= $\left \{ {{\frac{1}{10-0} }\\\ }} \right _0$ ; ( 0 ≤ y ≤ 10 ) $_{elsewhere$

f(y) = $\left \{ {{\frac{1}{10} }\\\ }} \right _0$ ; ( 0 ≤ y ≤ 10 ) $_{elsewhere$

Now,

a) the probability that it was produced during the first hour of operation during that period;

P( Y < 1 ) = $\int\limits^1_0 {f(y)} \, dy$

we substitute

= $\int\limits^1_0 {\frac{1}{10} } \, dy$

= $\frac{1}{10} [y]^1_0$

= $\frac{1}{10} [ 1 - 0 ]$

= $\frac{1}{10}$ or 0.1000

Therefore, the probability that the defective board was produced during the first hour of operation is $\frac{1}{10}$ or 0.1000

b) The probability that it was produced during the last hour of operation during that period.

P( Y > 9 ) = $\int\limits^{10}_9 {f(y)} \, dy$

we substitute

= $\int\limits^{10}_9 {\frac{1}{10} } \, dy$

= $\frac{1}{10} [y]^{10}_9$

= $\frac{1}{10} [ 10 - 9 ]$

= $\frac{1}{10}$ or 0.1000

Therefore, the probability that the defective board was produced during the last hour of operation is $\frac{1}{10}$ or 0.1000

no defective circuit boards were produced during the first five hours of operation.

probability that the defective board was manufactured during the sixth hour will be;

P( 5 < Y < 6 | Y > 5 ) = P[ ( 5 < Y < 6 ) ∩ ( Y > 5 ) ] / P( Y > 5 )

= P( 5 < Y < 6 ) / P( Y > 5 )

we substitute

$= (\int\limits^{6}_5 {\frac{1}{10} } \, dy) / (\int\limits^{10}_5 {\frac{1}{10} } \, dy)$

$= (\frac{1}{10} [y]^{6}_5) / (\frac{1}{10} [y]^{10}_5)$

= ( 6-5 ) / ( 10 - 5 )

= 0.2000

Therefore, the required probability is 0.2000

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3 years ago