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lidiya [134]
2 years ago
7

A machine can stamp 75 caps per minute. At this rate how long will it take to stamp 3000?

Mathematics
2 answers:
Leya [2.2K]2 years ago
6 0
<h2>Answer:</h2><h2>40 minutes</h2>

Step-by-step explanation:

75- caps ----------- 1 minute\\3000 caps ------------x - minutes\\\\Cross-Multiply\\75x = 3000 \times 1\\75x = 3000\\\frac{75}{75} =\frac{3000}{75} \\\\x = 40 minutes

NNADVOKAT [17]2 years ago
6 0
The answer is 40 mins hope This helped
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Sam and Bruno were computing how many kilometers they rode in the 3 bike trips they took last month. In order, they rode 45.7, 4
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Answer:

Correct solution:

45.7 + 40.9 + 38.0 = 124.6 miles. 124.6 divided by 0.621 = 200.64

Step-by-step explanation:

they rode 45.7, 40.9, and 38 miles

= (45.7 + 40.9 + 38) miles

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This is the correct solution because he divided 124.60 miles by 0.621

45.7 + 40.9 + 38.0 = 124.6 miles. 124.6 divided by 0.621 = 200.64

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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
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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]

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