The circumference of a circle is 28 pi inches. What is the length of the radius of this circle?

The graph that shows a negative linear relationship that has a correlation coefficient, r, that is closer to -1 is: B. graph B.

A graph that shows a negative linear relationship has a trendline that slopes downwards from your left to the right.

The closer the data points are to each other along the trendline in a negative linear graph, the closer the correlation coefficient, r, is to -1.

Therefore, the graph that shows a negative linear relationship and has a correlation coefficient, r, that is closer to -1 is: B. graph B.

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we know that
Figures can be proven similar if one, or more, similarity transformations (reflections, translations, rotations, dilations) can be found that map one figure onto another. 
In this problem to prove circle 1 and circle 2 are similar, a translation and a scale factor (from a dilation) will be found to map one circle onto another.

we have that
 

step 1
Move the center of the circle 1 onto the center of the circle 2
the transformation has the following rule
(x,y)--------> (x+6,y-4)
so
(-4,5)------> (-4+6,5-4)-----> (2,1)
so
center circle 1 is now equal to center circle 2 
The circles are now concentric (they have the same center)

step 2
A dilation is needed to increase the size of circle 1 to coincide with circle 2

scale factor=radius circle 2/radius circle 1-----> 6/2----> 3

radius circle 1 will be=2*scale factor-----> 2*3-----> 6 cm
radius circle 1 is now equal to radius circle 2 

A translation, followed by a dilation will map one circle onto the other, thus proving that the circles are similar

the answer is

The best point estimate for the true mean weight of all packages received by a parcel service, given a sample mean of 20.6 pounds and a standard deviation of 3.2 pounds, is 20.6 pounds.

In this problem, we are given that a sample of 41 packages were randomly selected. This sample provided a mean weight of 20.6 pounds and a standard deviation of 3.2 pounds. It is asked what the best point estimate for a confidence interval estimating the true mean weight of all packages is.

The best estimate of the true mean weight (µ) of all packages would be the mean of the sample that was taken, assuming that the sample was randomly selected and is representative of the entire population. This is due to the fact that in statistics, the sample mean is the best point estimate of the population mean.

Therefore, the best point estimate for the true mean weight of all the packages received by the parcel service is 20.6 pounds.

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Final answer:

The appropriate hypotheses to determine if graffiti incidents declined involve a null hypothesis where the mean difference is greater than or equal to zero and an alternative hypothesis where the mean difference is less than zero. The μd represents the mean difference in graffiti incidents before and after the institution of the community watch program.

Explanation:

The appropriate hypotheses to test whether the number of graffiti incidents declined in the suburban communities after the institution of a citizen community watch program are as follows:

Here, μd (mu-sub-d) represents the mean difference between the number of incidents before the program and after the program was implemented. To test these hypotheses, we would typically conduct a paired t-test if the samples are normally distributed or a Wilcoxon signed-rank test if the distribution is not normal.

Assuming μd is the mean difference in graffiti incidents before and after, the test will determine if there is statistically significant evidence to support that the community watch program resulted in a reduction of graffiti incidents.

Answer:

x-9,y+4,92

Step-by-step explanation:

The correct minimum number of students required to guarantee that at least three answer sheets must be identical is 5.

To solve this problem, we can use the pigeonhole principle. The pigeonhole principle states that if you have more pigeons than pigeonholes, at least one pigeonhole must contain more than one pigeon.

In this scenario, each question can be thought of as a pigeonhole, with the possible answers (a, b, c, d) as the pigeons. Since there are four possible answers for each question, we have four pigeonholes for each of the ten questions.

Now, let's consider the number of students (pigeons) and the number of unique answer sheets (pigeonholes). We want to find the minimum number of students such that at least three of them have identical answer sheets.

For two students, there are [tex]\(4^{10}\)[/tex] possible combinations of answers for the ten questions. For three students, the number of possible combinations of answers is [tex]\(4^{10} \times 4^{10}\)[/tex]. In general, for [tex]\(n\)[/tex] students, the number of possible combinations is [tex]\(4^{10} \times 4^{10} \times \ldots \times 4^{10}\) (\(n\) times)[/tex].

We need to find the smallest [tex]\(n\)[/tex] such that the number of combinations exceeds the number of ways to distribute the answer sheets without having three identical ones.

The number of ways to distribute the answer sheets without having three identical is given by the sum of the combinations of choosing 1, 2, ...,[tex]\(n-1\)[/tex] students out of [tex]\(n\)[/tex] to have unique answer sheets, and then multiplying by [tex]\(4^{10}\)[/tex] for each unique combination. This is because we are allowing for the possibility that some students have the same answer sheets, but not more than two.

The sum is given by:

[tex]\[ \binom{n}{1} \times 4^{10} + \binom{n}{2} \times 4^{10} + \ldots + \binom{n}{n-1} \times 4^{10} \][/tex]

We want to find the smallest [tex]\(n\)[/tex] such that:

[tex]\[ \binom{n}{1} \times 4^{10} + \binom{n}{2} \times 4^{10} + \ldots + \binom{n}{n-1} \times 4^{10} < 4^{10n} \][/tex]

For [tex]\(n = 4\)[/tex], we have:

[tex]\[ \binom{4}{1} \times 4^{10} + \binom{4}{2} \times 4^{10} + \binom{4}{3} \times 4^{10} = 4 \times 4^{10} + 6 \times 4^{10} + 4 \times 4^{10} = 14 \times 4^{10} \][/tex]

For [tex]\(n = 5\)[/tex], we have:

[tex]\[ \binom{5}{1} \times 4^{10} + \binom{5}{2} \times 4^{10} + \binom{5}{3} \times 4^{10} + \binom{5}{4} \times 4^{10} \][/tex]

[tex]\[ = 5 \times 4^{10} + 10 \times 4^{10} + 10 \times 4^{10} + 5 \times 4^{10} = 30 \times 4^{10} \][/tex]

Now, we compare [tex]\(30 \times 4^{10}\) with \(4^{10 \times 5}\)[/tex]. Since [tex]\(4^{10 \times 5}\)[/tex] is much larger than[tex]\(30 \times 4^{10}\)[/tex], we can see that for [tex]\(n = 5\)[/tex], the number of possible combinations exceeds the number of ways to distribute the answer sheets without having three identical ones.

Therefore, the minimum number of students required to guarantee that at least three answer sheets must be identical is 5.

Answer: 40 pi

Step-by-step explanation:

Answer: sinD = 0. 9459

cosD =0. 3243 tanD=2. 917

Step-by-step explanation:

From the figure above;

opposite = 35

adjacent = 12

hypotenuse = 37

sin D = opposite/hypotenuse

sinD = 35/37 = 0. 9459

cosD = adjacent/hypotenuse

= 12/37 = 0. 3243

tanD = adjacent/opposite

= 35/12 = 2. 9167

Area and volume are both used to determine the amount of space.

While area is used for two-dimensional shapes, volume is used for three-dimensional shapes.

The difference between the given parameters are:

For instance:

[tex]\mathbf{Length = b -a}[/tex]

[tex]\mathbf{Area = ab}[/tex]

[tex]\mathbf{Volume= abc}[/tex]

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The probability that the average mosquito count in a randomly chosen square meter of the pond is more than 68 mosquitoes is approximately 59.87%.

To find the probability that the average count of mosquitoes in a randomly chosen square meter of the pond is more than 68 mosquitoes per square meter, we can use the normal distribution properties.

Given:

We need to find P(X > 68). To do this, we convert the raw score to a Z-score using the formula:

Z = (X - μ) / σ

Substitute the given values:

Z = (68 - 70) / 8 = -2 / 8 = -0.25

Next, we need to find the probability corresponding to Z > -0.25 using the standard normal distribution table.

Looking up Z = -0.25, we find the cumulative probability (P(Z < -0.25)) is approximately 0.4013.

To find P(Z > -0.25):

P(Z > -0.25) = 1 - P(Z < -0.25) = 1 - 0.4013 = 0.5987

Therefore, the probability that the average mosquito count in a randomly chosen square meter is more than 68 mosquitoes per square meter is approximately 0.5987 or 59.87%.

The sample space contains three types of animals; 5 salamanders, 3 crayfish, and 12 minnows.

So population of sample space would be, n(S) = Salamanders + Crayfish + Minnows.

n(S) = 5 + 3 + 12 = 20.

She wants to randomly select an animal and check if it is Salamander or not.

population of favourable outcomes would be, n(salamander) = 5.

Now the probability for selecting a salamander would be :-

[tex]P(salamander) = \frac{n(salamander)}{n(S)} \\\\ P(salamander) = \frac{5}{20} \\\\ P(salamander) = \frac{1}{4} \;or\; 0.25 \;or\; 25\%[/tex]

Answer:

They are equal

Step-by-step explanation: