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Grace [21]
3 years ago
8

Hugh drove 774.4 miles. his car averaged 24.2 miles per gallon of gas. how many gallons of gas did the car use?

Mathematics
1 answer:
maksim [4K]3 years ago
5 0
774.4 miles divided by 24.2 mpg = 32 gallons of gas
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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
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
4 years ago
Read 2 more answers
SOLVE <br><br> x + 3y = -1<br><br> y = x - 7<br><br><br><br><br><br> Write the solution below.
ki77a [65]

Answer:

x= 5 ........................ .......

7 0
2 years ago
Simplify the expression completely.<br> (-3c-2d2e101-3)
GREYUIT [131]

Answer:

-3c-2d^(2)e^(101)-3

Step-by-step explanation:

8 0
3 years ago
Which expressions are equivalent when m = 1 and m = 4? 5 m minus 3 and 2 m 5 m 3 m 4 and m 4 2 m 2 m 7 and 3 m minus 3 m 5 m 3 a
emmasim [6.3K]

The expressions that are equivalent when m = 1 and m = 4 is;

Option B: 3m + 4 and m + 4 + 2m

We are given m = 1 and m =4;

A) 5m - 3 and 2m + 5 + m

B) 3m + 4 and m + 4 + 2m

C) 2m + 7 and 3m - 3 + m

D) 5m + 3 and 4m + 2 + 2m

For option B; 3m + 4 and m + 4 + 2m

Let's put m = 1

3(1) + 4 = 7

Also, 1 + 4 + 2(1) = 7

Similarly, let us put 4 for m to get;

3(4) + 4 = 16

Also, 4 + 4 + 2(4) = 16

In both cases, the expressions are equivalent and as such option B is the right one.

Read more about algebra simplifications at; brainly.com/question/4344214

4 0
3 years ago
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The fraction 4/10 can be written as<br><br> A) 0.0004 <br> B) 0.004 <br> C) 0.04 <br> D) 0.4
lara31 [8.8K]
You can simplify the fraction to get 2/5. If you convert this into a decimal you will get 0.4, so D is your answer.
6 0
4 years ago
Read 2 more answers
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