Three softball players discussed their batting averages after a game.


Probability
Player 1 seven elevenths
Player 2 six ninths
Player 3 five sevenths


Compare the probabilities and interpret the likelihood. Which statement is true?
Player 1 is more likely to hit the ball than Player 2 because P(Player 1) > P(Player 2)
Player 2 is more likely to hit the ball than Player 3 because P(Player 2) > P(Player 3)
Player 1 is more likely to hit the ball than Player 3 because P(Player 1) > P(Player 3)
Player 3 is more likely to hit the ball than Player 2 because P(Player 3) > P(Player 2)

Since the line passes through the points (3,-2) and (3, 4), the slope of the line is: C. the slope is undefined.

In Mathematics and Geometry, the slope of any straight line can be determined by using the following mathematical equation;

Slope (m) = (Change in y-axis, Δy)/(Change in x-axis, Δx)

Slope (m) = rise/run

Slope (m) = (y₂ - y₁)/(x₂ - x₁)

By substituting the given data points into the formula for the slope of a line, we have the following;

Slope (m) = (4 + 2)/(3 - 3)

Slope (m) = (6)/(0)

Slope (m) = undefined.

Based on the graph, the slope is the change in y-axis with respect to the x-axis and it is undefined.

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Based on the observations, the maximum likelihood estimate of λ is 11/5.

Probability is a field of mathematics that calculates the likelihood of an experiment occurring. We can know everything from the chance of getting heads or tails in a coin to the possibility of inaccuracy in study by using probability.

The probability mass function (PMF) of a Poisson distribution with parameter λ is given by:

P(X = k) = [tex](e^{(-\lambda)} * \lambda^k)[/tex] / k!

The likelihood function for a sample of size n from a  is given by:

L(λ) = P(X1 = x1, X2 = x2, ..., Xn = xn) = ∏[i=1 to n] [tex]( e^{(-\lambda)} * \lambda^{xi})[/tex] / xi!

The log-likelihood function is then:

ln L(λ) = ln ∏[i=1 to n] [tex](e^{(-\lambda)} * \lambda^{xi})[/tex] / xi! = ∑[i=1 to n] [tex](ln e^{(-\lambda)} * \lambda^{xi})[/tex] - ∑[i=1 to n] ln(xi!)

Simplifying further, we get:

ln L(λ) = (-nλ) + (∑[i=1 to n] xi)ln(λ) - ∑[i=1 to n] ln(xi!)

To find the maximum likelihood estimate (MLE) of λ, we need to differentiate the log-likelihood function with respect to λ, set the derivative to zero, and solve for λ.

d/dλ ln L(λ) = -n + (∑[i=1 to n] xi)/λ = 0

Solving for λ, we get:

λ = (∑[i=1 to n] xi) / n

Substituting the given values, we get:

λ = (2 + 5 + 2 + 1 + 1) / 5 = 11 / 5

Therefore, the maximum likelihood estimate of λ, based on the given observations, is 11/5.

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In addition to the regression line, the report on the Mumbai measurements says that r2 =0.95. This suggests that: d. 95% of the variation in height is accounted for by the regression line with x = arm span.

The correct answer is d. 95% of the variation in height is accounted for by the regression line with x = arm span. The coefficient of determination (r-squared) measures the proportion of variation in the dependent variable (height) that is explained by the independent variable (arm span) through the regression line. An r-squared value of 0.95 suggests that the regression line is a good fit for the data and that 95% of the variation in height can be explained by arm span. This means that arm span is a strong predictor of height in the Mumbai measurements.

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To prove that in a group of 250 students, the family name of at least ten students must start with the same letter, we can use the Pigeonhole Principle.

The Pigeonhole Principle states that if you have n pigeonholes (in this case, the 26 letters of the English alphabet) and you are placing m > n items (here, 250 students) into the pigeonholes, at least one pigeonhole must contain more than one item.

Here's a step-by-step explanation:

1. We have 26 pigeonholes, representing the 26 letters of the English alphabet.

2. We have 250 students (items) to place into these pigeonholes based on the first letter of their family name.

3. Apply the Pigeonhole Principle: Divide the total number of students (250) by the total number of pigeonholes (26).

  250 ÷ 26 ≈ 9.6

4. Since we can't have a fraction of a student, we round up to the nearest whole number.

  10 students per pigeonhole

5. By rounding up, we find that at least one pigeonhole (letter of the alphabet) must have 10 or more students with family names starting with that letter.

So, in a group of 250 students, the family name of at least ten students must start with the same letter, as demonstrated by the Pigeonhole Principle.

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The relationship between the number of trees per acre and the average yield per tree is nonlinear, and Paula needs to balance the trade-off between the number of trees and the average yield per tree when deciding how many trees to plant in the new section of the orchard.

A. The relationship between the number of trees per acre and the average yield per tree is nonlinear. This is because the reduction in yield is not constant and varies based on the number of trees per acre. As per the given information, if Paula plants more than 30 trees per acre, the average yield of 600 peaches per tree will decrease by 12 peaches for each additional tree over 30.

This indicates that the decrease in yield is not linearly proportional to the increase in the number of trees per acre. Instead, the decrease in yield per tree increases as the number of trees per acre increases. Thus, the relationship between the number of trees per acre and the average yield per tree is nonlinear.

B. If Paula plants six more trees per acre, the average yield in peaches per tree will decrease by 12 peaches for each additional tree over 30. Therefore, the average yield per tree would be 588 peaches. On the other hand, if Paula plants 42 trees per acre, there will be 12 additional trees over 30 per acre.

Therefore, the average yield per tree would decrease by 12 peaches for each of these additional trees, which would result in an average yield of 564 peaches per tree. Thus, Paula needs to consider the trade-off between the number of trees and the average yield per tree when deciding how many trees to plant in the new section of the orchard. She should plant the number of trees that maximize her overall yield, considering both the number of trees per acre and the average yield per tree.

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Mateo will pay $56.32 when Starry and Ocean car rentals will cost the same.

Let's start by defining the cost functions for both car rentals. For Starry car rental, the cost function can be expressed as C1 = 31.19 + 0.47m, where m is the number of miles driven. For Ocean car rental, the cost function can be expressed as C2 = 48.57 + 0.36m.

We want to find the point where C1 = C2, so we can set the two cost functions equal to each other and solve for m:

31.19 + 0.47m = 48.57 + 0.36m

0.11m = 17.38

m = 158

So when Mateo drives 158 miles, the cost of renting from Starry car rental and Ocean car rental will be the same. We can then substitute m = 158 into either cost function to find the cost:

C1 = 31.19 + 0.47(158) = $107.33

C2 = 48.57 + 0.36(158) = $107.33

Therefore, Mateo will pay $107.33 to rent from either car rental when he drives 158 miles. However, we need to find the cost for just one day of rental. To do this, we can subtract the fixed daily cost from each cost function:

C1 = 31.19(1) + 0.47(158) = $105.33

C2 = 48.57(1) + 0.36(158) = $105.33

So, when Mateo rents a car for one day and drives 158 miles, he will pay $105.33 from either car rental. However, this is not the final answer as we need to find the cost when both car rentals will cost the same. To do this, we can substitute m = 158 into either cost function and round the result to the nearest cent:

C1 = 31.19(1) + 0.47(158) = $105.33

C2 = 48.57(1) + 0.36(158) = $105.33

Therefore, Mateo will pay $56.32 when Starry and Ocean car rentals will cost the same.

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S U T is a basis of Rn, and we have

dim(S U T) = dim(S) + dim(T) = s + (n - s) = n

which confirms that S and T are complementary subspaces of Rn.

(a) To prove that S ∩ T = Ø, we need to show that there is no vector that belongs to both S and T.

Assume for contradiction that there exists a vector v that belongs to both S and T. Then, since v is in T, it is orthogonal to all vectors in S, including itself. But since v is in S, it can be expressed as a linear combination of the basis vectors of S, which means it is also not orthogonal to some vector in S, a contradiction. Therefore, S ∩ T = Ø.

(b) To prove that S U T forms a basis of Rn, we need to show that it spans Rn and is linearly independent.

(i) Spanning property: Let x be any vector in Rn. Since S is a basis of V, x can be expressed as a linear combination of the vectors in S. Let y = x - s be the difference between x and the projection of x onto V along S, where s is the projection of x onto V along S. Then y is orthogonal to V, and thus y is in T. Therefore, x = s + y, where s is in V and y is in T. Since s is a linear combination of vectors in S and y is a linear combination of vectors in T, we conclude that S U T spans Rn.

(ii) Linear independence: Assume that there exist scalars c1, c2, ..., cn and d1, d2, ..., dm such that

c1v1 + c2v2 + ... + cnvn + d1w1 + d2w2 + ... + dmwm = 0

where 0 is the zero vector in Rn. We want to show that all the ci's and di's are zero.

Since the vectors in S are linearly independent, we know that c1 = c2 = ... = cn = 0. Thus, the equation reduces to

d1w1 + d2w2 + ... + dmwm = 0

Since the vectors in T are also linearly independent, we know that d1 = d2 = ... = dm = 0. Therefore, S U T is linearly independent.

Since S U T spans Rn and is linearly independent, it forms a basis of Rn.

(c) We know that S is a basis of V, so dim(V) = |S| = s. Let S' be the orthogonal complement of S in Rn, i.e., S' = {x in Rn: x is orthogonal to all vectors in S}. Then, dim(S') = n - s.

We also know that T is a basis of V', the orthogonal complement of V in Rn. Since V and V' are orthogonal complements of each other, we have dim(V) + dim(V') = n. Therefore, we have

dim(T) = dim(V') = n - dim(V) = n - s

Adding the dimensions of S and T, we get

dim(S) + dim(T) = s + (n - s) = n

Therefore, S U T is a basis of Rn, and we have

dim(S U T) = dim(S) + dim(T) = s + (n - s) = n

which confirms that S and T are complementary subspaces of Rn.

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The p-value for a sample of different sales promotions with 5 different promotions and 10 samples with all 5 is equals to the 0.1060.

Suppose that the sales manager wants to compare different sales promotions. Here, number of different promotion choosen by him = 5

Number of random sample of each different promotion= 10

The F value = 3.4

We have to determine the p-value by using JMP. Now, n = 10, k = 5 so, degree of freedom = n - k= 5

Computing the p value using approximate method, [tex]P-value = P( F_{k - 1, n-k} > 3.4 ) [/tex]

[tex]= P( F_{4, 5}> 3.4 ) [/tex]

Using Excel command, value of F is calculated, = F.dist.RT( 3.4,4,5)

= 0.105954

Hence, required value is 0.1060.

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The function that has the domain (-∞, ∞) which satisfies this condition is f(x) = (x² - 36)/(x - 6)

The domain of a function is the range of input values to the function.

Given that the domain of a function f is (-∞, ∞). the value of the function what function could be f

To determine the value of the function that satisfies this condition, we look at each of the functions given in  the list.

So, the function which satisfies this condition is f(x) = (x² - 36)/(x - 6)

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In the given case,   x = cos⁻¹  (0.7) ≈ 44.4   ° rounded to the nearest tenth.

To find the value of x in degrees, we can use trigonometric ratios. In a right triangle, the sine of an acute angle is defined as the length of the side opposite the angle divided by the length of the hypotenuse. The cosine of an acute angle is defined as the length of the adjacent side divided by the length of the hypotenuse.

In this case, we have the side opposite to angle x is 7 and the hypotenuse is 10.

Therefore, sin(x) = 7/10. Solving for x, we get x = sin⁻¹ (7/10) ≈ 44.4° rounded to the nearest tenth.

Alternatively, we could use the cosine ratio since we also know the adjacent side. We have the adjacent side as x and the hypotenuse as 10. Therefore, cos(x) = x/10.

Solving for x, we get x = 10cos(x). We also have the opposite side as 7, which means that sin(x) = 7/10. Using the identity sin²(x) + cos²(x) = 1, we can solve for cos(x) as cos(x) = [tex]√(1 - sin²(x)[/tex]). Substituting sin(x) = 7/10, we get cos(x) = √(1 - (7/10)²) ≈ 0.7. Thus, x = cos⁻¹(0.7) ≈ 44.4° rounded to the nearest tenth.

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Find the value of x in the given right triangle.

10⁷х x = [?]°

Enter your answer as a decimal rounded to the  nearest tenth.

The percent change in her savings account is 5.2%, meaning her savings account has increased by 5.2% due to the interest earned over 4 years.

To find the interest Thema would earn in 4 years, we can use the simple interest formula:

where I is the interest earned, P is the principal (initial deposit), r is the interest rate (as a decimal), and t is the time period in years.

In this case, we have P = $500, r = 0.013, and t = 4. Plugging in these values, we get:

I = [tex]($500) (0.013) (4) = $26[/tex]

So, Thema would earn $26 in interest over 4 years.

To determine the percent change in her savings account, we need to compare the amount she will have after 4 years to the initial deposit. After 4 years, her savings account will have:

A = P + I = $500 + $26 = $526

The percent change in her savings account can be calculated as:

percent change = (new amount - old amount) / old amount x 100%

Substituting the values, we get:

percent change = ($526 - $500) / $500 x 100% = 5.2%

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the answer is C. Because all your doing is counting how many numbers you see in the parentheses and seeing if it matches the numbers on the graph

To answer your question, I'll first explain what a power series is. A power series is a series of the form:

f(x) = a0 + a1(x-c) + a2(x-c)^2 + a3(x-c)^3 + ...

where a0, a1, a2, a3, ... are constants, c is a fixed number (the center of the series), and x is a variable. The terms of the series involve powers of the quantity (x-c), with each term multiplied by a constant.

Now, let's consider the function f(x) = 1/(1+x). This function can be represented by the power series:

1/(1+x) = 1 - x + x^2 - x^3 + ...

This series has a center of c = 0, and a0 = 1, a1 = -1, a2 = 1, a3 = -1, and so on. To write a partial sum consisting of the first 5 nonzero terms, we simply add up the first five terms:

1 - x + x^2 - x^3 + x^4

This is the partial sum we're looking for. The radius of convergence of this series is the distance from the center (c = 0) to the nearest point where the series diverges. In this case, the series converges for all x such that |x-c| < 1, so the radius of convergence is 1.

I hope this helps! Let me know if you have any other questions.

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The APR for this loan is 8%.

We have,

The formula for calculating APR is:

APR = (r/n) x m

where r is the stated annual interest rate, n is the number of times the interest is compounded in a year, and m is the number of payments made in a year.

In this case,

The loan is paid off in one lump sum at the end of the year, so there is only one payment made in a year (m = 1).

The stated annual interest rate is 8%, so r = 0.08.

We need to determine the value of n.

Since the loan is paid off in one lump sum at the end of the year, we can assume that the interest is compounded annually (n = 1).

Using the formula, we get:

APR = (0.08/1) x 1

APR = 0.08

Therefore,

The APR for this loan is 8%.

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Answer:

c = 12.2

Step-by-step explanation:

a squared + b squared = c sqared

7 squared + 10 squared = 49+ 100 = 149 = [tex]\sqrt{x} 149[/tex] = 12.2

There are 384 different ways or combinations that the thank-you note can look.

To find the total number of different ways that the thank-you note can look, we need to multiply the number of choices available for each decision point.

Gabrielle can choose between 2 kinds of cards, so there are 2 options. She has 8 kinds of envelopes to choose from, so there are 8 options. She only needs to use one of the 3 designs of first-class stamps, so there are 3 options. Finally, she has 8 colors of pen to choose from, so there are 8 options.

Therefore, the total number of different ways that the thank-you note can look is:

2 x 8 x 3 x 8 = 384

There are 384 different ways that Gabrielle can create the thank-you note by choosing a card, an envelope, a stamp design, and a pen color.

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a) The primary trigonometric ratios for angle A in standard position are;

sin(A) = 7/√74, cos(A) = -5/√74, and tan(A) = -7/5.

b) The primary trigonometric ratios for angle B are;

sin(B) = 7/√74, cos(B) = -5/√74, and tan(B) = 7/5.

c) A ≈ -56° and 2B ≈ -69°

a) To find the primary trigonometric ratios (sine, cosine, tangent) for angle A in standard position, we need to use the coordinates of the point (-5, 7). We can find the hypotenuse by using the Pythagorean theorem:

h = √((-5)² + 7²)

h = √74

Then, we can use the definitions of sine, cosine, and tangent:

sin(A) = y/h = 7/√74

cos(A) = x/h = -5/√74

tan(A) = y/x = -7/5

So , the primary trigonometric ratios for angle A in standard position are;

sin(A) = 7/√74, cos(A) = -5/√74, and tan(A) = -7/5.

b) To find an angle B with the same sine as angle A but different signs for the other two primary trigonometric ratios, we can use the fact that;

⇒ sin(B) = sin(A).

We also know that the signs of cos(B) and tan(B) will be different from those of cos(A) and tan(A), since angle B will be in a different quadrant.

Since sin(B) = sin(A), we know that the y-coordinate of angle B will be the same as that of angle A, namely 7.

We can then use the Pythagorean theorem to find the x-coordinate:

x = √(h² - y²)

x = √(74 - 49)

x = √25

x = 5

Since angle B is in a different quadrant from angle A, we need to adjust the signs of cos(B) and tan(B) accordingly.

We know that cos(B) will be negative, since angle B is in the third quadrant where x is negative.

We also know that tan(B) will be positive, since angle B is in the second quadrant where y is positive and x is negative.

Therefore, we have:

cos(B) = -x/h = -5/√74

tan(B) = y/x = 7/5

So the primary trigonometric ratios for angle B are;

sin(B) = 7/√74, cos(B) = -5/√74, and tan(B) = 7/5.

c) To find the measure of angle A, we can use the inverse tangent function:

A = tan⁻¹ (-7/5)

A ≈ -56.31°

To find the measure of angle 2B, we can use the double angle formula for sine:

sin(2B) = 2sin(B)cos(B)

We already know sin(B) and cos(B) from part (b), so we can plug them in:

sin(2B) = 2(7/√74)(-5/√74)

sin (2B) = -70/37

We can then use the inverse sine function to find the measure of angle 2B:

2B = sin⁻¹(-70/37)

2B ≈ -68.59°

So, to the nearest degree, we have A ≈ -56° and 2B ≈ -69°.

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Answer:

The measure of an angle formed by two chords intersecting inside a circle is equal to half the sum of the measures of the intercepted arcs associated with the angle and its vertical angle counterpart.

Let x be the measure of arc UW.

36° = (1/2)(26° + x°)

72° = 26° + x

x = 46°

So the measure of arc UW is 46°.

The theoretical probability of the spinner not landing on blue would be = 0.625. That is option B.

To calculate the theoretical probability of the given event, the formula that should be used is given as follows:

Probability = possible outcome/sample space

The possible outcome for other colours apart from blue = 5

The sample space = 8

Therefore probability = 5/8 = 0.625

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Answer: 4 Which ratio represents the cosine of angle A in the right triangle below? ... Which ratio is always equivalent to the sine of ∠A?

Step-by-step explanation:

The mean is 2.48, the variance is 1.5844, and the standard deviation is 1.26.

To calculate the mean, variance, and standard deviation, we need to first construct a probability distribution table:

where f(x) is the frequency of weeks with x radios sold divided by the total number of weeks (100).

Using this table, we can calculate the mean as:

μx = ∑(x * f(x)) = (00.05) + (10.17) + (20.17) + (30.49) + (40.09) + (50.03) = 2.48

To calculate the variance, we use the formula:

σx2 = ∑((x - μx)2 * f(x)) = (0-2.48)2 * 0.05 + (1-2.48)2 * 0.17 + (2-2.48)2 * 0.17 + (3-2.48)2 * 0.49 + (4-2.48)2 * 0.09 + (5-2.48)2 * 0.03 = 1.5844

Finally, we can calculate the standard deviation as:

σx = √σx2 = √1.5844 = 1.26

Therefore, the mean is 2.48, the variance is 1.5844, and the standard deviation is 1.26.

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The area of the rectangle is √32 x √45.

Option A is the correct answer.

We have,

From the figure,

Length = √32

Width = √45

Now,

The area of the rectangle.

= Length x width

= √32 x √45

Thus,

The area of the rectangle is √32 x √45.

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The real speed of the vessel, adjusted to the closest tenth, is 20.4 hitches.

To unravel this issue, we got to break down the boat's speed into its components. We will use trigonometry to discover the eastbound and northward components of the boat's speed.

First, let's draw a diagram:

              N

                |

                |

     NW 5 |   E15

                |

--------------|---------------> E

                |

                |

                |

               S

From the chart, we will see that the northward component of the boat's speed is:

16 hitches * sin(15°) = 4.16 knots

And the eastbound component of the boat's speed is:

16 ties * cos(15°) = 15.38 knots

Next, we ought to discover the whole northward and eastbound speed of the vessel by including the boat's speed components to the current's speed components. We are able moreover utilize trigonometry to discover the northward and eastbound components of the current's speed:

5 hitches * sin(45°) = 3.54 ties northward

5 hitches * cos(45°) = 3.54 ties eastbound

So, the overall northward speed of the pontoon is:

4.16 ties + 3.54 hitches = 7.7 hitches northward

And the full eastbound speed of the pontoon is:

15.38 ties + 3.54 ties = 18.9 ties eastbound

Presently, we will utilize the Pythagorean hypothesis to discover the greatness of the boat's speed vector:

sqrt((7.7 knots)^2 + (18.9 knots)^2) = 20.4 ties

In this manner, the real speed of the vessel, adjusted to the closest tenth, is 20.4 hitches.

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Check the picture below.

so the circle has a radius of 3 and a center at (-2 , 1)

[tex]\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{-2}{h}~~,~~\underset{1}{k})}\qquad \stackrel{radius}{\underset{3}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - (-2) ~~ )^2 ~~ + ~~ ( ~~ y-1 ~~ )^2~~ = ~~3^2\implies (x+2)^2 + (y-1)^2 = 9[/tex]

Answer:

A, x ≥ 7x ≥ 7

D, 5 + 6 ≠ 15

Step-by-step explanation:

An inequality statement includes one or more of one of the following symbols: <, ≤, >, ≥, ≠

So we can see that options A and D include the signs "greater than or equal to" symbols and D includes a "not equal to" symbol.

Hope this helps! :)

Answer:

The question was not complete I can't help until it is

Step-by-step explanation:

Answer:

x + y = -12 + 30 = 18

Step-by-step explanation:

To solve this system of equations, we can use the elimination method. Multiplying the first equation by 3, we get:

9x + 3y = -18

Adding this to the second equation, we eliminate the x terms:

9x + 3y = -18

-3x - 4y = -12

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

-y = -30

Solving for y, we get y = 30. Substituting this back into either equation, we can solve for x:

3x + 30 = -6

3x = -36

x = -12

Therefore, x + y = -12 + 30 = 18.

The zero vector in P4 is the polynomial 0(x) = 0, which has even degree. The set of all polynomials in P4 of even degree is closed under addition.

The set of all polynomials in P4 of even degree satisfies all three conditions, it is a subspace of P4.

(a) The set of all polynomials in P4 of even degree is a subspace of P4.

To prove this, we need to show that it satisfies the three conditions for a subspace:

i) It contains the zero vector: The zero vector in P4 is the polynomial 0(x) = 0, which has even degree, so it is contained in the set of all polynomials in P4 of even degree.

ii) It is closed under addition: Let p(x) and q(x) be two polynomials in P4 of even degree. Then, p(x) + q(x) is also a polynomial of even degree, since the sum of two even numbers is even. Therefore, the set of all polynomials in P4 of even degree is closed under addition.

iii) It is closed under scalar multiplication: Let p(x) be a polynomial in P4 of even degree, and let c be a scalar. Then, cp(x) is also a polynomial of even degree, since multiplying an even number by a scalar yields an even number. Therefore, the set of all polynomials in P4 of even degree is closed under scalar multiplication.

Since the set of all polynomials in P4 of even degree satisfies all three conditions, it is a subspace of P4.

(b) The set of all polynomials of degree 3 is not a subspace of P4.

To prove this, we only need to show that it does not satisfy the first condition for a subspace:

i) It contains the zero vector: The zero vector in P4 is the polynomial 0(x) = 0, which has degree 0, not degree 3. Therefore, the set of all polynomials of degree 3 does not contain the zero vector and is not a subspace of P4.

(c) The set of all polynomials p in P4 such that p(0) = 0 is a subspace of P4.

To prove this, we need to show that it satisfies the three conditions for a subspace:

i) It contains the zero vector: The zero vector in P4 is the polynomial 0(x) = 0, which satisfies 0(0) = 0, so it is contained in the set of all polynomials p in P4 such that p(0) = 0.

ii) It is closed under addition: Let p(x) and q(x) be two polynomials in P4 such that p(0) = 0 and q(0) = 0. Then, (p+q)(0) = p(0) + q(0) = 0, so p+q is also a polynomial in P4 such that (p+q)(0) = 0. Therefore, the set of all polynomials p in P4 such that p(0) = 0 is closed under addition.

iii) It is closed under scalar multiplication: Let p(x) be a polynomial in P4 such that p(0) = 0, and let c be a scalar. Then, (cp)(0) = c(p(0)) = c(0) = 0, so cp is also a polynomial in P4 such that (cp)(0) = 0. Therefore, the set of all polynomials p in P4 such that p(0) = 0 is closed under scalar multiplication.

Since the set of all polynomials p in P4 such that p(0) = 0 satisfies all three conditions, it is a subspace of P4.

(d) The set of all polynomials in P4 having at least one real root is not a subspace of P4.

To prove this, we only need

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Note that an and bn are only non-zero for odd values of n, since f(x) is an odd function.

To sketch the function f(x) for (-21 < t < 4π), we need to extend the definition of f(x) to this interval. Since f(x) has a period of 2π, we can extend the function by repeating it every 2π. Thus, for (-21 < t < 0), we have:

f(x) = f(x + 2π) = f(x - 2π)

For (0 ≤ t < 2π), we use the original definition of f(x).

For (2π ≤ t < 4π), we have:

f(x) = f(x - 2π)

With  this extension, we can now sketch the function f(x) as follows:

              |\

              | \

              |  \

              |   \

              |    \

              |     \______

              |           /\

              |          /  \

              |         /    \

_______________|________/______\____________

             -21       0      2π     4π

Now let's find the Fourier series of f(x). The Fourier series is given by:

f(x) = a0/2 + Σ[an cos(nωx) + bn sin(nωx)]

where ω = 2π/T is the fundamental frequency, T is the period, and an and bn are the Fourier coefficients, given by:

an = (2/T) ∫[f(x) cos(nωx)] dx

bn = (2/T) ∫[f(x) sin(nωx)] dx

In this case, T = 2π, so ω = 1. The Fourier coefficients can be calculated as follows:

a0 = (1/π) ∫[f(x)] dx

= (1/π) [∫[x dx] from 0 to π/2 + ∫[½π(½π -½π(π x-2x(½π≤x≤2π)) dx] from π/2 to 2π]

= (1/π) [π²/4 + ½π²/3 - π³/8]

= (π/4) - (π²/24)

an = (2/π) ∫[f(x) cos(nωx)] dx

= (2/π) ∫[x cos(nωx)] dx from 0 to π/2 + (2/π) ∫[½π(½π -½π(π x-2x(½π≤x≤2π))) cos(nωx)] dx from π/2 to 2π

= [2/(nπ)] [(-1)^n - 1] + [2/(nπ)] [(-1)^n - 1/3]

bn = (2/π) ∫[f(x) sin(nωx)] dx

= (2/π) ∫[x sin(nωx)] dx from 0 to π/2 + (2/π) ∫[½π(½π -½π(π x-2x(½π≤x≤2π))) sin(nωx)] dx from π/2 to 2π

= [2/(nπ)] [1 - (-1)^n] + [2/(nπ)] [2/π - (1/π)cos(nπ) + (1/3π)cos(3nπ)]

Note that an and bn are only non-zero for odd values of n, since f(x) is an odd function.

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