look at image below/above

Answer:

4x + 3y >/= to 12

Step-by-step explanation

The required inequality is 4x + 3y ≤ 12 which represents the number of red peppers and tomatoes.

Inequality is defined as mathematical statements that have a minimum of two terms containing variables or numbers that are not equal.

Red peppers and tomatoes for salsa cost no more than $12. Red peppers are $4 per pound, while tomatoes are $3 per pound.

To write the inequality, we can let x represent the number of pounds of red peppers and y represents the number of pounds of tomatoes.

The total cost of the red peppers is 4x dollars and the total cost of the tomatoes is 3y dollars.

We know that the total cost of both the red peppers and tomatoes combined is 12 dollars.

Therefore, the inequality to represent is 4x + 3y ≤ 12.

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The set of all diagonal n × n matrices is a subspace of Mnn3.

Given, we need to use the Subspace Test to determine which of the sets are subspaces of Mnn3 and the sets are: The set of all diagonal n × n matrices.

The set of all n x n matrices A such that det(A) = 0.The set of all n x n matrices A such that tr(A) = 0.The set of all symmetric n x n matrices.

Subspace Test: A nonempty subset H of a vector space V is a subspace of V if for every u and v in H and every scalar c, the vector cu + v is in H.

The set of all diagonal n × n matrices.Here, we have to prove that the set of all diagonal matrices is a subspace of Mnn3.

Let A and B be two diagonal matrices. Then, A + B is also a diagonal matrix. Since the diagonal elements of A + B are equal to the sums of the corresponding diagonal elements of A and B. So, the set of all diagonal matrices is closed under addition. Let A be a diagonal matrix and c be a scalar.

Then, cA is also a diagonal matrix. Since the diagonal elements of cA are equal to the product of c and corresponding diagonal elements of A. So, the set of all diagonal matrices is closed under scalar multiplication.

Therefore, the set of all diagonal matrices is a subspace of Mnn3. Hence, the main answer is: a. The set of all diagonal n × n matrices is a subspace of Mnn3.

We have used the subspace test to prove that the set of all diagonal n × n matrices is a subspace of Mnn3. The subspace test is used to verify whether a given subset of a vector space is a subspace or not.

We have proved that the set of all diagonal matrices is closed under addition and scalar multiplication.

The diagonal elements of A + B are equal to the sums of the corresponding diagonal elements of A and B. And the diagonal elements of cA are equal to the product of c and corresponding diagonal elements of A.

Therefore, the set of all diagonal matrices satisfies all the three properties required for a subset to be a subspace.

Hence, the set of all diagonal n × n matrices is a subspace of Mnn3.

In conclusion, we can say that the set of all diagonal n × n matrices is a subspace of Mnn3 as it satisfies the subspace test.

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Answer: 12 Hershey bars is the better buy

Step-by-step explanation: If you get 24 Hershey bars (12 x 2) your price will be $6.24 and if you get 24 Butterfingers (8 x 3) your price will be $6.72

Answer:

he should chose compounded annually because he would have and extra 32.12 dollars at the end of the 4 years

Step-by-step explanation:

hope this helps

Answer:

8 lol

Step-by-step explanation:

Answer:

8

Step-by-step explanation:

The probability of getting a sample mean grand prize of exactly 84 thousand dollars is 2/6, which reduces to 1/3.

The mean of the four grand prizes is (34+54+74+94)/4 = 64. Therefore, the probability of selecting two envelopes and receiving an average of exactly 84 thousand dollars is the same as the probability of selecting two envelopes and getting a sum of 168 thousand dollars (since the sum of two values divided by two is equal to their average).

There are six possible combinations of two envelopes, as stated in the question. To find the probability of getting a sum of 168 thousand dollars, we need to find the number of combinations that add up to 168. These combinations are (74,94) and (94,74). Therefore, the probability of getting a sample mean grand prize of exactly 84 thousand dollars is 2/6, which reduces to 1/3.

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Answer:

13x25=455

Step-by-step explanation:

Answer:

x = -1, y = 1

Step-by-step explanation:

Answer:

720 cherries

Step-by-step explanation:

Answer:

720

Step-by-step explanation:

Cherries Cherie pie

240 = 3

X = 9

The known number of cherries will be given as x then u cross multiply

3×X=240×9

3X=2,160

Divide both sides by 3

3/3X=2,160/3

X=2,160/3

X=720

This implies that the number of cherries needed to make 9 cherry pie is 720

I hope this help

The measure of the central angle for the group that likes the program would be approximately 125 degrees in a circle graph.

A circle graph, also known as a pie chart, is a type of graph that is used to represent data as a circle divided into sectors. The size of each sector represents the proportion or percentage of the data that falls into that category.

To find the measure of the central angle for the group that likes the program, we first need to calculate the total number of students surveyed:

Students total:

190 + 135 + 220 = 545

The proportion of pupils who approve of the programme needs to be determined next:

Percentage of students who like the program = (190/545) x 100% ≈ 34.86%

To find the measure of the central angle, we need to multiply the percentage by 360 degrees:

Measure of central angle = 34.86% x 360° ≈ 125.49°

Rounding this answer to the nearest whole number gives:

Measure of central angle ≈ 125°

Therefore, the measure of the central angle for the group that likes the program would be approximately 125 degrees in a circle graph.

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Answer:

One student  ate 21 sticks of gum in math class.  Each package of gum has 3 sticks of gum. How many packages of gum did he use?

21 /3 = 7 packages of gum

Step-by-step explanation:

John was having a contest to see how many pies he could eat in 5 minutes.

Every minute John can eat 12 pies.  So how many times did he eat in 5 minutes?

12 pies x 5 mins = 60 pies in the 5 minutes

Answer:

n - 4 :)

Step-by-step explanation:

A number = n

So, 4 less than n is simply n - 4 :)

Based on the shipping rate and the price of the t-shirts, it can be concluded Sam ordered a total of 6 t-shirts.

5.99 + x12.50 = 80.99

x represents the number of t-shirts she bough, which is the focus of the equation.

Let's solve it:

Sam ordered 6 t-shirts.

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The general solution form for the recurrence tn = 120,-2 - 166n-3 + 2.

Given a recurrence relation tn = 120,-2 - 166n-3 + 2 we have to derive the general solution form for the recurrence sequence.

We have the recurrence relation tn = 120,-2 - 166n-3 + 2

We need to find the solution for the recurrence relation.

Associated Homogeneous Equation: First, we need to find the associated homogeneous equation.

                                     tn = -166n-3 …..(i)

The characteristic equation is given by the following:tn = arn. Where ‘a’ is a constant.

We have tn = -166n-3..... (from equation i)ar^n = -166n-3

                                Let's assume r³ = t.

Then equation i becomes ar^3 = -166(r³) - 3ar^3 + 166 = 0ar³ = 166

Hence r = ±31.10.3587Complex roots: α + iβ, α - iβ

Characteristics Polynomial:

                   So, the characteristic polynomial becomes(r - 31)(r + 31)(r - 10.3587 - 1.7503i)(r - 10.3587 + 1.7503i) = 0

The general solution of the Homogeneous equation:

Now we have to find the general solution of the homogeneous equation.

                  tn = C1(-31)n + C2(31)n + C3 (10.3587 + 1.7503i)n + C4(10.3587 - 1.7503i)

                        nWhere C1, C2, C3, C4 are constants.

Computing a Particular Solution:

                Now we have to compute the particular solution.

                                  tn = 120-2 - 166n-3 + 2

Here the constant term is (120-2) + 2 = 122.

The solution of the recurrence relation is:tn = A122Where A is the constant.

The General Solution of Non-Homogeneous Equation:

        The general solution of the non-homogeneous equation is given bytn = C1(-31)n + C2(31)n + C3 (10.3587 + 1.7503i)n + C4(10.3587 - 1.7503i)n + A122

Hence, we have derived the general solution form for the recurrence tn = 120,-2 - 166n-3 + 2.

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Answer: sorry I dont know!

Step-by-step explanation:

Answer:

350 miles per week

Step-by-step explanation:

That'd be 70 miles each day for 5 days, or 350 miles per week.

Answer:

36 minutes is the highest

Step-by-step explanation:

There are 7 motorcycles and 4 cars in the parking lot

An equation is an expression composed of variables and numbers linked together by mathematical operations.

Let x represent the number of motorcycle and y represent the car. There could be a 2 wheel motorcycle or a 4 wheel car.

There are 11 vehicles in the parking lot, hence:

x + y = 11    (1)

There are 30 wheels, hence:

2x + 4y = 30    (2)

From both equations:

x = 7, y = 4

There are 7 motorcycles

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Answer: 5 times

Step-by-step explanation: 11 goes into 50 four times. That means that the first four times the carousel runs, 11 kids will be on it. But since six ungrateful children were left out, the carousel will need to run one extra time so those brats can get a turn.

The factor form of the quadratic equations are listed below:

In this problem we find ten cases of quadratic equations that must be factorized by means of algebra properties, now proceed to determine the factor form of each case:

Case 1

8 · x² - 8 · x - 336

8 · (x² - x - 42)

8 · (x - 7) · (x + 6)

Case 2

36 · x² - 12 · x + 45 · x - 15

12 · x · (3 · x - 1) + 15 · (3 · x - 1)

(12 · x + 15) · (3 · x - 1)

3 · (4 · x + 5) · (3 · x - 1)

Case 3

- 12 · x² - 10 · x + 8

- 2 · (6 · x² + 5 · x - 4)

- 12 · [x² + (5 / 6) · x - 2 / 3]

- 12 · (x - 1 / 2) · (x + 4 / 3)

Case 4

16 · x² - 100

4 · (4 · x² - 25)

4 · (2 · x - 5) · (2 · x + 5)

Case 5

16 · x² - 9

(4 · x - 3) · (4 · x + 3)

Case 6

25 · x² + 10 · x - 75 · x - 30

5 · x · (5 · x + 2) - 15 · (5 · x + 2)

(5 · x - 15) · (5 · x + 2)

5 · (x - 3) · (5 · x + 2)

Case 7

6 · x² - 72 · x + 162

6  · (x² - 12 · x + 27)

6 · (x - 9) · (x - 3)

Case 8

- 21 · x² + 84 · x + 6 · x - 24

- 21 · x · (x - 4) + 6 · (x - 4)

(- 21 · x + 6) · (x - 4)

- 3 · (7 · x + 2) · (x - 4)

Case 9

36 · x² - 60 · x + 25

36 · [x² - (5 / 3) · x + 25 / 36]

36 · (x - 5 / 6)²

Case 10

490 · x² - 810

490 · (x² - 810 / 490)

490 · (x - 9 / 7) · (x + 9 / 7)

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Answer:

it is a downward

discline

Step-by-step explanation:

Answer:

Slope is how steep a line is. It  measured using the formula rise over run or Change in y/Change in X

Step-by-step explanation:

Answer:The correct awnser is D

Step-by-step explanation:

Answer:

its D

Step-by-step explanation:

(a) Since each game can result in a win, loss, or tie, the range of X is Sx = {0,1,2,3,4}. Let p be the probability of winning a game, so that the probability of losing a game is also p, and the probability of tying a game is 1-2p.

Then, the PMF of X is:

P(X = 0) = p^2

P(X = 1) = 2p(1-p)

P(X = 2) = (1-p)^2 + 2p^2

P(X = 3) = 2p(1-p)

P(X = 4) = p^2

(b) Let Y be the total number of points earned over n games, so that Y = X1 + X2 + ... + Xn. Then, E[X] = 2p + p + 0 = 3p, and Var[X] = 4p(1-p) + p(1-p) = 5p(1-p). Since X is an iid random variable, we have E[Y] = nE[X] = 3np and Var[Y] = nVar[X] = 5np(1-p). We can also write Y as a function of X as:

Y = 2X1 + 2X2 + ... + 2Xn + X1 + X2 + ... + Xn

= 3(X1 + X2 + ... + Xn)

= 3nX

(c) The MGF of X is given by:

Oxi(s) = E[e^(sX)] = p^2 e^(2s) + 2p(1-p) e^s + (1-p)^2 + 2p^2 e^(2s) + 2p(1-p) e^s + p^2 e^(2s)

= (p^2 + 2p(1-p) e^s + p^2 e^(2s))^2

= (1 - 2p + 4p^2 e^s)^n

The MGF of Y is given by:

Oyi(s) = E[e^(sY)] = E[e^(3nsX)]

= Oxi(3ns)

= (1 - 2p + 4p^2 e^(3ns))^n

(d) Let Z be the number of losses in n games, so that Z has a binomial distribution with parameters n and p. Then, we want to estimate the probability P(Z < n/2). By the CLT, Z can be approximated by a normal distribution with mean np and variance np(1-p), so that:

Z ~ N(np, np(1-p))

Using the continuity correction, we have:

P(Z < n/2) = P(Z < 50) ≈ P(Z < 50.5)

= P((Z - np)/sqrt(np(1-p)) < (50.5 - np)/sqrt(np(1-p)))

= Φ((50.5 - np)/sqrt(np(1-p)))

where Φ is the cumulative distribution function of the standard normal distribution. Substituting n = 100 and p = 1/3, we have:

P(Z < 50) ≈ Φ((50.5 - 100/3)/sqrt(100/3 * 2/3))

≈ Φ(-1.83)

≈ 0.0336

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Answer:

The diagram shows measures of some of the line segments in APQR. 4.5 /62.58° Q Length of QS P Area of APQR S ㅏ Use the given information to find the measures of the unknown angles and segment shown below. Round your answers to the nearest whole number. Length of PS units 6.4 -5.0- units R units²The diagram shows measures of some of the line segments in APQR. 4.5 /62.58° Q Length of QS P Area of APQR S ㅏ Use the given information to find the measures of the unknown angles and segment shown below. Round your answers to the nearest whole number. Length of PS units 6.4 -5.0- units R units²The diagram shows measures of some of the line segments in APQR. 4.5 /62.58° Q Length of QS P Area of APQR S ㅏ Use the given information to find the measures of the unknown angles and segment shown below. Round your answers to the nearest whole number. Length of PS units 6.4 -5.0- units R units²The diagram shows measures of some of the line segments in APQR. 4.5 /62.58° Q Length of QS P Area of APQR S ㅏ Use the given information to find the measures of the unknown angles and segment shown below. Round your answers to the nearest whole number. Length of PS units 6.4 -5.0- units R units²The diagram shows measures of some of the line segments in APQR. 4.5 /62.58° Q Length of QS P Area of APQR S ㅏ Use the given information to find the measures of the unknown angles and segment shown below. Round your answers to the nearest whole number. Length of PS units 6.4 -5.0- units R units²vvvThe diagram shows measures of some of the line segments in APQR. 4.5 /62.58° Q Length of QS P Area of APQR S ㅏ Use the given information to find the measures of the unknown angles and segment shown below. Round your answers to the nearest whole number. Length of PS units 6.4 -5.0- units R units²The diagram shows measures of some of the line segments in APQR. 4.5 /62.58° Q Length of QS P Area of APQR S ㅏ Use the given information to find the measures of the unknown angles and segment shown below. Round your answers to the nearest whole number. Length of PS units 6.4 -5.0- units R units²The diagram shows measures of some of the line segments in APQR. 4.5 /62.58° Q Length of QS P Area of APQR S ㅏ Use the given information to find the measures of the unknown angles and segment shown below. Round your answers to the nearest whole number. Length of PS units 6.4 -5.0- units R units²The diagram shows measures of some of the line segments in APQR. 4.5 /62.58° Q Length of QS P Area of APQR S ㅏ Use the given information to find the measures of the unknown angles and segment shown below. Round your answers to the nearest whole number. Length of PS units 6.4 -5.0- units R units²v

Step-by-step explanation:

its E and A

Answer:

The solutions to the inequality are B, E, and F: 25, 36, and 9.

Step-by-step explanation:

The solutions to the inequality √x < 10 are the values of x that make the inequality true when plugged in. To find these values, we can square both sides of the inequality:

√x < 10

x < 10^2 = 100

So, the values of x that make the inequality true are those that are less than 100. The options that satisfy this condition are:

A. 100 (not a solution)

B. 25 (solution)

C. -100 (not a solution)

D. 105 (not a solution)

E. 36 (solution)

F. 9 (solution)

Answer: y = 3x + 1

Step-by-step explanation:

3 is the slope and 1 is the y-intercept

The equation of the line parallel to 5x - 9y = 4 that passes through the point (1, -4) is y = (5/9)x - 31/9 in slope-intercept form and 5x - 9y + 31 = 0 in standard form.

To find the equation of a line parallel to a given line, we need to determine the slope of the given line and then use the point-slope form to derive the equation of the parallel line.
The given equation, 5x - 9y = 4, can be rewritten in slope-intercept form as y = (5/9)x - 4/9 by isolating y.
Since the parallel line has the same slope as the given line, which is 5/9, we can use the point-slope form:
y - y1 = m(x - x1),
where (x1, y1) is the given point (1, -4), and m is the slope.
Plugging in the values, we have:
y - (-4) = (5/9)(x - 1),
y + 4 = (5/9)x - 5/9,
y = (5/9)x - 31/9.
Therefore, the equation of the line parallel to 5x - 9y = 4 that passes through the point (1, -4) is y = (5/9)x - 31/9 in slope-intercept form.
Converting it to standard form, we move all the terms to one side:
(5/9)x - y + 31/9 = 0,
5x - 9y + 31 = 0.
Hence, the equation in standard form is 5x - 9y + 31 = 0.

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Answer:

When we multiply three or more whole numbers, the value of the product remains the same when they are grouped in any manner. The associative property of multiplication holds for whole numbers. Thus, if 'a', 'b', and 'c' are three whole numbers, then a × (b × c) = (a × b) × c = (a × c) × b.

Step-by-step explanation:

This property of the whole numbers tells that the order of addition does not change the value of the sum. Let a and b are two whole numbers, then a + b = b + a. Suppose a = 10 and b = 18 ⇒ 10 + 18 = 28 = 18 + 10.

Associative Property

When we add three or more whole numbers, the value of the sum remains the same. The order of addition of numbers is not important. Or, in other words, the numbers can be grouped in any manner. The sum remains the same. This is the associative property of addition.

If a, b, and c are three whole numbers, then a + (b + c) = (a + b) + c = (a + c) + b. For example, 10 + (5 + 12) = (10 + 5) + 12 = (10 + 12) + 5 = 27

Additive Identity

This is the property of zero by which the value of the whole number remains the same when added to any whole number. Zero is the additive identity of whole numbers. If w is a whole number, then w + 0 = w = 0 + w. For example, 0 + 7 = 7 = 7 + 0.

Properties of Subtraction

Closure Property

When one whole number is subtracted from another, the difference is not always a whole number. This means that the whole numbers are not closed under subtraction. If a and b are two whole numbers and a − b = c, then c is not always a whole number. Take a = 7 and b = 5, a − b = 7 − 5 = 2 and b − a = 5 − 7 = −2 (not a whole number).

Commutative Property

Subtraction of two whole numbers is not commutative. This means we cannot subtract two whole numbers in any order and get the same result. Let a and b be two whole numbers, then a − b ≠ b − a. Take a = 7 and b = 5, 7 − 5 = 2 ≠ 5 − 7 = −2.

Associative Property

An associative property does not hold for the subtraction of whole numbers. This means that we cannot group any two whole numbers and subtract them first. Order of subtraction is an important factor. If ‘a’, ‘b’, and ‘c’ are the three whole numbers then, a − (b − c) ≠ (a − b) − c. Consider the case when a = 8, b = 5 and c = 2, 8 − (5 − 2) = 5 ≠ (8 − 5) − 2 = 1.

Subtractive Property of Zero

When we subtract zero from a whole number, the value of the whole number remains the same. Take an example, a = 98, a − 0 = 98 − 0 = 98.

Properties of Multiplication

Closure Property

Multiplication of two whole numbers will result in a whole number. Suppose, a and b are the two whole numbers and a × b = c, then c is also a whole number. Let a = 10, b = 5, 10 × 5 = 50 (whole number). The whole number is closed under multiplication.

The left Riemann sum (L) for the given definite integral is 21.875, the right Riemann sum (R₁) is 43.125, and their average is 32.

The given definite integral is int_0^2 5(x^2+5)dx and the indicated value of n is 4.

Let's calculate Δx first.

Δx = (b-a)/nΔx = (2-0)/4

Δx = 0.5

Using Δx, we will find the values of xᵢ, which will be used to calculate Ln and Rn.

x₀ = 0x₁ = 0.5x₂ = 1x₃ = 1.5x₄ = 2

Using these values of xᵢ and Δx, we will calculate Ln, R₁, and their average.

Ln = Δx[f(x₀)+f(x₁)+f(x₂)+f(x₃)] = 0.5[f(0)+f(0.5)+f(1)+f(1.5)] = 21.875

R₁ = Δx[f(x₁)+f(x₂)+f(x₃)+f(x₄)] = 0.5[f(0.5)+f(1)+f(1.5)+f(2)] = 43.125

Average = (Ln + R₁)/2

Average = (21.875 + 43.125)/2

Average = 32

In the context of the given definite integral, the left Riemann sum (L) and the right Riemann sum (R₁) are both methods used to approximate the value of the integral.

The left Riemann sum (L) involves evaluating the function at the left endpoints of each subinterval and multiplying it by the width of the subinterval. In this case, with a partition of n=4, the left Riemann sum is calculated by summing the values of the function at x=0, x=0.5, x=1, and x=1.5, and multiplying each by the width of the subinterval, which is Δx=0.5. The resulting sum, 21.875, provides an approximation of the area under the curve using left endpoints.

On the other hand, the right Riemann sum (R₁) involves evaluating the function at the right endpoints of each subinterval and multiplying it by the width of the subinterval. In this case, the right Riemann sum is calculated by summing the values of the function at x=0.5, x=1, x=1.5, and x=2, and multiplying each by the width of the subinterval, which is Δx=0.5. The resulting sum, 43.125, provides an approximation of the area under the curve using the right endpoints.

These Riemann sums allow us to estimate the value of the definite integral by dividing the interval into subintervals and evaluating the function at specific points. The average of the left and right Riemann sums, which is 32 in this case, provides a more balanced approximation of the integral.

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Answer:

\(the \: third \: answer \: is \: correct\)

Answer:

5.2

Step-by-step explanation:

√27 = 5.1962 (5s.f.) = 5.2 (2s.f.)

Answer:

i hope 5.2 is the answer

Step-by-step explanation:

i m not sure