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2 years ago

B) The perimeter of a squared field is 60 m. Find its area.

Mathematics

1 answer:

Pani-rosa [81]2 years ago

6 0

$\qquad\qquad\huge\underline{{\sf Answer}}♨$

Let's solve ~

Perimeter of square is :

$\qquad \sf \dashrightarrow \:4 \times side \: length$

so, we can take side length as a, and solve for a ~

$\qquad \sf \dashrightarrow \:4a = 60$

$\qquad \sf \dashrightarrow \:a = 60 \div 4$

$\qquad \sf \dashrightarrow \:a = 15$

Now, since we know the side length. let's find Area of square :

$\qquad \sf \dashrightarrow \: {15}^{2}$

$\qquad \sf \dashrightarrow \:area = 225 \: m {}^{2}$

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2 years ago

A rectangular skate park is 60 yards long and 50 yards wide

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Here are summary statistics for randomly selected weights of newborn girls: nequals=174174, x overbarxequals=30.930.9 hg, seq

MaRussiya [10]

Answer:

The 95% confidence interval would be given by (29.780;32.020)

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=30.9$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=7.5 represent the sample standard deviation

n=174 represent the sample size

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=174-1=173$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,173)".And we see that $t_{\alpha/2}=1.97$ , this value is similar to the obtained with the normal standard distribution since the sample size is large to approximate the t distribution with the normal distribution.