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1 year ago

300-297+294-291+288-285+...+6-3

1 answer:

vladimir2022 [97]1 year ago

7 0

300 - 297 = 3

294 - 291 = 3

288 - 285 = 3

So, that is a sequence of sums of number 3: 3 + 3 + 3 + .... + 3

How may times will the number 3 be added?

Note that 300 is reduced in 6 units each time => 300 / 6 = 50 => 3 will be added 50 times.

=> 50 * 3 = 150

Answer: 150

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Mean of 7.2,8.5,7.0,8.1,6.7

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1 year ago

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= $\frac{\overline{Y}}{\overline{X}} .$

Therefore, the estimated total weight of grain in the population using ratio estimator is

$\widehat{Y}_R\: \: =r\times 8800\: \: =0.1\times 8800\: \: =880\,$ lbs

4) The variance of the ratio estimator is

var(r)= $\frac{N-n}{N}\frac{1}{n}\frac{1}{\mu_x^2}\frac{\sum_{i=1}^{5}(y_i-rx_i)^2}{n-1} , where \mu_x=8800/200=44lbs$

= $\frac{200-5}{200}\, \frac{1}{5}\: \frac{1}{44^2}\, \frac{0.2}{5-1}=0.000005$

Hence, the standard error of the estimate of the total population is

$\sigma_R=\sqrt{X^2 \: var(r)}\: \: \: =\sqrt{8800^2\times 0.000005}\: \: \:$ =21.556

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

$[\pm z_{0.025}\times \sigma_{R}]\: \: \: =[\pm 1.96\times 21.556]\: \: \: =[\pm 42.25]$

8 0

2 years ago

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What is the distance between (3, 5.25) and (3, –8.75)? 6 units 8.25 units 11.75 units 14 units

evablogger [386]

By using distance formula :

$\text{Distance formula,} \bold{ \boxed{ Distance = \sqrt{( x_{2} -x_{1})^{2}+(y_{2} - y_{1})^{2}) }}}$

Given points = ( 3 , 5.25 ) and ( 3 , - 8.75 )

$\bold{Taking \: \: \: x_{1}=3 \: \: , \: \: x_{2}= 3 \: \: , \: \: y_{1}= 5.25 \: \: , \: \: y_{2}= -8.75}$

On applying formula, we get

$Distance = \sqrt{ ( x_{2}-x_{1})^{2}+(y_{2}-y_{1})^2} \\ \\ \\ Distance = \sqrt{ ( 3 - 3 )^{2} + ( - 8.75 - 5.25 )^{2}} \\ \\ \\ Distance = \sqrt{ ( 0 )^{2} + ( - 14)^{2}} \\ \\ \\ Distance = \sqrt{ ( - 14 )^{2}} \\ \\ \\ Distance = \sqrt{ 14^{2}} \:\:\:\:\:\:\:\:\:\:\: \: \: \: \: \: \: \: \: | \bold{ ( - 14 )^{2} = 14^{2}} \\ \\ \\ Distance = {14}^{2 \times \frac{1}{2} } \\ \\ \\ Distance = {14}^{1} \\ \\ \\ Distance = 14 \: units$

Hence, Option D is correct.

4 0

1 year ago

Read 2 more answers