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tiny-mole [99]
2 years ago
11

Suppose that the light from the pinhole projects onto a screen 3.00 meters away. What is the radius r1 of the first dark ring on

that screen? Notice that the angle from part a is small enough that sinθ≈tanθ.
Mathematics
1 answer:
Fittoniya [83]2 years ago
3 0

Answer:

16 mm

Step-by-step explanation:

The next information is misssing: The first dark ring can be found at angle 0.305 degrees.

A rigth triangle is formed in wich one leg is the radius r1 of the first dark ring and the other leg is the distance between the screen and the laser (L), which is equal to 3 meters.

Using the relationship sinθ≈tanθ we get:

tanθ = r1/L

tan 0.305  = r1 / 3

r1 =  tan(0.305°) * 3 m

r1 = 0.016 m = 16 mm

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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
2 years ago
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Wetlands offer a diversity of benefits. They provide a habitat for wildlife, spawning grounds for U.S. commercial fish, and rene
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Question Completion:

How are the percentages distributed? Is the distribution skewed? Are there any gaps? (Select all that apply.)

Answer:

1. The percentages are concentrated from 20% to 60%.

2. These data are strongly skewed left.

3. There are no gaps in the data.

Step-by-step explanation:

1. Data

Percentage loss of wetlands per state

46 37 36 42 81 20 73 59 35 50

87 52 24 27 38 56 39 74 56 31

27 91 46 9 54 52 30 33 28 35

35 23 90 72 85 42 59 50 49

48 38 60 46 87 50 89 49 67

2. Re-arrangement of

Percentage loss of wetlands per state (in ascending order)

9    20  23  24  27  27   28  30

31   33  35  35  35  36   37  38  

38  39  42  42  46  46   46 48

49  49   50  50  50  52   52 54

56 56   59  59  60  67   72  73  

74  81  85  87   87   89  90  91  

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1 year ago
Jody bought 20 shares of amazon at the close price of $121.00. She bought 20 more shares a year later at the price of $127.00. T
hjlf

Answer:

Jody made $210 after all of these transactions.

Step-by-step explanation:

Jody bought 20 shares of amazon at the price of $121.00

= 121.00 × 20 = $2,420

After one year she bought 20 more shares at the price of $127.00

= 127.00 × 20 = $2,540

Her broker charges $50.00 for each transaction = 50 × 2 = $100

Total cost of all shares = 2,420 + 2,540 + 100 = $5,060

She sold all of her shares at the price of $133.00

= 133 × 40 = $5,320

She gave broker charge for this transaction = $50.00

Total selling amount  = 5,320 - 50 = $5,270

Profit = Selling amount - cost

Profit = 5,270 - 5,060 = $210

Jody made $210 after all of these transactions.

8 0
2 years ago
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Karen finished watching a movie at 1:00 PM. The movie lasted 1 hour 38 minutes. at what time Karen started watching the movie. -
Andrew [12]

The correct answer would be 11:32am. i hope this helps!

6 0
2 years ago
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Triangle RST was dilated by a scale factor of One-half. The image, triangle R'S'T', is an isosceles triangle, with each leg meas
andrew11 [14]

Answer:

The length side of the pre-image is 16 units

Step-by-step explanation:

we know that

The length side of the image is equal to the length side of the pre-image multiplied by the scale factor

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The length side of the pre-image is equal to the length side of the image divided by the scale factor

in this problem we have that

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The length side of the image is 8 units

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