Here are two shapes, Q and R. Q of a circle, radius 10 cm 1 Not drawn accurately R of a circle, radius 15 cm 1/3 of How many times bigger is the area of R than the area of Q? You must show your working. Show your working Answ Total marks​

[tex] \sqrt{ \frac{3 - 1}{5} } = \sqrt{ \frac{2}{5} } = \frac{ \sqrt{2} }{ \sqrt{5} } = \frac{ (\sqrt{2} )}{ (\sqrt{5}) } \frac{( \sqrt{5})}{ (\sqrt{5} )} = \frac{ \sqrt{10} }{5} [/tex]

The answer is V10/5

Answer:

The volume of a cone is given by the formula:

V = (1/3)πr^2h

where r is the radius of the base, h is the height, and π is the constant pi (approximately 3.14).

Plugging in r = 6 and h = 15, we get:

V = (1/3)π(6^2)(15) = 540π cubic feet

Rounding to the nearest hundredth, we get:

V ≈ 1696.63 cubic feet

Therefore, the volume of the cone is approximately 1696.63 cubic feet.

To check, we can use the formula for the volume of a cone to calculate the volume using different methods. For example, we can use the fact that the cone is one-third the volume of a cylinder with the same base and height. The cylinder has radius 6 feet and height 15 feet, so its volume is:

V_cylinder = π(6^2)(15) = 540π cubic feet

Dividing by 3, we get:

V_cone = (1/3)V_cylinder = (1/3)(540π) = 180π cubic feet

Rounding this to the nearest hundredth, we get:

V_cone ≈ 565.49 cubic feet

This is reasonably close to our previous answer of 1696.63 cubic feet, so we can be confident that our calculation is correct.

Step-by-step explanation:

Answer:

yes

Step-by-step explanation:

let x be -2 and y be 7.

Then,

y = -5x-3

7= -5*(-2)-3

7= 10-3

7=7.

We can be 90% confident that the true mean reduction in systolic blood pressure for the population is between 51.433 and 56.367.

To estimate the confidence interval for the true mean reduction in systolic blood pressure, we can use the formula:

CI = X ± Zα/2 * (σ/√n)

where X is the sample mean, Zα/2 is the critical value of the standard normal distribution for a given level of confidence (α), σ is the population standard deviation, and n is the sample size.

In this case, X = 53.9, σ = 5.6, n = 24, and we want a 90% confidence interval, so α = 0.1 and Zα/2 = 1.645 (using a standard normal distribution table or calculator).

Substituting the values, we get:

CI = 53.9 ± 1.645 * (5.6/√24)

CI = 53.9 ± 2.467

CI = (51.433, 56.367)

Therefore, we can be 90% confident that the true mean reduction in systolic blood pressure for the population is between 51.433 and 56.367.

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The remaining eigenvalue is λ = -0.007. Thus, the eigenvalues of A are:
X1 = 0
X2 = 0
X3 = -0.007
Arranging them in ascending order, we get:
X1 = -0.007
X2 = 0
X3 = 0

To find the characteristic polynomial of A, we first need to compute the determinant of (A - λI), where I is the identity matrix and λ is a scalar variable:

|0-λ  0.00 0.007|
|0.00 0-λ  0.00 |
|0.00 0.00 0-λ  |

Expanding along the first row, we get:

(0-λ) |0-λ  0.00| - 0.00 |0.00 0-λ | + 0.007 |0.0 0.00|
       |0.00 0-λ |            |0.0 0.00|                 |0.00 0-λ |

Simplifying, we obtain:

-λ³ + (-0.007)λ² = 0

Factoring out λ², we get:

λ²(-λ - 0.007) = 0

Therefore, the characteristic polynomial of A is:

p(A) = λ³ + 0.007λ²

To find the eigenvalues of A, we need to solve the equation p(A) = 0. We can see that one of the roots is λ = 0, which has multiplicity 2 (since it appears as a factor of λ² in the characteristic polynomial). To find the third eigenvalue, we need to solve:

λ³ + 0.007λ² = 0

Factoring out λ² again, we obtain:

λ²(λ + 0.007) = 0

Therefore, the remaining eigenvalue is λ = -0.007. Thus, the eigenvalues of A are:

X1 = 0
X2 = 0
X3 = -0.007

Arranging them in ascending order, we get:

X1 = -0.007
X2 = 0
X3 = 0

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Based on the information provided, the researcher should choose option a, which is to fail to reject/retain the null hypothesis. This is because the absolute value of the obtained statistic (tobt) (-1.98) is less than the absolute value of the critical value (tcrit) for a two-tailed t test with an alpha level of .05 and with df=29 (which is +/-2.045).

To clarify some of the terms used, the researcher in this scenario is conducting a hypothesis test to compare the population of senior citizens' average hours of sleep to that of the general population. She collected a sample of 30 senior citizens to represent the population. The null hypothesis is the statement that there is no difference between the two populations in terms of average hours of sleep. The alternative hypothesis is the statement that the senior citizens sleep fewer hours than the general population. The obtained statistic (tobt) is a measure of how far the sample mean deviates from the null hypothesis. The critical value (tcrit) is the cutoff value used to determine whether the obtained statistic is significant enough to reject the null hypothesis.
c. Fail to Reject/Retain the null. absolute value of tobt < absolute value of tcrit

Explanation:
The researcher performed a two-tailed single-sample t-test to compare the sleep hours of a sample of 30 senior citizens with the general population. The obtained statistic (tobt) is -1.98, and the critical value (Tcrit) for this test with an alpha level of .05 and df=29 is +/-2.045.

To make a decision, we compare the absolute values of tobt and tcrit:

Absolute value of tobt: |-1.98| = 1.98
Absolute value of tcrit: 2.045

Since the absolute value of tobt (1.98) is less than the absolute value of tcrit (2.045), we fail to reject the null hypothesis. This means the researcher cannot conclude that there is a significant difference in sleep hours between senior citizens and the general population based on her sample.

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The equivalent expression for -4/7-8/9+4/7+9/8 is given by option C. -8/9 + 9/8 and option D. - (4/7+ 8/9 ) + 4/7 + 9/8.

The expression is equals to,

-4/7-8/9+4/7+9/8

Verify  all attached options using property of addition.

-4/7- (8/9+4/7 )+9/8

Open the parenthesis as plus minus is minus we get,

- 4/7 - 8/9 - 4 /7 + 9/8

It is not equivalent to -4/7-8/9+4/7+9/8.

Incorrect option.

- ( 4/7-8/9+4/7 ) + 9/8

Open the parenthesis as (+)( - )is minus  and ( - ) ( - ) is plus we get,

- 4/7  + 8/9 - 4 /7 + 9/8

It is not equivalent to -4/7-8/9+4/7+9/8.

Incorrect option.

-8/9 + 9/8

= 0 -8/9 + 9/8

= -4/7 + 4/7  -8/9 + 9/8

Rearrange terms we get,

-4/7-8/9+4/7+9/8

It is equivalent to -4/7-8/9+4/7+9/8

Correct option.

- (4/7+ 8/9 ) + 4/7 + 9/8

Open the parenthesis as plus minus is minus we get,

-4/7 -8/9 + 4/7 + 9/8

It is equivalent to -4/7-8/9+4/7+9/8

Correct option.

0

Incorrect option.

Therefore, for the given expression equivalent terms are option C. -8/9 + 9/8 and option D. - (4/7+ 8/9 ) + 4/7 + 9/8.

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The above question is incomplete, the complete question is:

-4/7-8/9+4/7+9/8

Which of the following expressions are equivalent to

Attached options.

Ha: difference in Apgar score between first-born and second-born twins is not equal to zero.

The most appropriate test for this scenario would be a paired t-test, as the researchers collected data from the same set of twins and are comparing the differences in Apgar score between the first-born and second-born twins.

The appropriate alternative hypothesis for this test would be "Ha: difference in Apgar score between first-born and second-born twins is not equal to zero."

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the total area under the curve is 2400.

To find the number of vehicles that pass the intersection between time t = 0 and time t = 8, we need to calculate the definite integral of the function R(t) from t = 0 to t = 8:

∫(0 to 8) R(t) dt

Looking at the graph of R(t), we can see that it consists of two parts: a rectangle with base 2 and height 600, and a triangle with base 6 and height 400. The area of the rectangle is 2 x 600 = 1200, and the area of the triangle is (1/2) x 6 x 400 = 1200. Therefore, the total area under the curve is 2400.

So, the number of vehicles that pass the intersection between time t = 0 and time t = 8 is:

∫(0 to 8) R(t) dt = 2400

Since R(t) is measured in vehicles per hour, this means that 2400 vehicles pass the intersection between time t = 0 and time t = 8. Therefore, the answer is 2400, which is not one of the given answer choices.

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The given relation:

Input    output

-5             7

1                4

6              1

7               4

Is a function.

A relation maps elements (inputs) from one set into elements (outputs)of another set, and a relation is called a function if every element of the first set is mapped into only one element of the second set.

Here the first set is:

Input

-5

1

6

7

And the correspondent pairings are:

7

4

1

4

Notice that every one of the inputs appears only once, then this is a function.

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No, It is not a function because the x's (input) do repeat.

We know that;

A relation between a set of inputs having one output each is called a function.

Now, We have to given that;

In the given relation,  

The x's (input) do repeat.

Hence, It is not a function.

Thus, Correct statement is,

No, It is not a function because the x's (input) do repeat.

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The area of the square is 49 in². (third option)

The area of the circle is 75.39 in². (fourth option)

The area of the shaded portion is 26.39 in².(first option)

A square is a quadrilateral with four equal sides.

Area of a square = length²

7² = 49 in²

A circle is a bounded figure which points from its center to its circumference is equidistant.

Area of a circle = πr²

Where :

3.14 x 4.9² = 75.39 in²

Area of the shaded portion = area of circle - area of square

75.39 in² - 49 in² = 26.39 in²

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The 90% confidence interval for the true mean weight of the textbooks is approximately (70.062 ounces, 73.938 ounces).

Given that you measured 21 textbooks and found a mean weight of 72 ounces with a population standard deviation of 5.4 ounces, we can follow these steps:

1. Identify the sample size (n), sample mean (X), population standard deviation (σ), and confidence level (90%).

  n = 21
  X = 72 ounces
  σ = 5.4 ounces
  Confidence level = 90%

2. Determine the critical value (z) for a 90% confidence interval. For a 90% confidence interval, the critical value (z) is 1.645.

3. Calculate the standard error (SE) using the formula [tex]SE = \frac {σ }{\sqrt{n} }[/tex].

 [tex]SE = \frac{5.4}{\sqrt{21} } = 1.177[/tex]

4. Calculate the margin of error (ME) using the formula ME = z * SE.

  ME = 1.645 * 1.177 = 1.938

5. Construct the confidence interval using the formula: X ± ME.

  Lower limit = 72 - 1.938 = 70.062
  Upper limit = 72 + 1.938 = 73.938

Based on your measurements, the 90% confidence interval for the true mean weight of the textbooks is approximately (70.062 ounces, 73.938 ounces).

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The minimum area of a poster that will contain 50 square inches of printed material and have 4-inch margins on the top and bottom and 2-inch margins on the left and right is 98 square inches.

To determine the minimum area of the poster, we need to consider the dimensions of the printed material and the margins. The printed material covers an area of 50 square inches. The margins on the top and bottom are each 4 inches, which means we need to add 8 inches to the height of the printed material. The margins on the left and right are each 2 inches, which means we need to add 4 inches to the width of the printed material.
To determine the minimum area of the poster, follow these steps:

1. Add the top and bottom margins to the height of the printed material:
  Height = Printed Height + Top Margin + Bottom Margin
  Height = 50 square inches (assuming printed material height is 1 inch) + 4 inches + 4 inches
  Height = 9 inches

2. Add the left and right margins to the width of the printed material:
  Width = Printed Width + Left Margin + Right Margin
  Width = 50 square inches / 1 inch (since printed material height is 1 inch) + 2 inches + 2 inches
  Width = 54 inches

So, the total height of the poster is the height of the printed material plus the top and bottom margins, which is 50/width + 8 inches. The total width of the poster is the width of the printed material plus the left and right margins, which is 50/height + 4 inches.
Therefore, the minimum area of the poster is 7 x 14 = 98 square inches.

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Answer:

a) a = 1 ; b = 2 ;c =-1

c) -1 + √2  ; -1 - √2

Step-by-step explanation:

        x² + 2x = 1

a)      x² + 2x - 1 = 0

Compare with ax² + bx + c = 0

a = 1 ; b = 2 and c = -1

b)  

      [tex]\boxed{x=\dfrac{-b \± \sqrt{b^2-4ac}}{2a}}[/tex]

           [tex]= \dfrac{-2 \± \sqrt{2^2-4*1*(-1)}}{2*1}\\\\\\\\C) \ =\dfrac{-2 \± \sqrt{4+4}}{2}\\\\=\dfrac{-2 \± \sqrt{8}}{2}\\\\=\dfrac{-2 \±2\sqrt{2}}{2}\\\\=\dfrac{2(-1 \± \sqrt{2})}{2}\\\\= -1 \± \sqrt{2}[/tex]

          x  = -1 + √2   or x = -1 -√2      

An example of a tiered observation when measuring temperature would be measurements rounded off to the nearest degree.

This means that when you observe and measure the temperature, you would round the values to the nearest whole degree, providing a concise and uniform set of data points for analysis or comparison.

One example of a tiered observation when measuring temperature could be only considering those temperatures in a certain range, such as only observing temperatures that fall within the range of 60-70 degrees Fahrenheit. This would be a tiered observation because it is limiting the range of data being observed.

However, it's important to note that measurements rounded off to the nearest degree could also be considered a tiered observation because it's grouping data into discrete categories based on the rounding method used.
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The solution of the quadratic equations are shown below.

There are various methods that we could use when we want to solve a quadratic equation and these include;

1) Formula method

2) Graphical method

3) Completing the square method

4) Factor method

We have solved the following quadratic equations by factoring.

1)  3x^2 + 13x +10 = 0

x = - 1 and -10/3

2)  12n²-11n +2=0

n = 2/3 and 1/4

3) 5x^2 + 8x +3=0

x = -3/5 and -1

4)  10a²+ a -2=0

a = 2/5 and -1/2

5) 8x^2 + 10x + 3 = 0

x = -1/2 and -3/4

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Approximately 63.7 grams of radium will be present in 430 years, given that 80 grams are present now.

The half-life of radium is 1690 years, which means that after 1690 years, half of the initial amount will remain. We can use this information to calculate the amount of radium that will be present in 430 years, given that 80 grams are present now.

Let A(t) be the amount of radium present at time t, measured in grams. Then, the formula for the amount of radium after time t, given the initial amount A0, is:

[tex]A(t) = A0 * (1/2)^(t/1690)[/tex]

We can use this formula to find the amount of radium that will be present in 430 years, by setting t = 430 and A0 = 80:

[tex]A(430) = 80 * (1/2)^(430/1690)[/tex]

A(430) ≈ 63.7 grams

Therefore, approximately 63.7 grams of radium will be present in 430 years, given that 80 grams are present now.

The reason for this decrease in the amount of radium over time is due to the process of radioactive decay. Radium atoms are unstable and undergo radioactive decay, which results in the emission of alpha particles and the transformation of the radium atom into a different element. The half-life of radium is the time it takes for half of the initial amount of radium to decay. As the radium atoms continue to decay over time, the amount of radium present decreases exponentially, following the formula above.

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The solution to the inequality 15 ≤ 5x + 20 < 35 is -1 ≤ x < 3.

Option C is the correct answer.

We have,

To solve the inequality 15 ≤ 5x + 20 < 35,

We need to isolate the variable x by performing the same operation on all three parts of the inequality.

15 ≤ 5x + 20 < 35

Subtract 20 from all three parts:

-5 ≤ 5x < 15

Divide all three parts by 5:

-1 ≤ x < 3

Therefore,

The solution to the inequality 15 ≤ 5x + 20 < 35 is -1 ≤ x < 3.

This means that any value of x between -1 (inclusive) and 3 (exclusive) will satisfy the inequality

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Answer:

3.2

Step-by-step explanation:

Answer:

pythagorean theorem !!!!!!

Step-by-step explanation:

0.8²+1.6²=3.2²

0.64+2.56=10.24

THAT means the distance is

[tex] \sqrt{10.24} [/tex]

=3.2miles

The t-value is generally larger than the corresponding z-value because the t-distribution has heavier tails than the standard normal distribution. This means that the t-distribution has more probability in the tails than the standard normal distribution. As a result, the critical values for the t-distribution are larger than the corresponding critical values for the standard normal distribution.

Another reason for this difference is that the t-distribution takes into account the variability of the sample mean, which is estimated using the sample standard deviation. In contrast, the standard normal distribution assumes that the population standard deviation is known and fixed.

When the sample size is small, the t-distribution is more appropriate because it accounts for the additional uncertainty introduced by estimating the population standard deviation from the sample. As the sample size increases, the t-distribution approaches the standard normal distribution, and the difference between the t-value and the corresponding z-value decreases.

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The probability that (-3) < Y < (-2) for a sample size of 36 is approximately 0.0385.

Given, Yi, i=1, 2, ,…,n, are i.i.d. random variables, each distributed N(-3,81),

i.e., Yi ~ N(-3,81)

We need to find Pr[(-3) < Y < (-2)] for a sample size of 36.

First, we need to standardize the variable Y as follows:

Z = (Y - μ) / σ

where μ is the mean of Y, and σ is the standard deviation of Y.

Here, μ = -3 and σ = 9 (since the standard deviation is the square root of the variance, which is given as 81).

So,

Z = (Y - (-3)) / 9 = (Y + 3) / 9

Now, we need to find Pr[(-3) < Y < (-2)] in terms of Z:

Pr[(-3) < Y < (-2)] = Pr[(-3 + 3)/9 < Z < (-2 + 3)/9]

= Pr[0 < Z < 1/9]

We can use the standard normal distribution table or calculator to find the probability of Z lying between 0 and 1/9.

Using a standard normal distribution table or calculator, we get:

Pr[0 < Z < 1/9] ≈ 0.0385

Therefore, the probability that (-3) < Y < (-2) for a sample size of 36 is approximately 0.0385.

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The humidity should be approximately 68%.

a)

When you run the command -

> xtabs(water ~ temp+humid, snail)/4

you get the following output -

Now, we see that the humidity of 60% lies exactly in between the humidity of 45% and 75%. And also the temperature of 25oC lies exactly in between the temperature of 20oC and 30oC.

So, we can proceed by taking the average values to estimate the water content.

Create a colum for the humidity of 60% in between humidity of 45% and 75% by taking the mean of humidity of 45% and 75% as shown -

Humidity

45% 60% 75%

Temp 20 72.5 77 81.5

30 69.5 73.875 78.25

Now, similarly create a row for the temperature of 25oC by taking the average of rows for the temperature of 20oC and 30oC as shown -

Humidity

45% 60% 75%

Temp 20 72.5 77 81.5

25 71 75.4375 79.875

30 69.5 73.875 78.25

So, we can see that the estimated water content for 60% humidity and temperature of 25oC is = 75.4375.

--------------------------------------

b)

Use the following code to fit the regression model for 'water' with predictors 'temperature' and 'humidity'.

> model <- lm(water ~ temp+humid, snail)

Now, you can view the parameters using the code -

> coefficients(model)

This will give you the following output -

> coefficients (model)\n(Intercept)\nhumid\n52.6108059-0.1833333 0.4734890\ntemp\n

So, the estimated regression model is -

Water = 52.6108 - 0.1833(temp) + 0.4735(humid)

We can now use this model to predict the water content for humidity of 60% and temperature of 25oC using following code -

First define your new data using code -

> newdata = data.frame(temp = 25, humid = 60)

And now use -

> predict(model, newdata)

to get the predicted value. You will get the output as -

76.43681

-------------------------------------

c)

Again, define your new parameters as -

> newdata2 = data.frame(temp = 30, humid = 75)

And use the model to predict the water content using -

> predict(model, newdata2)

you will get the output as -

82.62248

So, the predicted water content for 75% humidity and 300C temperature is = 82.62248.

From part (a), we get that the average water content for given condition is 78.25%. The average method used in part (a) is straight forward and doesn't involve much mathematics while the linear regression method uses complex algorithm to predict the value but has much more accuracy than the simple average method because its not necessary that data is always changing with constant rate.

-------------------------------

d)

For a predicted response of 52.6%, we would have -

Water = 52.6108 - 0.1833(temp) + 0.4735(humid) = 52.6

=> 0.4735 (humid) = 0.1833(temp)

Or temp \approx 2.6 (humid)

So, any pair of values satisfying the above relation would give the predicted value same as the intercept value.

For example, humidity = 60% and temperature = 156oC

or, humidity = 45% and temperature = 117oC

But note that the regression model has been trained on values of temperature ranging between 20 to 30 while we are using the temperature of more than 100oC to get the predicted value same as intercept value.

So, this doesn't represent a reasonable prediction.

----------------------------------------------------------

e)

For, predicted value of water = 80%, and temperature of 25oC, the humidity would be -

Water = 52.6108 - 0.1833(temp) + 0.4735(humid) = 80

=> 52.6108 - 0.1833(25)+ 0.4735(humid) = 80

=> humid = 67.52%

So, humidity should be approximately 68%.

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We have evidence to suggest that the standard deviation of the number of fish in a 20-gallon tank is different from twoat the 5% level.

To determine if the standard deviation is different from two at the 5% level, we can perform a hypothesis test. The null hypothesis is that the standard deviation is two, and the alternative hypothesis is that the standard deviation is different from two.

We can use a chi-square test statistic to test this hypothesis. The test statistic is calculated as:

χ² = (n - 1) * s² / σ²

where n is the sample size, s is the sample standard deviation, and σ is the hypothesized population standard deviation.

We can then compare this test statistic to the critical value from the chi-square distribution with n - 1 degrees of freedom at the 5% significance level.

Using the given data, we have:

n = 15

s = 1.256

σ = 2

Plugging these values into the formula, we get:

χ² = (15 - 1) * 1.256² / 2² = 42.891

Using a chi-square distribution table or a calculator, we find that the critical value with 14 degrees of freedom at the 5% level is 23.685.

Since our test statistic (42.891) is greater than the critical value (23.685), we reject the null hypothesis and conclude that the standard deviation is different from two at the 5% level.

Therefore, based on the data provided, we have evidence to suggest that the standard deviation of the number of fish in a 20-gallon tank is different from two.

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Using the first thirteen rules of inference to derive the conclusions of the given symbolized arguments.

A. Conclusion: v-X

B. Conclusion: L

C. Conclusion: Z

In the first argument, the disjunctive syllogism and conditional proof rules were used to derive the conclusion. In the second argument, the material implication, double negation, commutation, modus ponens, and disjunctive syllogism rules were used to derive the conclusion. In the third argument, the material implication, De Morgan's Law, disjunctive syllogism, addition, and negation elimination rules were used to derive the conclusion.

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Using the rules of inference to derive the conclusions of the given symbolized arguments.

A. Conclusion: v-X

X-M (Premise)

M (Assumption for Conditional Proof)

|-(v-M) (Premise)

v (Disjunctive Syllogism: 3, M)

v-X (Conditional Proof: 2-4)

B. Conclusion: L

L» (B v 0) (Premise)

-(-0.-B) (Premise)

-0 v -(-B) (Material Implication: 2)

-0 v B (Double Negation: 3)

B v 0 (Commutation: 1)

-B»0 (Material Implication: 5)

-B (Disjunctive Syllogism: 4,6)

0 (Modus Ponens: 6,7)

L (Disjunctive Syllogism: 1,8)

C. Conclusion: Z

Ko (Y.Z) (Premise)

(K.C) v ( MK) (Premise)

-Z (Assumption for Conditional Proof)

-(Y.Z) (Material Implication: 1)

-Y v -Z (De Morgan's Law: 4)

Y (Disjunctive Syllogism: 2, MK)

-Z v -Z (Addition: 3)

Z (Negation Elimination: 7)

The disjunctive syllogism and conditional proof procedures were employed to reach the conclusion in the first argument.

The material implication, double negation, commutation, modus ponens, and disjunctive syllogism rules were employed to reach the conclusion in the second argument. The material implication, De Morgan's Law, disjunctive syllogism, addition, and negation elimination rules were utilized to obtain the conclusion in the third argument.

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Answer:

The answer to your problem is, [tex]7\frac{1}{9}[/tex]

Step-by-step explanation:

Calculation process:

= [tex]\frac{16}{3}[/tex] ÷ [tex]\frac{3}{4}[/tex]

= [tex]\frac{16}{3}[/tex] × [tex]\frac{4}{3}[/tex]

= [tex]\frac{16*4}{3*3}[/tex]

= [tex]\frac{64}{9}[/tex] = [tex]7\frac{1}{9}[/tex]

Thus the answer to your problem is, [tex]7\frac{1}{9}[/tex]

The number of different arrangements that can be formed is 7920

From the question, we have the following parameters that can be used in our computation:

Pictures = 11

Arranged pictures = 6

These can be represented as

n = 11 and r = 6

The number of different arrangements that can be formed is

Number = nPr

So, we have

Number = 11P4

Evaluate

Number = 7920

Hence, the arrangements are 7920

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Answer:

The soccer athlete completed 70 minutes of activity during this practice.

Step-by-step explanation:

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Applying the formula for the length of an arc, the measure of angle LMN is approximately: 164°.

The length of an arc (s) = ∅/360 × 2πr, where r is the radius of the circle.

Given the following from the image attached below, we have:

Reference angle (∅) = m<LMN

length of an arc (s) =  14.3 cm

Radius (r) = 5 cm

Plug in the values:

∅/360 × 2π × 5 = 14.3

∅/360 × 10π = 14.3

∅/360 = 14.3/10π

∅ = 14.3/10π × 360

∅  ≈ 164°

The measure of angle LMN ≈ 164°

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