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1 year ago

9

Dave walked to his friend's house at a rate of 4mph and returned back biking at a rate of 10mph. If it took him 18 minutes longe

r to walk than to bike, what was the total distance of the round trip

1 answer:

schepotkina [342]1 year ago

4 0

Answer:

4 miles

Step-by-step explanation:

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The lucas sequence is the sequence 1, 3, 4, 7, 11, $\ldots$ where the first term is 1, the second term is 3 and each term after

nexus9112 [7]

Mod 8, the sequence repeats after 12 terms:
1 3 4 7 3 2 5 7 4 3 7 2

Thus the 100th term, mod 8, will be the 4th term of this sequence: 7

7 0

2 years ago

Decrease £19064.67 by 9.5%<br> Give your answer rounded to 2 DP.

alina1380 [7]

Answer:

New amount (decrease) = £17253.53

Step-by-step explanation:

<u>Given the following data;</u>

Amount = £19064.67

Percentage = 9.5%

To find the percentage decrease;

$Percentage \; cost = \frac {9.5}{100} * 19064.67$

$Percentage \; cost = 0.095 * 19064.67$

Percentage cost = £1811.14

New amount (decrease) = Original amount - percentage cost

Substituting into the equation, we have;

New amount (decrease) = £19064.67 - £1811.14

<em>New amount (decrease) = £17253.53</em>

6 0

2 years 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

The graph shows the distance in miles of a runner over x hours. What is the average rate of speed over the interval [9, 11]?

JulsSmile [24]

For this case what you should see is that for the interval [9, 11] the behavior of the function is almost linear.
Therefore, we can find the average rate of change as follows:
m = (y2-y1) / (x2-x1)
m = (11-6) / (11-9)
m = (5) / (2)
m = 5/2
Answer:
the average rate of speed over the interval [9, 11] is:
D. 5 / 2

8 0

2 years ago

Read 2 more answers

The shape below is constructed from six squares, and the lengths of the sides of the three smallest squares are written inside o

Maurinko [17]

Answer:

78

Step-by-step explanation:

Smallest square: 1

Small Square: 2, Area of 4

Mid Square: 3, Area of 9

Mid White Square: 5, Area of 25

Big Square: 8, Area of 64

1 + 4 + 9 + 64 = 78

5 0

2 years ago