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2 years ago

Miranda bought a square frame that has an area of 30 square inches what is the approximate side lenght of the frame

Mathematics

2 answers:

Whitepunk [10]2 years ago

4 0

<u>Answer</u>:

The side length of the square frame is 5.5 inches approximately.

Step-by-step explanation:

Given:

Area of the square frame bought by Miranda= 30 square inches

To find:

Side length of the frame=?

Solution:

we know that,

$\text{{Area of a square}} = \text{( side length )}}^2$

Here, area of square = 30 square inches .

$\text{( side length )}}^2 = 30$ square inches

Side length = $\sqrt30$ inches

Side length = 5.4772 inches

Alexxandr [17]2 years ago

3 0

Area of a square equals side squared

(A = s²)

30 = s²

√30 = s

You can use a calculator to find out that √30 ≈ 5.48

or you can use estimation (√25 < √30 < √36) which means that √30 is between 5 and 6.

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The table represents an exponential function.

Minchanka [31]

we know that

The multiplicative rate of change of the exponential function between two points is equal to

$[f(b) / f(a) ] / (b-a)$

Let

$A(1,6)\\B(2,4)$

we have that

$f(a)=6\\f(b)=4\\a=1\\b=2$

substitute in the formula

$[4 / 6 ] / (2-1)=4/6=2/3$

Let

$A(3,8/3)\\B(4,16/9)$

we have that

$f(a)=8/3\\f(b)=16/9\\a=3\\b=4$

substitute in the formula

$[(16/9) / (8/3) ] / (4-3)=48/72=2/3$

therefore

<u>the answer is the option</u>

B.) 2/3

8 0

2 years ago

Read 2 more answers

A wheat farmer cuts down the stalks of wheat and gathers them in 200 piles. The 200 gathered piles will be put on a truck. The t

denpristay [2]

Answer:

Check the explanation

Step-by-step explanation:

We want to estimate the total weight of grain on the field based on the data on a simple random sample of 5 piles out of 200. The population and sample sizes are N=200 & n=5 respectively.

1) Let Y_1,Y_2,...,Y_{200} be the weight of grain in the 200 piles and y_1,y_2,...,y_{5} be the weights of grain in the pile from the simple random sample.

We know, the sample mean is an unbiased estimator of the population mean. Therefore,

$\widehat{\mu}=\overline{y}=\frac{1}{5}\sum_{i=1}^{5}y_i=\frac{1}{5}(3.3+4.1+4.7+5.9+4.5)=4.5$

where \mu is the mean weight of grain for all the 200 piles.

Hence, the total grain weight of the population is

$\widehat{Y}=Y_1+Y_2+...+Y_{200}$

= $200\times \widehat{\mu}\: \: \: =200\times 4.5\: \: \: =900\, lbs$

2) To calculate a bound on the error of estimates, we need to find the sample standard deviation.

The sample standard deviation is

S= $\sqrt{\frac{1}{5-1}\sum_{i=1}^{5}(y_i-\overline{y})^2}\: \: \: =0.9486$

Then, the standard error of $\widehat{Y} is$

$\sigma_{\widehat{Y}} =\sqrt{\frac{N^2S^2}{n}\bigg(\frac{N-n}{N}\bigg)}\: \: \:$ =83.785

Hence, a 95% bound on the error of estimates is

$[\pm z_{0.025}\times \sigma_{\widehat{Y}}]\: \: \: =[\pm 1.96\times 83.875]\: \: \: =[\pm 164.395]$

3) Let x_1,x_2,...,x_5 denotes the total weight of the sampled piles.

Mean total weight of the sampled piles is

$\overline{x}=\frac{1}{5}\sum_{i=1}^{5}x_i=45$

The sample ratio is

$r=\frac{\overline{y}}{\overline{x}}=\frac{4.5}{45}$ =0.1 , this is also the estimate of the population ratio R= $\frac{\overline{Y}}{\overline{X}} .$