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2 years ago

In the expression 4 (2m-n), determine which of the following best describes the role of (2m-n).

Mathematics

1 answer:

Dmitrij [34]2 years ago

4 0

Answer:

Step-by-step explanation:

(2m-n) is a factor of 4(2m-n)

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We want to estimate the population mean within 5, with a 99% level of confidence. the population standard deviation is estimated

pashok25 [27]

A sample size of 60 is required.

We use the formula
$n=(\frac{z*\sigma}{E})^2$

We first find the z-score associated with this level of confidence:
Convert 99% to a decimal: 99/100 = 0.99
Subtract from 1: 1-0.99 = 0.01
Divide by 2: 0.01/2 = 0.005
Subtract from 1: 1-0.005 = 0.995

Using a z-table (http://www.z-table.com) we see that this value is equally distant from 2.57 and 2.58; therefore we will use 2.575:

$n=(\frac{15(2.575)}{5})^2=59.68\approx60$

3 0

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

Read 2 more answers

Harrison has 30 cds in his music collection. If 40% of the cds are country and 30% are pop, how many are other types of music

Pavel [41]

Well you have to find out how many cd is 40% and 30%
30% = 30x.3=9
40% = 30x.4=16
now add 16 and 9
16+9=25 now subtract 25 from 30 and see that you have 5 cd of a different type of music to find that as a % divide 5 by 35 to get .14 I rounded it to the nearest hundreth now move the decimal over 2 spots to get 14%

8 0

2 years ago

Read 2 more answers

Which congruence theorem can be used to prove △BDA ≅ △DBC? Triangles B D A and D B C share side D B. Angles C B D and A D B are

Tomtit [17]

Answer:

A. Hypotenuse-leg (HL) congruence.

HL, when you have 2 right triangles and their hypotenuses are congruent you are able to say HL

Step-by-step explanation:

We know that the hypotenuse-leg theorem states that if the hypotenuse and one leg of a right triangle are congruent to hypotenuse and corresponding leg of another right triangle, then the triangles are congruent.

hypotenuse(AB) of △BDA equals to hypotenuse (CD) of △DBC.

BDA and DBC share a common side DB.

Using Pythagorean theorem we will get,

$CD^{2}=DB^{2}+BC^{2}...(1) \\\\AB^{2}=DB^{2}+AD^{2}...(2)$