A manufacturer measures the number of cell phones sold using the binomial 0. 015c+2. 81. She also measures the wholesale price on these phones using a binomial 0. 011c+3. 52. Calculate her revenue if she sells 100,000 cell phones. Revenue = (numberofcellphones)(wholesaleprice) = (0. 015c+2. 81)(0. 011c+3. 52)

The regular price of the camera is $227.

Given that, the total amount paid=$201.75 and the cost of the memory card=$12.35.

Cost of camera=201.75-12.35=$189.4

A discount is a deduction from the usual price of something. To discount means to deduct an amount from the price.

Discount is given to the camera=20%.

The regular price of the camera=189.4+20% of 189.4

=189.4+0.2×189.4=189.4+37.88=$227.28

The estimation to the nearest the whole number is $227.

Therefore, the regular price of the camera is $227.

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answer

yes do it your self

Answer: The answer is B i just did this test

Step-by-step explanation:

The only number on the list that matches the rounding requirement is 0.579. Hence, the correct option is B.

Rounding some number to a specific value is making its value simpler (therefore losing accuracy), mostly done for better readability or accessibility.

Rounding to some place keeps it accurate on the left side of that place but rounded or sort of like trimmed from the right in terms of exact digits.

We can consider what the offered numbers are when they are rounded to hundredths:

 0.574 ⇒ 0.57

 0.579 ⇒ 0.58 . . . matches the requirement

 0.585 ⇒ 0.59

 0.681 ⇒ 0.68

The only number on the list that matches the rounding requirement is 0.579.

Hence, the correct option is B.

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To solve the initial value problem (IVP) dy/dt - y = 2e + 24e^(9t) with y(0) = 9, we can use an integrating factor and then apply the appropriate integration techniques.

The differential equation can be written in the standard form as:

dy/dt - y = 2e + 24e^(9t)

Comparing this to the standard form, we have:

P(t) = -1

Q(t) = 2e + 24e^(9t)

The integrating factor (IF) is given by the exponential of the integral of P(t), i.e., IF = e^(∫-1 dt) = e^(-t).

Now, multiply the entire equation by the integrating factor:

e^(-t) * (dy/dt - y) = e^(-t) * (2e + 24e^(9t))

This simplifies to:

e^(-t) * dy/dt - e^(-t) * y = 2e^(1-t) + 24e^(8t)

Using the product rule, we can rewrite the left side as the derivative of the product:

d(y * e^(-t)) / dt = 2e^(1-t) + 24e^(8t)

Integrating both sides with respect to t, we get:

y * e^(-t) = -2e^(1-t) / (1 - (-1)) + 24e^(8t) / (8 - (-1)) + C

y * e^(-t) = -e^(1-t) + 3e^(8t) + C

Now, solve for y:

y = (-e^(1-t) + 3e^(8t) + C) * e^t

y = -e + e^t + 3e^(9t) + Ce^t

y = e^t + 3e^(9t) - e + Ce^t

To determine the value of C, we use the initial condition y(0) = 9:

9 = e^0 + 3e^(9*0) - e + Ce^0

9 = 1 + 3 - e + C

9 = 4 - e + C

C = 9 - 4 + e

C = 5 + e

Therefore, the solution to the initial value problem dy/dt - y = 2e + 24e^(9t) with y(0) = 9 is:

y = e^t + 3e^(9t) - e + (5 + e)e^t

Simplifying:

y = (6 + 2e)e^t + 3e^(9t) - e

Note that the term (6 + 2e) in front of e^t accounts for the constant term arising from the integration of the homogeneous solution.

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Since y_1(0) = 1, it satisfies the initial condition. However, y_2(0) = 0 does not satisfy the initial condition, as it should be y(0) = 1. Therefore, only y_1(t) is a solution of the initial value problem.

To verify that both y_1(t) = 1 - t and y_2(t) = -t^2/4 are solutions of the initial value problem, we first need to understand what the problem is. An initial value problem is a differential equation that includes an initial condition. In this case, we can assume that the initial condition is y(0) = 1.
Now, let's substitute both y_1(t) and y_2(t) into the differential equation and see if they satisfy the initial condition. The differential equation is not provided, but assuming it is y'(t) = -t/2, we have:
y_1'(t) = -1
y_2'(t) = -t/2
Substituting y_1(t) and y_2(t) into the differential equation gives:
y_1'(t) = -1

= -t/2 (when t = 2)
y_2'(t) = -t/2

= -t/2 (for all t)
Thus, both y_1(t) and y_2(t) satisfy the differential equation. Now, let's check if they satisfy the initial condition.
y_1(0) = 1 - 0

= 1
y_2(0) = -0^2/4

= 0
In conclusion, y_1(t) = 1 - t is the only solution that satisfies the differential equation and initial condition, while y_2(t) = -t^2/4 is not a solution since it does not satisfy the initial condition.

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Answer:C

Step-by-step explanation:

To find the product of 6x and [4-21 730], we need to simplify the expression first.

To simplify, we perform the subtraction first and then multiply.  

So, [4-21 730] can be simplified as follows: [4-21 730] = 4 - 21730 = -21726  

Now, we can find the product of 6x and -21726 as follows: 6x(-21726) = -130356  

Therefore, the product of 6x and [4-21 730] is -130356.

Answer:

$5.40

Step-by-step explanation:

A tax is a compulsory sum levied by the government or an agency of the government on goods and services. Taxes increase the price of good. The items bought would increase by 8%

Tax paid by Dave = 0.08 x $15 = $1.2

total price paid by Dave = 1.2 + $15 = $16.2

Tax paid by Mel = 0.08 x $20 = $1.6

total price paid by Mel = $20 + $1.6 = $21.60

Difference in prices

21.6 -16.2 = $5.40

If log 7 equals a and log 8 equals b then log 224 equals: E: none of the above.

Simplify the expression for log (224) by using:

log (ab) = log (a) + log (b) and log (a/b) = log (a) - log (b):

So,

log (224) = log ( 7 × 8 × 4)

= log (7) + log (8) + log (4)

Since log(4) = 0 (since 4 = 2^2)

Simplify

log(224) = log (7) + log (8) + log (4)

= log 7 + log 8

= a + b

Therefore the correct option is E.

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Answer:

options: 83/90 92/99 83/9 83/100. 0.92 = 92 / 100

Step-by-step explanation:

Chelsea needs at least 190 cm² of paper to cover the box.

To find the minimum amount of paper Chelsea needs to cover the rectangular prism-shaped box, we need to calculate the surface area of the box.

Surface Area = 2(lw + lh + wh)

Where,

L is length, W is width, aH nd f f is height.

So, to find the minimum amount of paper Chelsea needs, we need to know the box's surface area of the box. Once we have the dimensions, we can plug them into the formula and calculate the surface area.

For example, if the box has dimensions of length of 10 cm, width 5 cm, and height 30 cm, the surface area would be:

Surface Area = 2(50 + 30 + 15)

Surface Area = 2(95)

Surface Area = 190 cm²

Therefore, Chelsea needs at least 190 cm² of paper to cover the box.

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Answer:

6 is a answer

Step-by-step explanation:

if it is x 3+3 is equal to 6

if it is 10+10 is equal to 20

Given an arithmetic sequence, an = {1, 40, 47, 54}, create the explicit formula and list any restrictions to the domain.Solution:

The difference between consecutive terms is the same and is called the common difference (d).The explicit formula for an arithmetic sequence is:an = a1 + (n - 1)dwhere an represents the nth term, a1 is the first term, n is the number of terms, and d is the common difference.

To find d, take the difference between any two consecutive terms. Here, d = 7 – 0 = 7.To find a1, use the formula a1 = a2 – d = 40 – 7 = 33. Therefore, the explicit formula for the given arithmetic sequence is:an

= 33 + 7(n - 1)where n ≥ 1.List any restrictions to the domain:n is the number of terms.

Since the first term corresponds to n = 1, the domain of the sequence starts at n = 1.

Therefore, the restriction on the domain is n ≥ 1.The correct answer is:an = 33 + 7(n - 1) where n ≥ 1.

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Answer:

We can use the formula for simple interest:

I = P*r*t

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

Plugging in the given values:

$1,440 = $3,000 * r * 4

Simplifying:

r = $1,440 / ($3,000 * 4)

r = 0.12

Multiplying by 100 to convert to a percentage:

r = 12%

Therefore, the interest rate on this savings account was 12%.

Answer:

16 sqrt(30)

Step-by-step explanation:

2√20 × 4√6

Multiply

2*4  sqrt(20*6)

8 sqrt(120)

Simplifying

8 sqrt(4 *30)

We know sqrt(ab) = sqrt(a) sqrt(b)

8 sqrt(4) sqrt(30)

8 *2* sqrt(30)

16 sqrt(30)

To minimize the amount of tin used, the height of the can must be 4 inches (option D).

The volume of a cylindrical tin can is given by the formula V = πr²h, where V is the volume, r is the radius, and h is the height. To minimize the amount of tin used, we need to minimize the surface area, which is given by the formula A = 2πrh + 2πr².

Given the volume is 16π cubic inches, we have:

16π = πr²h

Now, we can find the relationship between r and h:

h = 16/r²

Now, substitute this into the surface area formula:

A = 2πr(16/r²) + 2πr²
A = 32π/r + 2πr²

To minimize the surface area, we can take the derivative with respect to r and set it to 0:

dA/dr = -32π/r² + 4πr

0 = -32π/r² + 4πr

Solving for r:

r³ = 8
r = 2 (since r > 0)

Now, substituting r back into the relationship between r and h:

h = 16/(2²)
h = 16/4
h = 4

Therefore, the height of the can must be 4 inches (option D) to minimize the amount of tin used.

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A aeroplane intersects a prism parallel to the base, the performing  sampling will have the same shape as the base of the prism.

The performing  sampling will have the same shape as the base of the prism.    When a aeroplane intersects a prism parallel to the base, the performing  sampling is a shape that has the same  figure as the base of the prism. This is because the aeroplane

            intersects the prism along a  resemblant aeroplane and as a result, the  sampling will have the same shape and  confines as the base.    For  illustration, if the base of the prism is a cube, the  sampling will also be a cube. also, if the base is a triangle, the  sampling will be a triangle. This holds for any polygonal base,  similar as a forecourt, pentagon, hexagon,etc.    The size of the  sampling will depend on the position of the aeroplane

            within theprism.However, the  sampling will be  lower, If the aeroplane is  near to the top or bottom of theprism.However, the  sampling will be larger, If the aeroplane

            is  near to the middle of the prism.    In summary, when a aeroplane intersects a prism parallel to the base When a aeroplane intersects a prism parallel to the base, the performing  sampling is a shape that has the same  figure as the base of the prism. This is because the aeroplane

            intersects the prism along a  resemblant aeroplane and as a result, the  sampling will have the same shape and  confines as the base.    For  illustration, if the base of the prism is a cube, the  sampling will also be a cube. also, if the base is a triangle, the  sampling will be a triangle. This holds for any polygonal base,  similar as a forecourt, pentagon, hexagon,etc.    The size of the  sampling will depend on the position of the aeroplane

            within theprism.However, the  sampling will be  lower, If the aeroplane is  near to the top or bottom of theprism.However, the  sampling will be larger, If the aeroplane is  near to the middle of the prism.    In summary, when a aeroplane intersects a prism parallel to the base, the performing  sampling will have the same shape as the base of the prism.

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Answer:

1234 Get On The Dance Floor

Both axiom 1 and axiom 7 hold for all vectors in R³ while considering the set R³ with standard addition and scalar multiplication.

Vector space axioms:

1. Closure under addition: For any vectors u, v in R³, the sum u + v is also in R³.

7. Scalar multiplication: For any scalar c and any vector v in R³, the product cv is also in R³.

Proof that axioms 1 and 7 hold for all vectors in R³:

To prove that axiom 1 holds for all vectors in R³, we need to show that for any two vectors u and v in R³, their sum u + v is also in R³. Since R³ is defined as the set of all ordered triples of real numbers (x, y, z), we can write:

u = (u₁, u₂, u₃)

v = (v₁, v₂, v₃)

where u₁, u₂, u₃, v₁, v₂, and v₃ are real numbers. Then the sum of u and v is:

u + v = (u₁ + v₁, u₂ + v₂, u₃ + v₃)

Since each component of the sum is a real number and there are three components, we see that u + v is an ordered triple of real numbers. Therefore, u + v is an element of R³ and axiom 1 holds.

To prove that axiom 7 holds for all vectors in R³, we need to show that for any scalar c and any vector v in R³, the product cv is also in R³. Again using the definition of R³ as the set of all ordered triples of real numbers (x, y, z), we can write:

v = (v₁, v₂, v₃)

where v₁, v₂, and v₃ are real numbers. Then the product of c and v is:

cv = (cv₁, cv₂, cv₃)

Since each component of the product is a real number and there are three components, we see that cv is an ordered triple of real numbers. Therefore, cv is an element of R³ and axiom 7 holds.

Therefore, we have shown that both axiom 1 and axiom 7 hold for all vectors in R³.

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The addition of the polynomial \(x^{3}+3xy-2xy^{2} +y^{3}\) with \(2x^{3}-5x^{2} y-3xy^{2}-2y^{3}\) is  \(3x^{3}+3xy-5x^{2} y-5xy^{2}-y^{3}\).

What is a polynomial?

⇒ A polynomial is an expression consisting of indeterminates and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables.

⇒ In the addition of polynomials, the like terms are added while in subtraction, the like terms are subtracted.

Calculation;

We have been given two polynomial which we have to add   \(x^{3}+3xy-2xy^{2} +y^{3}\) and \(2x^{3}-5x^{2} y-3xy^{2}-2y^{3}\)

The sign after addition or subtraction will always be of the variable having more value.

\((x^{3}+3xy-2xy^{2} +y^{3} )+(2x^{3}-5x^{2} y-3xy^{2}-2y^{3})\)

On adding like terms with each other

⇒ \((x^{3} +2x^{3})+ 3xy-5x^{2} y-(2xy^{2}+3xy^{2})+(y^{3}-2x^{3})\)

⇒ \(3x^{3}+3xy-5x^{2} y-5xy^{2}-y^{3}\)

Hence the addition of the polynomial\(x^{3}+3xy-2xy^{2} +y^{3}\) and \(2x^{3}-5x^{2} y-3xy^{2}-2y^{3}\) is \(3x^{3}+3xy-5x^{2} y-5xy^{2}-y^{3}\).

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To answer the questions about group G with order 77: (1) Show that there are normal subgroups H and K of the group with order 7 and 11, respectively.

Since the order of G is 77, by the Sylow theorems, there exist Sylow 7-subgroups and Sylow 11-subgroups in G.

Let H be a Sylow 7-subgroup of G and K be a Sylow 11-subgroup of G. Since Sylow subgroups are conjugate to each other, H and K are both normal subgroups of G.

(2) Show that HK={hk|h∈H, k∈K} is an Abelian subgroup of group G.

Since H and K are normal subgroups of G, we have that HK is a subgroup of G. To show that HK is an Abelian subgroup, we need to prove that for any elements hk and h'k' in HK, their product is commutative.

Let hk and h'k' be arbitrary elements in HK. Since H and K are normal subgroups, we have that h'khk' = kh'h. Thus, the product hk h'k' is equal to kh'h, which implies that HK is an Abelian subgroup.

(3) Show that HK=G.

To show that HK=G, we need to prove that every element g in G can be expressed as a product hk, where h∈H and k∈K.

Since H and K are normal subgroups of G, their intersection H∩K is also a normal subgroup of G. By Lagrange's theorem, the order of H∩K divides both the order of H (which is 7) and the order of K (which is 11). Since 7 and 11 are coprime, the only possible order for the intersection is 1.

Thus, H∩K={e}, where e is the identity element of G. This implies that every element g in G can be uniquely expressed as g = hk, where h∈H and k∈K. Therefore, HK=G.

(4) Show that G is a cyclic group.

Since HK=G, and HK is an Abelian subgroup, we have that G is an Abelian group. Every Abelian group of prime order is cyclic. Since the order of G is 77, which is not prime, G cannot be cyclic.

Therefore, G is not a cyclic group.

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To answer the questions about group G with order 77: (1) Show that there are normal subgroups H and K of the group with order 7 and 11, respectively.

Since the order of G is 77, by the Sylow theorems, there exist Sylow 7-subgroups and Sylow 11-subgroups in G.

Let H be a Sylow 7-subgroup of G and K be a Sylow 11-subgroup of G. Since Sylow subgroups are conjugate to each other, H and K are both normal subgroups of G.

(2) Show that HK={hk|h∈H, k∈K} is an Abelian subgroup of group G.

Since H and K are normal subgroups of G, we have that HK is a subgroup of G. To show that HK is an Abelian subgroup, we need to prove that for any elements hk and h'k' in HK, their product is commutative.

Let hk and h'k' be arbitrary elements in HK. Since H and K are normal subgroups, we have that h'khk' = kh'h. Thus, the product hk h'k' is equal to kh'h, which implies that HK is an Abelian subgroup.

(3) Show that HK=G.

To show that HK=G, we need to prove that every element g in G can be expressed as a product hk, where h∈H and k∈K.

Since H and K are normal subgroups of G, their intersection H∩K is also a normal subgroup of G. By Lagrange's theorem, the order of H∩K divides both the order of H (which is 7) and the order of K (which is 11). Since 7 and 11 are coprime, the only possible order for the intersection is 1.

Thus, H∩K={e}, where e is the identity element of G. This implies that every element g in G can be uniquely expressed as g = hk, where h∈H and k∈K. Therefore, HK=G.

(4) Show that G is a cyclic group.

Since HK=G, and HK is an Abelian subgroup, we have that G is an Abelian group. Every Abelian group of prime order is cyclic. Since the order of G is 77, which is not prime, G cannot be cyclic.

Therefore, G is not a cyclic group.

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The range of ages may be 39 years, it does not provide sufficient information to determine the actual distribution or spread of ages within that range.

Andy's calculation of the range of ages as 39, based on subtracting the minimum age from the maximum age (50 - 11), is incorrect. The range of a set of values represents the difference between the maximum and minimum values, but it does not necessarily reflect the actual span or dispersion of the values within that range.

In the context of ages, the range refers to the difference between the highest and lowest ages in a given group. However, this calculation alone does not provide any information about the distribution or spread of ages within that range.

For example, if a group consists of individuals with ages 11, 12, 13, 14, ..., 49, 50, the range would still be 39 years (50 - 11). However, the ages within that range are evenly distributed and not widely spread.

On the other hand, if the group has ages 11, 11, 12, 13, ..., 49, 50, the range would still be 39 years, but the ages are tightly clustered around the lower end of the range.

Therefore, while the range of ages may be 39 years, it does not provide sufficient information to determine the actual distribution or spread of ages within that range.

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Answer:

1.6

2. Honestly, I think it is, because it repeats at regular intervals. The period is -2.

Step-by-step explanation:

Answer:

B. 4x-6

Step-by-step explanation:

I TOOK THE TEST!!!

The line passes through the points (0,-6), (-2,1),The equation of the line will be y= -3.5x -6

An equation is a statement that two expressions, which include variables and/or numbers, are equal. It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.

The coordinate of points obtained from the graph,

(0,-6), (-2,1), (2,2)

The slope of the line for the graph is,

m = (1-(-6) / (-2-0)

m = 7/-2

m= - 3.5

The standard equation of the line is,

y-y= m(x-x₁)

y-(-6) = -3.5(x-0)

y+6=-3.5x

y= -3.5x -6

Thus, the line passes through the points (0,-6), (-2,1), The equation of the line will be y= -3.5x -6

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The smallest integer n so that the condition number of the n x n Hilbert Matrix is greater than 107 is 4.

The given 4x4 Hilbert matrix can be represented as below:

H = [1/1 1/2 1/3 1/4;1/2 1/3 1/4 1/5;1/3 1/4 1/5 1/6;1/4 1/5 1/6 1/7]

In order to find the smallest integer n so that the condition number of the n x n Hilbert Matrix is greater than 107, first we find the condition number of the matrix for each value of n and then compare the values of the condition numbers.

Let's solve for n = 2, 3, 4...

Using MATLAB, we can find the condition number of the matrix as:

cn4 = cond(hilb(4))

cn3 = cond(hilb(3))

cn2 = cond(hilb(2))

cn1 = cond(hilb(1))

We get the following values:

cn4 = 15513.7387389294

cn3 = 524.056777586064

cn2 = 19.2814700679036

cn1 = 1

As we can see, for n = 4, the condition number of the matrix is greater than 107.

Hence, the smallest integer n so that the condition number of the n x n Hilbert Matrix is greater than 107 is 4.

Therefore, the value of n is 4.

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Writing linear equations given two points can be a useful skill when graphing linear equations.

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Earning 2% on $1,000 for two years would result in interest of $80 if the interest rate were to double to 4%.

According to the data, if someone earns 2% interest on $1,000 for two years, the interest rate will be: = 2% 2 = 4%.

The sales for the two years are divided by two to arrive at the midpoint formula, which will be used to predict sales (in billions of dollars) for 2017 as follows:

[Sales from 2016 and 2018] / 2

It should be remembered that the formula for simple interest is:

Interest is calculated as follows: Principal x Rate x Time / 100

Interest is equal to 1000 4 2 /100.

Interest equals 8000/100.

$80 plus interest

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Answer:

a. (-4,8)

Step-by-step explanation:

the two lines intersect at this point