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Sidana [21]
2 years ago
9

Find the missing side.

Mathematics
1 answer:
Alona [7]2 years ago
3 0

Answer:

9

Step-by-step explanation:

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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
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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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A coordinate grid with 2 lines. The first line is labeled y equals 0.5 x plus 3.5 and passes through (negative 3, 1), (negative
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Answer:

d

Step-by-step explanation:

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2 years ago
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Sanya noticed that the temperature was falling at a steady rate of 1.4 degrees every hour after the time that she first checked
kakasveta [241]

Answer:A not b or c or d A!!

Step-by-step explanation:

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2 years ago
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If y varies directly as x, and y is 400 when x is r and y is r when x is 4, what is the numeric constant of variation in this
otez555 [7]

Answer:

10

Step-by-step explanation:

If there is a direct relation between two variables x and y then it can be represented as

y = kx ,

where y is dependent variable

x is  independent variable

k is constant of variation

_____________________________

First condition

y = 400

x = r

using y = kx then relationship will be

400 = kr

finding k here

k = 400/r

Second condition

y = r

x = 4

using y = kx then relationship will be

r = 4k

finding k

k = r/4

since  in both condition equation is same

thus, value of k will also be same

thus,

400/r = r/4

=> 400*4 = r*r

=> 1600 = r^2

\sqrt{r^{2} }  = \sqrt{1600} \\r = 40

Thus, 40 is the value of r

k = r/4 = 40/4 = 10

Thus, constant of variation is 10 which is correct choice.

To cross validate

k = 400/r = 400/40 = 10

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223.61 Meters???? Don't quote me
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