All categories

2 years ago

Why is it important to measure perimeter and area

2 answers:

7 0

<span>There are so many ways that landscapers use the formula for area. One of the most common ways they use the formula is to calculate the area of a yard. This helps landscapers figure out how much grass they need to cover a yard. </span>

blagie [28]2 years ago

4 0

Perimeter and area are two important and fundamental mathematical topics. They help you to quantify physical space and also provide a foundation for more advanced mathematics found in algebra, trigonometry, and calculus. Perimeter is a measurement of the distance around a shape and area gives us an idea of how much surface the shape covers.

Knowledge of area and perimeter is applied practically by people on a daily basis, such as architects, engineers, and graphic designers, and is math that is very much needed by people in general. Understanding how much space you have and learning how to fit shapes together exactly will help you when you paint a room, buy a home, remodel a kitchen, or build a deck.

You might be interested in

Sophia scored 92% in a math test. If the test had 50 questions, how many did she get right? *

zloy xaker [14]

Answer:

46 questions correct

Step-by-step explanation:

100% = 50 questions
2% = 1 question
92% = 46 questions

6 0

2 years ago

If it is 5 degrees outside and the temperature will drop 17 degrees in the next 6 hours how cold will it get?

ser-zykov [4K]

5 - 17 = 12 so the answer is that it will be -12 degrees outside. Fahrenheit or Celsius??

Hope This Helps!

7 0

3 years ago

Read 2 more answers

A rectangle is increased by 25% and the width is decreased by 20%, what is the change in the area of the rectangle? The change i

faltersainse [42]

Answer:

Step-by-step explanation:

Please refer to the previously asked question:

brainly.com/question/24265969?referrer=searchResults

3 0

2 years ago

In the number 2145857, how does the digit 5 in the thousands place compare to the digit 5 in the tens place?

fgiga [73]

It is 1,000 bigger than the one in the tens place. 5,000 compared to 50

7 0

3 years ago

A simple random sample of 28 Lego sets is obtained and the number of pieces in each set was counted.The sample has a standard de

Karo-lina-s [1.5K]

Answer:

Step-by-step explanation:

Given that:

A simple random sample n = 28

sample standard deviation S = 12.65

standard deviation $\sigma$ = 11.53

Level of significance ∝ = 0.05

The objective is to test the claim that the number of pieces in a set has a standard deviation different from 11.53.

The null hypothesis and the alternative hypothesis can be computed as follows:

Null hypothesis:

$H_0: \sigma^2 = \sigma_0^2$

Alternative hypothesis:

$H_1: \sigma^2 \neq \sigma_0^2$

The test statistics can be determined by using the following formula in order to test if the claim is statistically significant or not.

$X_0^2 = \dfrac{(n-1)S^2}{\sigma_0^2}$

$X_0^2 = \dfrac{(28-1)(12.65)^2}{(11.53)^2}$

$X_0^2 = \dfrac{(27)(160.0225)}{132.9409}$

$X_0^2 = \dfrac{4320.6075}{132.9409}$

$X_0^2 = 32.5002125$

$X^2_{1- \alpha/2 , df} = X^2_{1- 0.05/2 , n-1}$

$X^2_{1- \alpha/2 , df} = X^2_{1- 0.025 , 28-1}$

From the chi-square probabilities table at 0.975 and degree of freedom 27;

$X^2_{0.975 , 27}$ = 14.573

$X^2_{\alpha/2 , df} = X^2_{ 0.05/2 , n-1}$

$X^2_{\alpha/2 , df} = X^2_{0.025 , 28-1}$

From the chi-square probabilities table at 0.975 and degree of freedom 27;

$X^2_{0.025 , 27}=$ 43.195

Decision Rule: To reject the null hypothesis if $X^2_0 \ > \ X^2_{\alpha/2 , df} \ \ \ or \ \ \ X^2_0 \ < \ X^2_{1- \alpha/2 , df}$ ; otherwise , do not reject the null hypothesis:

The rejection region is $X^2_0 \ > 43.195 \ \ \ or \ \ \ X^2_0 \ < \ 14.573$

Conclusion:

We fail to reject the null hypothesis since test statistic value 32.5002125 lies between 14.573 and 43.195.

6 0

2 years ago