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3 years ago

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Mathematics

2 answers:

ASHA 777 [7]3 years ago

8 0

The length of the side of the square is $\boxed{35{\text{ meter}}}.$

Further Explanation:

The formula of area of square can be expressed as follows,

$\boxed{Area = \left( s \right) \times \left( s\right)}$

The formula of perimeter of the rectangle can be expressed as follows,

$\boxed{{\text{Perimeter}} = 4 \times s}$

Here, s represents the side of square.

Given:

The cost of plugging in a square field is Rs 2450.

The cost for plugging per square meter is Rs 2.

Explanation:

The total area that plugged can be obtained as follows,

$\begin{aligned}{\text{Area}}&= \frac{{{\text{Cost of plugging}}}}{{{\text{cost of per square meter}}}}\\&= \frac{{2450}}{2}\\&= 1225\\\end{aligned}$

The length of the side of the square can be obtained as follows,

$\begin{aligned}{\text{Area}}&= {\left( {\text{S}} \right)^2}\\1225&= {\left( {\text{S}} \right)^2}\\\sqrt {1225} &= {\text{S}}\\35&= {\text{S}}\\\end{aligned}$

The length of the side of the square is $\boxed{35{\text{ meter}}}.$

Learn more:

Learn more about inverse of the functionhttps://brainly.com/question/1632445.
Learn more about equation of circle brainly.com/question/1506955.
Learn more about range and domain of the function brainly.com/question/3412497

Answer details:

Grade: Middle School

Subject: Mathematics

Chapter: Mensuration

Keywords: length, square, area, side of square, width, breadth, area, dimension, measure of sides, dimensions of rectangle, expression, plugging, square field, Rs 2450, 2 per square meter, length of the side.

Alenkinab [10]3 years ago

4 0

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$t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37$

$p_v =P(t_{4}>2.37)=0.038$

If we compare the p value and a significance level assumed $\alpha=0.1$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.

Step-by-step explanation:

Data given and notation

$\bar X=250000$ represent the sample mean

$s=37500$ represent the standard deviation for the sample

$n=5$ sample size

$\mu_o =210250$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

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Null hypothesis: $\mu \leq 210250$

Alternative hypothesis: $\mu > 210250$

Compute the test statistic

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37$

Now we need to find the degrees of freedom for the t distirbution given by:

$df=n-1=5-1=4$

Conclusion

Since is a one right tailed test the p value would be:

$p_v =P(t_{4}>2.37)=0.038$

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