All categories

2 years ago

Find the area to the nearesy square foot of the shaded region below, consisting of a square with a circle cut out of it.

Mathematics

1 answer:

frutty [35]2 years ago

4 0

Based on the calculations, the area of the shaded region is equal to 22 square foot.

<h3>How to calculate the area of the shaded region?</h3>

By critically observing the geometric shapes, the area of the shaded region can be calculated by using this formula:

Area of shaded region = Area of square - Area of circle.

<u>Note:</u> The diameter of this circle is equal to the side length of the square because the circumference of this circle lies on the circumference of the square.

Radius = diameter/2 = 10/2 = 5 feet.

For the area of square, we have:

Area of square = S²

Area of square = 10²

Area of square = 100 square foot.

For the area of circle, we have:

Area of circle = πr²

Area of circle = 3.14 × 5²

Area of circle = 78.5 square foot.

Substituting the parameters into the formula, we have;

Area of shaded region = Area of square - Area of circle.

Area of shaded region = 100 - 78.5

Area of shaded region = 21.5 ≈ 22

Area of shaded region = 22 square foot.

Read more on area of circle here: brainly.com/question/2278809

#SPJ1

<h3>Complete Question:</h3>

Find the area to the nearest square foot of the shaded region below, consisting of a square with a circle cut out of it. Use 3.14 as an approximation for π.

You might be interested in

2.13m + 8.6 round to least precise measurement

kodGreya [7K]

The answer would be 10.19m. If you want to round it of to a less precise measurement it would be 10 m.

4 0

3 years ago

Read 2 more answers

I wasn’t at school when we learned this. Could someone he

Ksenya-84 [330]

Answer:

a= (1,4)

b=(4,2)

c=(2,6)

d=(2,-1)

Step-by-step explanation:

All I did was looked at the regions the points began and ended. I'm not 100% this is correct but I hope I helped!

6 0

3 years ago

A 16-digit binary pattern converts to a ____ digit hexadecimal value.

Doss [256]

Solution -

In order to convert a binary number to hexadecimal ,we need to chunk the binary number into groups of 4 digits, adding 0 if necessary.

As the given number number is a 16 digit binary number, so when chunk in 4 digit groups, it will form 4 groups. Each group will represent a single digit in hexadecimal code.

So, a 16-digit binary pattern converts to a 4 digit hexadecimal value.

7 0

3 years ago

Thirty subtracted from four times a number is greater than the opposite of 10

Arada [10]

30-4x>-10

Hope I helped

5 0

3 years ago

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x overbar, is found

joja [24]

Answer:

(a) 80% confidence interval for the population mean is [109.24 , 116.76].

(b) 80% confidence interval for the population mean is [109.86 , 116.14].

(c) 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed.

Step-by-step explanation:

We are given that a simple random sample of size n is drawn from a population that is normally distributed.

The sample mean is found to be 113 and the sample standard deviation is found to be 10.

(a) The sample size given is n = 13.

Firstly, the pivotal quantity for 80% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = 113

s = sample standard deviation = 10

n = sample size = 13

$\mu$ = population mean

<em>Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

<u>So, 80% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-1.356 < $t_1_2$ < 1.356) = 0.80 {As the critical value of t at 12 degree

of freedom are -1.356 & 1.356 with P = 10%}

P(-1.356 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.356) = 0.80

P( $-1.356 \times }{\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

<u>80% confidence interval for </u> $\mu$ = [ $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.356 \times }{\frac{10}{\sqrt{13} } }$ , $113+1.356 \times }{\frac{10}{\sqrt{13} } }$ ]

= [109.24 , 116.76]

Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].

(b) Now, the sample size has been changed to 18, i.e; n = 18.

So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.

<u>80% confidence interval for </u> $\mu$ = [ $\bar X-1.333 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.333 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.333 \times }{\frac{10}{\sqrt{18} } }$ , $113+1.333 \times }{\frac{10}{\sqrt{18} } }$ ]

= [109.86 , 116.14]

Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].

(c) Now, we have to construct 98% confidence interval with sample size, n = 13.

So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.

<u>98% confidence interval for </u> $\mu$ = [ $\bar X-2.681 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+2.681 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-2.681 \times }{\frac{10}{\sqrt{13} } }$ , $113+2.681 \times }{\frac{10}{\sqrt{13} } }$ ]

= [105.56 , 120.44]

Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed because t test statistics is used only when the data follows normal distribution.

6 0

3 years ago