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Zigmanuir [339]

2 years ago

11

Find the area of quadrilateral ABCD. Round the area to the nearest whole number, if necessary. A(-5, 4) 4 B(0, 3) 2. F(-2, 1) -2

2. 6 x -2 24 E(2, -3) D( 45) 6 X The area is X x square units

1 answer:

valentina_108 [34]2 years ago

5 0

Answer:26

Step-by-step explanation:

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sertanlavr [38]

Answer:

Area of the rectangle: 24 cm²

Area of the circle: 380.13 cm²

Area of between the circle and the rectangle: 356.13 cm²

Step-by-step explanation:

8 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

Chuck has a gross pay of $815.70. By how much will Chuck’s gross pay be reduced if he has the following items withheld?

Alex787 [66]

Answer:

273.38

Step-by-step explanation:

815.70 * 6.2% = 50.57

815.70 * 1.45% = 11.83

815.70 * 19% = 154.95

Add those 3 totals to the additional $56 in federal tax

56+50.57+11.83+154.95 = 273.35

5 0

2 years ago

Read 2 more answers

On a road map, the locations A, B and C are collinear. Location C divides the road from location A to B, such that AC:CB = 1:2.

koban [17]

Answer:

C. (-1,-2)

Step-by-step explanation:

Since C internally divides AB in the ratio AC/CB = 1/2 = m/n where m = 1 and n = 2, we use the formula for internal division.

Let A = (x₁, y₁) = (5, 16), B = (x₂, y₂) and C = (x, y) = (3, 10)

So x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

Substituting the values of the coordinates, we have

x = (mx₂ + nx₁)/(m + n)

3 = (1 × x₂ + 2 × 5)/(2 + 1)

3 = (x₂ + 10)/3

multiplying through by 3, we have

9 = x₂ + 10

x₂ = 9 - 10

x₂ = -1

y = (my₂ + ny₁)/(m + n)

10 = (1 × y₂ + 2 × 16)/(2 + 1)

10 = (x₂ + 32)/3

multiplying through by 3, we have

30 = y₂ + 32

y₂ = 30 - 32

y₂ = -2

So, the coordinates of B are (-1, -2)

3 0

1 year ago

"Majesty Video Production Inc. wants the mean length of its advertisements to be 30 seconds. Assume the distribution of ad lengt

Westkost [7]

Answer:

a) $\bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})$

b) $Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5$

c) $P(\bar X >31.25)=0.006=0.6\%$

d) $P(\bar X >28.25)=0.9997=99.97\%$

e) $P(28.25$

Step-by-step explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variabl length of advertisements produced by Majesty Video Production Inc. We know from the problem that the distribution for the random variable X is given by:

$X\sim N(\mu =30,\sigma =2)$

We take a sample of n=16 . That represent the sample size.

a. What can we say about the shape of the distribution of the sample mean time?

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is also normal and is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\bar X \sim N(\mu=30, \frac{2}{\sqrt{16}})$

b. What is the standard error of the mean time?

The standard error is given by this formula:

$Se=\frac{\sigma}{\sqrt{n}}=\frac{2}{\sqrt{16}}=0.5$

c. What percent of the sample means will be greater than 31.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula given by:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$

And we want to find this probability:

$P(\bar X >31.25)=1-P(\bar X$

d. What percent of the sample means will be greater than 28.25 seconds?

In order to answer this question we can use the z score in order to find the probabilities, the formula is given by:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$

And we want to find this probability:

$P(\bar X >28.25)=1-P(\bar X$

e. What percent of the sample means will be greater than 28.25 but less than 31.25 seconds?"

We want this probability:

$P(28.25$

3 0

2 years ago