‘write down £1 as a fraction of £8’

19. Bella is putting down patches of sod

Fofino [41]

Answer:

The dimensions of the two different rectangular regions are;

1st Arrangement:

W = 4 yards and L = 5 yards or W = 5 yards and L = 4 yards

2nd Arrangement:

W = 2 yards and L = 10 yards or W = 10 yards and L = 2 yards

The perimeter of the two different rectangular regions are;

1st Arrangement:

P₁ = 18 yards

2nd Arrangement:

P₂ = 24 yards

Step-by-step explanation:

Bella is putting down patches of sod to start a new lawn.

She has 20 square yards of sod.

We are asked to provide the dimensions of two different rectangular regions that she can cover with the sod.

Recall that a rectangle has an area given by

Area = W*L

Where W is the width of the rectangle and and L is the length of the rectangle.

Since Bella has 20 square yards of sod,

20 = W*L

There are more than two such possible rectangular arrangements.

Out of them, two different possible arrangements are;

1st Arrangement:

20 = (4)*(5) = (5)*(4)

Width is 4 yards and length is 5 yards or width is 5 yards and length is 4 yards

2nd Arrangement:

20 = (2)*(10) = (10)*(2)

Width is 2 yards and length is 10 yards or width is 10 yards and length is 2 yards

Therefore, the dimensions of two different rectangular regions are;

1st Arrangement:

W = 4 yards and L = 5 yards or W = 5 yards and L = 4 yards

2nd Arrangement:

W = 2 yards and L = 10 yards or W = 10 yards and L = 2 yards

What is the perimeter of each region?

The perimeter of a rectangular shape is given by

P = 2(W + L)

Where W is the width of the rectangle and and L is the length of the rectangle.

The perimeter of the 1st arrangement is

P₁ = 2(4 + 5)

P₁ = 2(9)

P₁ = 18 yards

The perimeter of the 2nd arrangement is

P₂ = 2(2 + 10)

P₂ = 2(12)

P₂ = 24 yards

So the perimeter of the 1st arrangement is 18 yards and the perimeter of the 2nd arrangement is 24 yards.

Note:

Another possible arrangement is,

20 = (1)*(20) = (20)*(1)

Width is 1 yard and length is 20 yards or width is 20 yards and length is 1 yard.

3 0

1 year ago

Last Saturday they were 1486 people at the Cineplex. There were about the same number of people in each of the six theaters. Bet

padilas [110]

When you do 1486x6 it equals 8,916

4 0

1 year ago

If T(n) = 2n-7, what is the fifth term?

charle [14.2K]

$T(5)=2\cdot5-7=3$

8 0

2 years ago

Alina is able to swim at a rate of about 2.5 miles per hour. Talia is able to swim at a rate of 1.8 meters per second. Which sta

Inessa05 [86]

Answer: A/ alina swims about 0.7 miles per hour

Step-by-step explanation:

7 0

2 years ago

During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

Brilliant_brown [7]

Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

A

The sample proportion is $\r p = 0.47$

B

$E =4.4 \%$

C

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

Ci

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

D

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

E

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

F

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

3 0

1 year ago