Find the Surface Area?

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

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

Answer:  16.8 feet

Note: your teacher may not want you to enter "feet" and instead may just want the number only.

===================================================

Work Shown:

tan(angle) = opposite/adjacent

tan(40) = h/20

20*tan(40) = h

h = 20*tan(40)

h = 16.7819926 which is approximate

h = 16.8

The streetlamp is approximately 16.8 ft tall.

When using your calculator, make sure it's in degree mode. One way to check is to compute something like tan(45) and you should get 1 as a result.

Step-by-step explanation:

The inverse of matrix A is:

A^-1 = |-3 -2|

|-2 -1|

Using Adjoint of a Matrix A=2 -1 0

0 1 2

1 1 0

The first step is to find the determinant of the matrix A:

|A| = 2(10 - 21) - (-1)(10 - 12) + 0(11 - 10)

= 4 + 2 + 0

= 6

Next, we need to find the adjoint of matrix A, which is the transpose of its cofactor matrix. The cofactor matrix is obtained by taking the determinant of the submatrix obtained by removing each element of the original matrix in turn and multiplying it by (-1)^(i+j), where i and j are the row and column indices of the removed element, respectively.

Cofactor matrix of A is

| 1 2 -1|

|-2 0 2|

|-1 2 1|

Taking the transpose of the cofactor matrix, we get the adjoint matrix of A as follows:

A^T = | 1 -2 -1 |

| 2 0 2 |

|-1 2 1 |

To find the inverse of A, we use the formula:

A^-1 = (1/|A|) A^T

Substituting the values, we get:

A^-1 = (1/6) | 1 -2 -1 |

| 2 0 2 |

|-1 2 1 |

Using Gauss Jordan A= |1 3|

|2 5|

We can find the inverse of a matrix using Gauss-Jordan elimination method as follows:

|1 3|1 0| |1 3|0 1|

|2 5|0 1|-> |0 1|-2/3 -1/3|

Therefore, the inverse of matrix A is:

A^-1 = |-2/3 -1/3|

| 1/3 1/3|

Using Equations and Identity Matrix A= |1 -2|

|2 -3|

We can find the inverse of a matrix A using the equations AX=I, where I is the identity matrix and X is the matrix that represents the inverse of A. The solution is given by:

|1 -2| |x11 x12| |1 0|

|2 -3| |x21 x22| = |0 1|

Multiplying the matrices, we get:

x11 = -3

x12 = -2

x21 = -2

x22 = -1

Therefore, the inverse of matrix A is:

A^-1 = |-3 -2|

|-2 -1|

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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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Car A can travel 1.25 times the distance car B can travel when both cars use 1 gallons of gas.

In Mathematics and Geometry, a proportional relationship refers to a type of relationship that produces equivalent ratios and it can be modeled or represented by the following mathematical equation:

y = kx

Where:

Next, we would determine the constant of proportionality (k) for car A by using various data points as follows:

Constant of proportionality, k = y/x

Constant of proportionality, k = 50/2

Constant of proportionality, k = 25.

y = 25x = 25(1) = 25 miles.

For car B, the constant of proportionality (k) is given by;

Constant of proportionality, k = 60/3

Constant of proportionality, k = 20.

y = 20x = 20(1) = 20 miles.

Difference = 25/20 = 1.25.

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Missing information:

The question is incomplete and the complete question is shown in the attached picture.

a.   (-∞,3) - f(x) is decreasing

   (3,∞) - f(x) is rising.

b. Local minima at x=3. No local Minima

c. The function f(x)=x²-6x is always concave upwards.

d. Concave up and does not change concavity, so, No Inflection points.

f(x)= x²-6x

f'(x) = 2x-6

f"(x) = 2

Critical point f'(x)=0

2x-6=0

x=3

Thus, we have two sub intervals over the entire number line. (-∞,3) , (3,∞)

a) sub-interval      x-value           f'(x)                           verdict

         (-∞,3)                1             2(1)-6=-4<0             f(x) is decreasing  

          (3,∞)                4            2(4)-6=2>0              f(x) is increasing

b) At x=3; Before x=3, f(x) decreasing and after x=3, f(x)

is increasing, thus Local minima at x=3.

No local Minima

c) Since f"(x)=2 Always, the function f(x)=x²-6x is always concave upwards

d) Inflection points

Since graph of function f(x)=x²-6x have only been concave up and does not change concavity,

No Inflection points.

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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]

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 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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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 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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As per the given situation, it will take 15 weeks for Marlon and Cassandra to have the same balance in their accounts is 600−25x=45+12x. The correct option is A.

We can start by setting up an equation that represents the balance of each person after x weeks.

After x weeks, Marlon's balance would be:

600 - 25x

After x weeks, Cassandra's balance would be:

45 + 12x

To find the number of weeks it will take until they have the same balance, we can set these two expressions equal to each other:

600 - 25x = 45 + 12x

Simplifying this equation, we can combine like terms:

600 = 45 + 37x

Subtracting 45 from both sides:

555 = 37x

Dividing both sides by 37:

x = 15

Therefore, it will take 15 weeks for Marlon and Cassandra to have the same balance in their accounts.

Thus, the correct equation is A) 600−25x=45+12x.

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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: 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:

a)An equation of / is y = 2x + 1.

b) The exact length of AC is 2√17.

c) The exact length of BM is √17.

d) The coordinates of B are (5, 4).

(a) Since / is the perpendicular bisector of AC, it passes through the midpoint M of AC. The coordinates of M can be found by taking the average of the x-coordinates and the average of the y-coordinates of A and C, respectively:

x-coordinate of M = (-3 + 5)/2 = 1

y-coordinate of M = (4 + 2)/2 = 3

Therefore, the coordinates of M are (1, 3). Since / is perpendicular to AC, its slope is the negative reciprocal of the slope of AC:

slope of AC = (2 - 4)/(5 - (-3)) = -1/2

slope of / = 2 (negative reciprocal of -1/2)

Using the point-slope form of the equation of a line with the point M, we get:

y - 3 = 2(x - 1)

Simplifying and rearranging, we get:

y = 2x + 1

Therefore, an equation of / is y = 2x + 1.

(b) The length of AC can be found using the distance formula:

AC = √[(5 - (-3))^2 + (2 - 4)^2] = √64 + 4 = √68 = 2√17

Therefore, the exact length of AC is 2√17.

(c) Since BM is a median of triangle ABC, it divides AC into two equal parts. Therefore, BM has length half of AC:

BM = AC/2 = (1/2)(2√17) = √17

Therefore, the exact length of BM is √17.

(d) Since B lies on line /, its x-coordinate is 5 (since / passes through point (5, 2)). To find the y-coordinate, we can substitute x = 5 into the equation of /:

y = 2x + 1 = 2(5) + 1 = 11

Therefore, the coordinates of B are (5, 11).

Alternatively, we can use the fact that B lies on the circumcircle of triangle ABC (since angle ABC is a right angle) to find its coordinates. The circumcenter of triangle ABC is the midpoint of AC (which is also the intersection of the perpendicular bisectors of AC), so the coordinates of the circumcenter are (1, 3). The radius of the circumcircle is half of AC, so it is √17. Therefore, the equation of the circumcircle is:

(x - 1)^2 + (y - 3)^2 = 17

Since B lies on the circumcircle, its coordinates satisfy this equation. Substituting x = 5, we get:

(5 - 1)^2 + (y - 3)^2 = 17

16 + (y - 3)^2 = 17

(y - 3)^2 = 1

y - 3 = ±1

Solving for y, we get y = 2 or y = 4. Since B lies on line /, its y-coordinate must be greater than 2, so y = 4. Therefore, the coordinates of B are (5, 4).

(e) There are two possible positions of B, which are reflections of each other across the line /. One position has coordinates (5, 11), as found in part (d), and the other has coordinates (-3, 2). To find the coordinates of the second position, we can reflect point A across line / to get the point B':

B' has the same x-coordinate as A, which is -3. To find its y-coordinate, we can use the equation

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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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The size of the angle X is calculated to be equal to 47.4° to the nearest tenth of degree using trigonometric ratios cosine

The trigonometric ratios is concerned with the relationship of an angle of a right-angled triangle to ratios of two side lengths.

The basic trigonometric ratios includes;

sine, cosine and tangent.

Given the triangle STU;

cos X = ST/SU {adjacent/hypotenuse}

cos X = 8.8/13

X = cos⁻¹(8.8/13) {cross multiplication}

X = 47.3963°

Therefore, the measure of the angle X is calculated to be equal to 47.4° to the nearest tenth of degree using trigonometric ratios cosine

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Answer:

I beleive c. i had the same question last year, and i believe i got the answer right. so sorry if its wrong. hope this helps.

Step-by-step explanation:

Adeline has 60 notebooks and 4 children in the given case.

Let's assume that Adeline has x notebooks and y children.

According to the problem, if she gives 13 notebooks to each child, she will have 8 left. This can be expressed as:

x - 13y = 8 --- equation 1

Also, if she gives 15 notebooks to each child, she will have zero left. This can be expressed as:

x - 15y = 0 --- equation 2

We can solve these equations simultaneously to find the values of x and y.

Multiplying equation 1 by 15 and equation 2 by 13, we get:

15x - 195y = 120 --- equation 3

13x - 195y = 0 --- equation 4

Subtracting equation 4 from equation 3, we get:

2x = 120

x = 60

Substituting the value of x in equation 2, we get:

60 - 15y = 0

y = 4

Therefore, Adeline has 60 notebooks and 4 children.

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Adeline has a box full of notebooks. If she gives 13 notebooks to each child, she will have 8 left. If she gives 15 notebooks to each child, she will have zero left. How many notebooks and how many children does Adeline have?

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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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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Answer:

h = -1.25

Step-by-step explanation:

isolate the variable by dividing each side by factors that don't contain the variable.

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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The cash payment equivalent to making monthly payments of $1074 for 5 years with an interest rate of 11% per annum compounded semi-annually is $49,701.

To find the cash payment, we can use the Present Value of the Annuity formula:

PV = PMT * [(1 - (1 + r)⁻ⁿ) / r]

Where:
PV = Present Value (cash payment)
PMT = Monthly payment ($1074)
r = Monthly interest rate
n = Number of payments (5 years * 12 months = 60)

First, we need to find the monthly interest rate. Since the interest is compounded semi-annually, we'll divide the annual interest rate by 2, and then convert it to a monthly rate:

Semi-annual interest rate = 11% / 2 = 5.5% = 0.055
[tex](1 + 0.055)^{(1/6)} - 1[/tex] ≈ 0.008944 (rounded to 6 decimal places)

Now we can plug the values into the Present Value of Annuity formula:

PV = 1074 * [(1 - (1 + 0.008944)⁻⁶⁰) / 0.008944]
PV ≈ 1074 * [1 - (0.5861) / 0.008944]
PV ≈ 1074 * 46.2768
PV ≈ 49,701

The cash payment equivalent is approximately $49,701 (rounded to the nearest dollar).

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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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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 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.

[tex]$f(x) = q_a$[/tex] for all [tex]$x \in (a-1, a+1)$[/tex]. Since a was arbitrary, it follows that f is constant on any open interval. Since f is continuous, it follows that f is constant on [tex]$\mathbb{R}$[/tex].

Let a be any real number in R. Since f is continuous, the intermediate value theorem implies that the image of any closed interval under f is also an interval. Therefore, [tex]$f([a-1, a+1])$[/tex] is an interval in [tex]$\mathbb{Q}$[/tex]. Since the only intervals in [tex]$\mathbb{Q}$[/tex] are single points, [tex]$f([a-1, a+1]) = {q_a}$[/tex] for some rational number [tex]$q_a$[/tex].

Now let b be any real number with [tex]$b > a+1$[/tex]. By the intermediate value theorem, there exists some [tex]$x \in [a, b]$[/tex] such that [tex]$f(x) = \frac{q_a+q_b}{2}$[/tex]. But since f takes only rational values, [tex]$f(x) = q_a$[/tex]. This argument applies to all real numbers b with [tex]$b > a+1$[/tex], so [tex]$f(x) = q_a$[/tex] for all [tex]$x > a+1$[/tex]. Similarly, we can show that [tex]$f(x) = q_a$[/tex] for all [tex]$x < a-1$[/tex].

Therefore, [tex]$f(x) = q_a$[/tex] for all [tex]$x \in (a-1, a+1)$[/tex]. Since a was arbitrary, it follows that f is constant on any open interval. Since f is continuous, it follows that f is constant on [tex]$\mathbb{R}$[/tex].

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