​Joe's annual income has been increasing each year by the same dollar amount. The first year his income was ​$17 comma 90017,900​, and the 44th year his income was ​$20 comma 30020,300. In which year was his income $ 30 comma 700 question mark

Answer:

[tex]x=\frac{-31\pm \sqrt{1000.2}}{4}[/tex]

Step-by-step explanation:

Given quadratic equation,

[tex]2x^2+31x-4.9=0[/tex]

Since, by the quadratic formula,

The solution of a quadratic equation [tex]ax^2+bx+c=0[/tex] is,

[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

Here, a = 2, b = 31, c = -4.9,

Thus, by the quadratic formula,

[tex]x=\frac{-31\pm \sqrt{31^2-4\times 2\times -4.9}}{2\times 2}[/tex]

[tex]=\frac{-31\pm \sqrt{961+39.2}}{4}[/tex]

[tex]=\frac{-31\pm \sqrt{1000.2}}{4}[/tex]

[tex]\implies x = \frac{-31+\sqrt{1000.2}}{4}\text{ or }x=\frac{-31- \sqrt{1000.2}}{4}[/tex]

Answer:  80

Step-by-step explanation:

Given : A restaurant serves a luncheon special .

The number of choices for appetizers= 4

The number of choices for salad = 4

The number of choices for desert = 5

Then, the total number of different possible meals are given by :_

[tex]4\times4\times5=80[/tex]

Therefore, the total number of different possible meals are =80

Answer:

Neither

Step-by-step explanation:

We are given that a sequence

5,8,13,21,34,55,....

We have to determine the sequence is an arithmetic , geometric or neither.

[tex]a_1=5,a_2=8,a_3=13,...a_6=55[/tex]

We know that if a sequence is an arithmetic then the difference between consecutive terms are equal.

If a sequence is geometric then the ratio of consecutive terms is constant.

[tex]d_1=a_2-a_1=8-5=3[/tex]

[tex]d_2=a_3-a_2=13-8=5[/tex]

[tex]d_1\neq d_2[/tex]

Hence, the difference between the consecutive terms is not equal .Therefore, sequence is not an arithmetic sequence.

[tex]r_1=\frac{a_2}{a_1}=\frac{8}{5}[/tex]

[tex]r_2=\frac{a_3}{a_2}=\frac{13}{8}[/tex]

[tex]r_1\neq r_2[/tex]

Hence, the ratio of consecutive terms is not equal .Therefore, given sequence is not geometric sequence.

Therefore, given sequence is not an arithmetic nor geometric sequence.

Smallest number of eggs in the basket = 119

In the question,

We know that the number of eggs in the basket should be a a multiple of 7.

But it can not be a multiple of 2, 3, 4, 5 and 6 because every time we pick up the eggs 2, 3, 4, 5 or 6 at a time we are left with some eggs with us.

Therefore, the number of eggs can not be a multiple of these numbers.

Now,

Let us say the number of eggs in the basket be 7x.

So,

Let us take the LCM of 2, 3, 4, 5 and 6.

So,

LCM = 60

Now, the number would be greater than 60 and the multiple of 7.

So, checking on all the multiples of 7 above 60 and checking the condition that, the remainder left on dividing by,

2 is 1. (2x + 1)

by 3 is 2. (3x + 2)

by 4 is 3. (4x + 3)

by 5 is 4. (5x + 4)

by 6 is 5. (6x + 5)

So,

On Checking the multiples of 60 which are divisible by 7 are ,

60 + 1 , 120 + 1, 60 - 1, 120 - 1.

So,

For 61 it is not satisfying all the conditions.

121 is also not satisfying all the conditions.

59 is also not satisfying all the conditions.

But,

The number 119 on checking its divisibility by 2. Leaves remainder 1 as, (2(59) + 1).

Divisibility by 3 leaves remainder 2 as, {3(39) + 2}.

Divisibility by 4 leaves remainder 3 as, {4(29) + 3}.

Divisibility by 5 leaves remainder 4 as, {5(23) + 4}.

Divisibility by 6 leaves remainder 5 as, {6(19) + 5}.

Divisibility by 7 leaves remainder 0 as, {7(17) + 0}.

Therefore, the minimum number of eggs in the basket are 119.

Answer: 119 is the smallest number of eggs in the  basket

Step-by-step explanation:

let k be the number

so k is less than a multiple of 2 by 1 so we have k = 2A - 1 for a positive integer  A

so k is 2 more and thus less with 1, for a multiple of 3, k = 3B - 1, for some positive integer B

so k is 3 more and thus less with 1, with a multiple of 4, so k = 4C -1 for some positive integer

so k is 4 more and thus less with 1, with a multiple of 5, so k = 5D -1 for some positive integer

so k is 5 more and thus less with 1, with a multiple of 6, so k = 6E -1 for some positive integer

thus k is a multiple of 7, so k = 7F for some positive integer F

THUS K = 2A - 1 = 3B - 1 = 4C -1 = 5D -1 = 6E -1 = 7F

ADD 1 to k above gives k + 1 = 2A = 3B = 4C = 5D = 6E = 7F + 1

of k + 1 = 7F + 1 has to be a common multiple of 2, 3, 4, 5, 6. The least common multiple  of 2, 3, 4, 5, 6 is 60

thus k + 1 is a multiple of 60 so k = 60 -1 = 59

again  find the least multiple of 60 that is 1 greater  than a multiple of  60

the first multiple of 60 is 60 and 60 - 1 = 59, but 59 is not a multiple of 7

the second multiple of 60 is 120 and 120 -1 = 119

and 7 is a multiple of 119 because 7 * 17 = 119

Answer:

You should make 150 quarts of Creamy Vanilla and 50 quarts of Continental Mocha.

Step-by-step explanation:

This problem can be solved by a system of equations.

The largest profit is going to earned when all the eggs and cups of cream in stock are used.

I am going to call x the number of quarts of Creamy Vanilla and y the number of quarts of Continental Mocha.

The problem states that each quart of Creamy Vanilla uses 2 eggs and each quart of Continental Mocha uses 1 egg. There are 350 eggs in stock, so:

[tex]2x + y = 350[/tex]

Each quart of Creamy Vanilla uses 3 cups of cream, as does each quart of Continental Mocha. There are 600 cups of cream in stock.

So:

[tex]3x + 3y = 600[/tex] Simplifying by 3.

[tex]x + y = 200[/tex]

We have the following system

[tex]1)2x + y = 350[/tex]

[tex]2)x + y = 200[/tex]

I am going to multiply 2) by (-1) and then add with 1), so i can eliminate y

[tex]1)2x + y = 350[/tex]

[tex]2)-x - y = -200[/tex]

[tex]2x - x + y - y = 350 - 200[/tex]

[tex]x = 150[/tex]

[tex]x + y = 200[/tex]

[tex]y = 200 - 150 = 50[/tex]

You should make 150 quarts of Creamy Vanilla and 50 quarts of Continental Mocha.

The question regarding the number of quarts of Creamy Vanilla and Continental Mocha for maximum profit can be solved using a method in mathematics called linear programming. It involves setting up inequalities to represent the constraints (number of eggs and cups of cream) and a profit function which we want to maximize. The specific solution will depend on the particular set of constraints and profit per flavor.

This problem can be approached as a linear programming problem. Let x represent the number of quarts of Creamy Vanilla and y represent the number of quarts of Continental Mocha. The constraints from the problem can be represented with the following inequalities, considering the number of eggs and cups of cream:

2x + y ≤ 350 (eggs)

3x + 3y ≤ 600 (cups of cream)

We want to maximize the profit function P which is given by P = 3x + 2y. This can be solved graphically by plotting the constraints on a graph, finding the feasible region and then determining which point in that region provides the maximum value of P.

The solution to this problem can vary depending on the specific set of supplies and the profit of each flavor.

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Answer:

Chris lives farther from school than Alex lives, and Chris walks to school at a faster rate than Alex

Step-by-step explanation:

Chris walked 4 miles.

Alex walked 2.

Alex takes 25 minutes to walk one mile. (50÷2) and Chris takes 15 minutes to walk one mile (60÷4), meaning Chris walks at a faster rate although he lives farther away.

Answer:

Chris lives farther from the school than Alex lives, and Chris walked to school at a faster rate than Alex walked.

Answer:

the patient receiving is 15 mg/hr

Step-by-step explanation:

given data

infusing = 30 mL/hr

drug mixed = 250 mg

D5W = 500ml

to find out

How many mg/hr is the patient receiving

solution

we know infusing rate is = 30 mL/hr   ...................1

and we know mixed is = [tex]\frac{250mg}{500ml}[/tex]   ..............2

so

now we multiply both equation 1 and 2

patient receiving = [tex]\frac{250mg}{500ml}[/tex]  × [tex]\frac{30ml}{hr}[/tex]

solve we get here

patient receiving = [tex]\frac{15mg}{hr}[/tex]

so the patient receiving is 15 mg/hr

The patient is receiving 15 mg/hr of aminophylline.

To calculate the mg/hr that the patient is receiving, we need to find the concentration of aminophylline in the infusion and multiply it by the infusion rate. The concentration of aminophylline is 250 mg in 500 ml of D5W. This means that there are 250 mg of aminophylline in 500 ml, or 0.5 mg/ml. The infusion rate is 30 ml/hr. So, to find the mg/hr, we multiply the concentration by the infusion rate: 0.5 mg/ml * 30 ml/hr = 15 mg/hr. Therefore, the patient is receiving 15 mg/hr of aminophylline.

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Step-by-step explanation:

Consider the provided information.

Integers: Integers are the set of whole number and the negatives of the natural numbers, i.e, … ,-2, -1, 0, 1, 2,...

Rational number: A number   is said to be rational, if it is in the form   of p/q. Where p and q are integer and   denominator is not equal to 0.The decimal expansion of rational numbers may terminate or become periodic.

Irrational   number: A number is irrational if it cannot   be expressed be expressed by dividing two   integers. The decimal expansion of   Irrational numbers are neither terminate nor   periodic.

Now consider the provided statement "Negative numbers are integers."

The statement is incorrect as all negative numbers are not integers they might be rational number or irrational numbers.

For example: -2.5 is a negative number but it is not an integer. The number is a rational number.

Hence, her statement is incorrect.

She can say that all negative whole numbers are integers.

Answer:

The linear equation relating price to demand is [tex]p=-0.05x+360.2[/tex]

Step-by-step explanation:

A Linear Demand Function expresses demand p (the number of items demanded) as a function of the unit price x (the price per item).

From the information given we know two facts:

Let x be the price and p the number of cards sold.

The fact 1. is the slope of the function because is the change in demand per unit change in price.

We can use fact 2. to find the value of b in the equation

[tex]p=-0.05x+b\\b=p+0.05x[/tex]

[tex]b=360+0.05\cdot 4\\b=360+0.2\\b=360.2[/tex]

The linear equation relating price to demand is [tex]p=-0.05x+360.2[/tex]

Final answer:

The correct negation of the statement is option C: 'This triangle has two 45 degree angles and it is not a right triangle,' which follows the 'P and not Q' format for negating 'If P, then Q' statements.

Explanation:

The negation of the statement 'If this triangle has two 45 degree angles then it is a right triangle' is 'This triangle has two 45 degree angles and it is not a right triangle.' To negate a conditional statement like the one given, you would state that while the condition holds true, the conclusion does not. This is represented in option C. When negating an 'if-then' statement, the format usually follows 'If P, then Q' to 'P and not Q.' Thus, the correct answer is C, 'this triangle has two 45 degree angles and it is not a right triangle.'

Answer with Step-by-step explanation:

1.We are given

2783 and 7283

We have to compare the values of the underline digits

Place value of 2 in 2783 =2000 because 2 is at thousand place

2 in 7283 is at hundred place

Therefore, the place value of 2 in 7283=200

Therefore, the value of 2 in 2783 is 10 times the value of 2 in 7283.

2.We have to compare the values of 7  in both numbers

Place of 7 in 503497=one's

Therefore, place value of 7 =7

7 in 26475 is at tens place

Therefore, the place value of 7 in 26475=70

Hence, the value of 7 in 26475 is 10 times the value of 7 in 503497.

3.We have to compare the values of 4 in both given numbers

Place of 4 in 34258=Thousand

Place value of 4 in 34258=4000

Place of 4 in 47163=Ten thousand

Place value of 4 in 47163=40000

Hence, the value of 4 in 47163 is 10 times the value of 4 in 34258.

Answer:

Every line in [tex]\mathbb{R}^{3}[/tex] is a function of the form [tex]\gamma (t)={\bf p}+t {\bf v} [/tex], where [tex]{\bf p}[/tex] is point where the line passes and [tex]{\bf v}[/tex] is a nonzero vector which is called the direction vector of the line. Then, if we derive the function [tex]\gamma[/tex] we obtain [tex]\gamma'(t)={\bf v} \neq (0,0,0)[/tex], so [tex]\gamma(t)={\bf p}+t {\bf v}[/tex] is a regular curve.

Step-by-step explanation:

Every line in [tex]\mathbb{R}^{3}[/tex] can be parametrized by

[tex]\gamma (t)={\bf p}+t{\bf v}=(p_{1},p_{2},p_{3})+t(v_{1},v_{2},v_{3})=(p_{1}+tv_{1},p_{2}+tv_{2},p_{3}+tp_{3})[/tex], where [tex]t\in \mathbb{R}[/tex]. To derivate the function [tex]\gamma [/tex] we only need to derive each component. Then we have that

[tex]\gamma'(t)=(\frac{d}{dt}(p_{1}+tv_{1}),\frac{d}{dt}(p_{2}+tv_{2}),\frac{d}{dt}(p_{3}+tv_{3}))=(v_{1},v_{2},v_{3})={\bf v}\neq (0,0,0).[/tex]

Now, remember that a a parametrized curve is said to be regular if [tex]\gamma'\neq 0[/tex] for all [tex]t[/tex].

Answer:

No. of juniors = 14

No. of seniors = 16

Total students = 30

A) From the 30 members, how many ways are there to arrange 5 members of the club in a line?

Since we are asked about arrangement so we will use permutation

Formula : [tex]^nP_r=\frac{n!}{(n-r)!}[/tex]

n = 30

r = 5

[tex]^{30}P_5=\frac{30!}{(30-5)!}[/tex]

[tex]^{30}P_5=17100720[/tex]

So, From the 30 members, there are 17100720 ways to arrange 5 members of the club in a line?

B) How many ways are there to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line?

Out of 16 seniors 2 will be selected

So, 3 places are vacant

Remaining students = 30-2 = 28

So, out of 28 students 3 students will be selected

No. of ways = [tex]^{16}P_2 \times ^{28}P_3[/tex]

No. of ways = [tex]\frac{16!}{(16-2)!}\times\frac{28!}{(28-3)!}[/tex]

                   = [tex]4717440[/tex]

There are 4717440 ways to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line.

C)If the club sends 2 juniors and 2 seniors to the tournament, how many possible groupings are there?

Since we are not asked about arrangement so we will use combination

Out of 16 seniors 2 will be selected

Out of 14 juniors 2 will be selected

Formula : [tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]

So, No. of possible groupings = [tex]^{16}C_2 \times ^{14}C_2[/tex]

                                                  = [tex]\frac{16!}{2!(16-2)!} \times \frac{14!}{2!(14-2)!}[/tex]

                                                  = [tex]10920[/tex]

If the club sends 2 juniors and 2 seniors to the tournament, there are 10920 possible groupings

D) If the club sends either 4 juniors or 4 seniors, how many possible groupings are there?

Out of 16 seniors 4 will be selected

or

Out of 14 juniors 4 will be selected

So, No. of possible groupings = [tex]^{16}C_4 + ^{14}C_4[/tex]

                                                  = [tex]\frac{16!}{4!(16-4)!} + \frac{14!}{4!(14-4)!}[/tex]

                                                  = [tex]2821[/tex]

So,If the club sends either 4 juniors or 4 seniors, there are 2821 possible groupings .

Answer:

0.2449 or 24.49%

Step-by-step explanation:

Calculate the IRR if the power company gets a fixed feed-in tariff of $0.25/kWh.

IRR means Internal Rate of Return is given by

[tex]IRR=(\frac{FV}{PV})^{\frac{1}{n}}-1[/tex]

Where,  FV = Final value ($20)

             PV = Present value ($0.25)

               N = 20 years

Now put the values

[tex]IRR=(\frac{20}{0.25})^{\frac{1}{20}}-1[/tex]

= [tex]IRR=(80)^{\frac{1}{20}}-1[/tex]

= 1.24495742 - 1

= 0.24495742

Converting in percentage :

0.24495742 × 100 = 24.49%

IRR = 0.2449 or 24.49%

Answer:

IRR = 0.24495

Step-by-step explanation:

Given data:

Tariff =$0.25

So, present value = $0.25

N = 20 year

salvage value after 20 year is $20 M

final value is $20 M

IRR means internal rate of return and it is given as

[tex]IRR =[\frac{FV}{PV}]^{1/n} -1[/tex]

Where FV is final value and PV is present value

[tex]IRR = [\frac{20}{0.25}]^{1/20} -1[/tex]

IRR = 0.24495

Answer:

Inverse Laplace of [tex]\frac{1}{(S+4)^2}[/tex] will be [tex]te^{-4t}[/tex]

Step-by-step explanation:

We have to find the inverse Laplace transform of [tex]H(S)=\frac{1}{(S+4)^2}[/tex]

We know that of [tex]\frac{1}{s+4}[/tex] is [tex]e^{-4t}[/tex]

As in H(s) there is square of [tex]s+4[/tex]

So i inverse Laplace there will be multiplication of t

So the inverse Laplace of [tex]\frac{1}{(s+4)^2}[/tex]  will be [tex]te^{-4t}[/tex]

[tex]L^{-1}\frac{1}{(S+4)^2}=te^{-4t}[/tex]

The  difference in the cost for a one-night stay with and without breakfast is:

                             $ 23

Total cost to stay one night at sleepyhead Motel without breakfast is: $119.

and cost to stay one night at sleepyhead Motel with breakfast is: $142.

Hence, the difference in the cost for a one-night stay with and without breakfast is calculated by:

               $ (142-119)

                 =   $ 23

Hence, the answer is: $ 23

Answer:

Step-by-step explanation:

We can work with these values as a set value, and build a Venn Diagram from them.

I am going to say the set A are those that have the health insurance plan.

Set B are those that have the life insurance plan

Set C are those that have the investment plan.

We have that:

[tex]A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)[/tex]

In which a is the number of employees that only have the health insurance plan, [tex]A \cap B[/tex] is the number of employees that have both the health and the life insurance plans, [tex]A \cap C[/tex] is the number of employees that have both the health insurance and the investment plans. and [tex]A \cap B \cap C[/tex] is the number of employees that have all three of those plans.

By the same logic, we have that:

[tex]B = b + (A \cap B) + (B \cap C) + (A \cap B \cap C)[/tex]

[tex]C = c + (B \cap C) + (A \cap C) + (A \cap B \cap C)[/tex]

The problem states that:

An employee cannot have both life insurance and investment plans. So:

[tex]B \cap C = 0, A \cap B \cap C = 0[/tex]

45 employees have only the life insurance plan. So:

[tex]b = 45[/tex]

There are 20 more employees that have both health and life plans than those that have both health and investment plans

[tex]A \cap B = A \cap C + 20[/tex]

320 employees have the health insurance plan.

[tex]A = 320[/tex]

450 employees have at least one plan

[tex]a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 450[/tex]

330 employees have only one plan

[tex]a + b + c = 330[/tex]

How many people have the investment plan?

We have to find the value of C.

Now we solve:

[tex]a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 450[/tex]

Applying what we have

-----------

[tex]a + b + c + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 450[/tex]

[tex]330 + A \cap C + 20 + A \cap C = 450[/tex]

[tex]2(A\capC) = 100[/tex]

[tex]A \cap C = 50[/tex]

[tex]A \cap B = A \cap C + 20 = 50 + 20 = 70[/tex]

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

[tex]A = 320[/tex]

[tex]A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)[/tex]

[tex]a + 70 + 50 = 320[/tex]

[tex]a = 200[/tex]

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

[tex]b = 45[/tex]

[tex]a + b + c = 330[/tex]

[tex]245 + c = 330[/tex]

[tex]c = 85[/tex]

The number of people that have the investment plan is:

[tex]C = 85 + 50 = 135[/tex]

135 people have the investment plan

Answer:

The solution for the initial value problem is:

[tex]y(x) = -5e^{\frac{x^{2}}{2} - 2x} + 10[/tex]

Step-by-step explanation:

We have the following initial value problem:

[tex]\frac{dy}{dx} = (x-2)(y-10)[/tex]

The first step is solving the differential equation. We can do this by the variable separation method. It means that every term with y in on one side of the equality, every term with x on the other side. So:

[tex]\frac{dy}{dx} = (x-2)(y-10)[/tex]

[tex]\frac{dy}{y-10} = (x-2)dx[/tex]

To find y in function of x, we integrate both sides.

[tex]\frac{dy}{y-10} = (x-2)dx[/tex]

[tex]\int {\frac{1}{y-10}} \, dy = \int {(x-2)} \, dx[/tex]

Solving each integral separately

[tex]\int {\frac{1}{y-10}} \, dy[/tex]

This one we solve by substitution

[tex]u = y-10, du = dy[/tex]

[tex]\int {\frac{1}{y-10}} \, dy = \int {\frac{1}{u}} \, du = \ln{u} = \ln{(y-10)}[/tex]

[tex] \int {(x-2)} \, dx = \frac{x^{2}}{2} - 2x + K[/tex]

Now we have that:

[tex]\int {\frac{1}{y-10}} \, dy = \int {(x-2)} \, dx[/tex]

[tex]\ln{(y-10) = \frac{x^{2}}{2} - 2x + K}[/tex]

To solve for y, we apply the exponential to both sides, since the exponential and ln are inverse operations:

[tex]e^{\ln{(y-10)} = e^{\frac{x^{2}}{2} - 2x + K}[/tex]

[tex]y - 10 = Ke^{\frac{x^{2}}{2} - 2x}[/tex]

[tex]y(x) = Ke^{\frac{x^{2}}{2} - 2x} + 10[/tex]

[tex]y(0) = 5[/tex] means that when [tex]x = 0, y(x) = 5[/tex]. So:

[tex]5 = Ke^{\frac{0^{2}}{2} - 2*0} + 10[/tex]

[tex]Ke^{0} = -5[/tex]

[tex]K = -5[/tex]

So, the solution for the initial value problem is:

[tex]y(x) = -5e^{\frac{x^{2}}{2} - 2x} + 10[/tex]

Answer:

y = 2

Step-by-step explanation:

Answer:

y=2

Step-by-step explanation:

Answer:

88

Step-by-step explanation:

We are given that in arithmetic sequence , the nth term [tex]a_n[/tex] is given by the formula

[tex]A_n=a_1+(n-1)d[/tex]

Where [tex]a_1=first term[/tex]

d=Common difference

In an geometric sequence, the nth term is given by

[tex]a_n=a_1r^{n-1}[/tex]

Where r= Common ratio

1,4,7,10,..

We have to find 30th term.

[tex]a_1=1,a_2=4,a_3=7,a_4=10[/tex]

[tex]d=a_2-a_1=4-1=3[/tex]

[tex]d=a_3-a_2=7-4=3[/tex]

[tex]d=a_4-a_3=10-7=3[/tex]

[tex]r_1=\frac{a_2}{a_1}=\frac{4}{1}=4[/tex]

[tex]r_2=\frac{a_3}{a_2}=\frac{7}{4}[/tex]

[tex]r_1\neq r_2[/tex]

Therefore, given sequence is an arithmetic sequence because the difference between consecutive terms is constant.

Substitute n=30 , d=3 a=1 in the given formula of arithmetic sequence

Then, we get

[tex]a_{30}=1+(30-1)(3)=1+29(3)=1+87=88[/tex]

Hence, the 30th term of sequence is 88.

Area = Length time width.

The area of the shaded area is Length x width of the entire shape, minus the length x width of the unshaded area.

Full area: 27 x 10 = 270 square m.

Unshaded area: 17 x 5 = 85 square m.

Area of shaded region: 270 - 85 = 185 square m.

Answer:

The equation of trend line is [tex]y = 0.500000x + 16.857143[/tex].

The linear trend forecast for period 8 is about 20.86.

Step-by-step explanation:

The given data table is

Period            Sales

1                      19

2                     18

3                     15

4                     20

5                     18

6                     22

7                     20

We need to find the linear trend forecast for period 8.

The general form of linear regression is

[tex]y=a+bx[/tex]             .... (1)

where, a is y-intercept and b is slope.

[tex]b=\frac{\sum_{i=1}^nx_iy_i-n\overline{x}\overline{y}}{\sum_{i=1}^nx_i^2-n\overline{x}^2}[/tex]

[tex]a=\overline{y}-b\overline{x}[/tex]

Using the graphing calculator we get

[tex]a=16.857143[/tex]

[tex]b=0.500000[/tex]

Substitute these values in equation (1).

[tex]y = 0.500000x + 16.857143[/tex]

The equation of trend line is [tex]y = 0.500000x + 16.857143[/tex].

Substitute x=8 to find the linear trend forecast for period 8.

[tex]y = 0.500000(8) + 16.857143[/tex]

[tex]y =20.857143[/tex]

[tex]y \approx 20.86[/tex]

Therefore the linear trend forecast for period 8 is about 20.86.

Answer:

The sequence of functions [tex]\{x^{n}\}_{n\in \mathbb{N}}[/tex] converges to the function

[tex]f(x)=\begin{cases}0&0\leq x<1\\1&x=1\end{cases}[/tex].

Step-by-step explanation:

The limit [tex]\lim_{n\to \infty }c^{n}[/tex] exists and converges to zero whenever [tex]\lvert c \rvert <1 [/tex]. But, if [tex]c=1[/tex] the sequence [tex]\{c^{n}\}[/tex] is constant and all its terms are equal to [tex]1[/tex], then converges to [tex]1[/tex]. Using this result, consider the sequence of functions [tex]\{f_{n}\}[/tex] defined on the interval [tex][0,1][/tex] by [tex]f_{n}(x)=x^{n}[/tex]. Then, for all [tex]0\leq x<1[/tex] we have that [tex]\lim_{n\to \infty}x^{n}=0[/tex]. Now, if [tex]x=1[/tex], then [tex]\lim_{n\to \infty }x^{n}=1[/tex]. Therefore, the limit function of the sequence of functions is

[tex]f(x)=\begin{cases}0&0\leq x<1\\1&x=1\end{cases}[/tex].

To show that the convergence is not uniform consider [tex]0<\varepsilon<1[/tex]. For any [tex]n>1[/tex] choose [tex]x\in (0,1)[/tex]  such that [tex]\varepsilon^{1/n}<x<1[/tex]. Then

[tex]\varepsilon <x^{n}=\lvert f(x)-f_{n}(x)\rvert[/tex]

This implies that the convergence is not uniform.

Answer:

Maximize C =[tex]9x + 7y[/tex]

[tex]8x + 10y \leq 17[/tex]

[tex]11x + 12y\leq 25[/tex]

and x ≥ 0, y ≥ 0

Plot the lines on graph

[tex]8x + 10y \leq 17[/tex]

[tex]11x + 12y\leq 25[/tex]

[tex]x\geq 0[/tex]

[tex]y\geq 0[/tex]

So, boundary points of feasible region are (0,1.7) , (2.125,0) and (0,0)

Substitute the points in  Maximize C

At  (0,1.7)

Maximize C =[tex]9(0) + 7(1.7)[/tex]

Maximize C =[tex]11.9[/tex]

At  (2.125,0)

Maximize C =[tex]9(2.125) + 7(0)[/tex]

Maximize C =[tex]19.125[/tex]

At   (0,0)

Maximize C =[tex]9(0) + 7(0)[/tex]

Maximize C =[tex]0[/tex]

So, Maximum value is attained at   (2.125,0)

So, the optimal value of x is 2.125

The optimal value of y is 0

The maximum value of the objective function is 19.125

There are several ways to calculate optimal solution; one of them, is by using graphs.

The given parameters are:

[tex]\mathbf{Max\ C = 9x + 7y}[/tex]

[tex]\mathbf{8x + 10y \le 17}[/tex]

[tex]\mathbf{11x + 12y \le 25}[/tex]

[tex]\mathbf{x,y\ge 0}[/tex]

See attachment for the graphs of [tex]\mathbf{8x + 10y \le 17}[/tex] and [tex]\mathbf{11x + 12y \le 25}[/tex]

From the graph, the feasible regions are:

[tex]\mathbf{(x,y) = (0,1.7), (2.125,0)}[/tex]

Test these values in the objective function

[tex]\mathbf{Max\ C = 9x + 7y}[/tex]

[tex]\mathbf{C = 9 \times 0 + 7 \times 1.7 = 11.9}[/tex]

[tex]\mathbf{C = 9 \times 2.125 + 7 \times 0 = 19.125}[/tex]

So, the value of C is maximum at: [tex]\mathbf{(x,y) = (2.125,0)}[/tex]

So, we have:

Read more about maximizing functions at:

https://brainly.com/question/11212148

Final answer:

To calculate the sustainable growth rate of Mercoxit Energy Inc., the retention ratio is multiplied by the calculated Return on Equity (ROE), which is derived from the ROA and EM figures given. The result yields a sustainable growth rate of 5.21% for Mercoxit Energy Inc.

Explanation:

The sustainable growth rate (SGR) is a company's maximum growth rate in sales using internal financial resources while maintaining a constant debt to equity ratio. To calculate the SGR for Mercoxit Energy Inc., one needs to use the following formula:

SGR = Retention Ratio x Return on Equity (ROE)

While we are given the retention ratio and the Return on Assets (ROA), we are not directly given the ROE. However, we can calculate ROE using the provided Equity Multiplier (EM), since ROE can be written as:

ROE = ROA x EM

Considering the given figures:

Retention Ratio = 0.79

ROA = 5.94%

EM = 1.11

Now calculate ROE:

ROE = 0.0594 x 1.11 = 0.065934 (or 6.5934%)

Finally, we calculate the SGR:

SGR = 0.79 x 0.065934 = 0.05208786 (or 5.208786%)

If we round to two decimal places, Mercoxit Energy Inc. has a sustainable growth rate of 5.21%.

Answer:

Mean = 3640

Mode = 4100

Median = 3830.

Step-by-step explanation:

We are given the following data in the question:

Strength of casts (in psi):

3970,4100,3100,3200,2950,3830,4100,4050,3460

Formula:

[tex]Mean = \displaystyle\frac{\text{Sum of all observation}}{\text{Total number of observations}}[/tex]

[tex]\displaystyle\frac{3970+4100+3100+ 3200+ 2950+ 3830+4100+ 4050+ 3460}{9} = \displaystyle\frac{32760} {9} = 3640[/tex]

Mode is the entry with most frequency. Thus, for the given sample mode = 4100.

Median

Since n = 9 is odd,

Formula:

[tex]Median = \displaystyle\frac{n+1}{2}th~term[/tex]

Data in ascending order:

2950,3100,3200,3460,3830,3970,4050,4100,4100

Median = 5th term = 3830.

The mean strength of the concrete is 3640.

The mode strength of the concrete is 4100.

The median strength of the concrete is 3830.

Given

A concrete mix is designed to withstand 3000 pounds per square inch​ (psi) of pressure.

The following data represent the strength of nine randomly selected casts​ (in psi).

3970​, 4100​, 3100​, 3200​, 2950​, 3830​, 4100​, 4050​, 3460

The mean value of the data is given by;

[tex]\rm Mean = \dfrac{Sum \ of \ all \ observation}{Total \ number \ of \ observation}[/tex]

The mean of the strength of the concrete is;

[tex]\rm Mean = \dfrac{Sum \ of \ all \ observation}{Total \ number \ of \ observation}\\\\Mean =\dfrac{3970+4100+3100+3200+2950+3830+ 4100+4050+3460 }{9}\\\\Mean =\dfrac{32760}{9}\\\\Mean=3640[/tex]

The mean strength of the concrete is 3640.

Mode is the entry with the most frequency.

Thus, for the given sample mode = 4100.

The mode strength of the concrete is 4100.

The mid-value of the data is called the median.

The median strength of the concrete is 3830.

To know more about Mean click the link given below.

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[tex]\Rightarrow[/tex]

Suppose first that [tex]H\subset G[/tex] is a normal subgroup. Then by definition we must have for all [tex]a\in H[/tex], [tex]xax^{-1} \in H[/tex] for every [tex]x\in G[/tex]. Let [tex]a\in G[/tex] and choose [tex](ab)\in aH[/tex] ([tex]b\in H[/tex]). By hypothesis we have [tex]aba^{-1} =abbb^{-1}a^{-1}=(ab)b(ab)^{-1} \in H[/tex], i.e. [tex]aba^{-1}=c[/tex] for some [tex]c\in H[/tex], thus [tex]ab=ca \in Ha[/tex]. So we have [tex]aH\subset Ha[/tex]. You can prove [tex]Ha\subset aH[/tex] in the same way.

[tex]\Leftarrow[/tex]

Suppose [tex]aH=Ha[/tex] for all [tex]a\in G[/tex]. Let [tex]h\in H[/tex], we have to prove  [tex]aha^{-1} \in H[/tex] for every [tex]a\in G[/tex]. So, let [tex]a\in G[/tex]. We have that [tex]ha^{-1} =a^{-1}h'[/tex] for some [tex]h'\in H[/tex] (by the hypothesis). hence we have [tex]aha^{-1}=h' \in H[/tex]. Because [tex]a[/tex] was chosen arbitrarily  we have the desired .

Hi here´s a way to solve it

Let m and n be odd integers. Then,  we can express m as 2r + 1 and n as 2s + 1, where r and s are integers.

This means that any odd number can be written as the sum of some even integer and one.

Substituting, we have that m + n = (2r + 1) + 2s + 1 = 2r + 2s + 2.

As we defined r and s as integers, 2r + 2s + 2 is also an integer.

Now It is clear that 2r + 2s + 2 is an integer divisible by 2 becasue we have 2 in each of the integers.

Therefore,  2r + 2s + 2 = m + n is even.

So, the sum of two odd integers is even.

Answer:  Option 'c' is correct.

Step-by-step explanation:

Since we have given that

Number of ounces steaks = 4.5 oz

Number of guests = 180

Percentage of waste = 25%

Number of ounces of fresh steak is given by

[tex]4.5\times 180\\\\=810\ oz[/tex]

Let the number of ounces of raw steak be x.

According to question,

[tex]\dfrac{100-25}{100}\times x=810\\\\\dfrac{75}{100}\times x=810\\\\0.75\times x=810\\\\x=\dfrac{810}{0.75}\\\\x=1080\ oz[/tex]

As we need in pounds, and we know that

1 pound = 16 ounces

So, Number of pounds of raw steak that Chef could order is given by

[tex]\dfrac{1080}{16}=67.5\ lbs[/tex]

Hence, Option 'c' is correct.

To determine how many pounds of raw steak Chef should order for 180 guests with 25% waste, calculate the weight per guest and multiply by the number of guests. Convert the weight to pounds by dividing by 16. The answer is approximately B. 63.3 lbs.

To determine how many pounds of raw steak Chef should order, we need to calculate the total weight of steak required for 180 guests, taking into account the 25% waste.

First, we calculate the weight of steak for one guest by multiplying 4.5 oz by 1.25 (to account for the waste), which equals 5.625 oz.

Then, we multiply the weight per guest by the number of guests to get the total weight of steak required: 5.625 oz x 180 = 1012.5 oz.

Finally, we convert the ounces to pounds by dividing by 16, since there are 16 ounces in a pound.

Thus, Chef should order 1012.5 oz ÷ 16 = 63.28125 lbs, which rounds to approximately 63.3 lbs (option B).

Answer: 0.002718

Step-by-step explanation:

Given : The population mean annual salary for environmental compliance specialists is about ​$62,000.

i.e. [tex]\mu=62000[/tex]  

Sample size : n= 32

[tex]\sigma=6200[/tex]

Let x be the random variable that represents the annual salary for environmental compliance specialists.

Using formula [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex], the z-value corresponds to x= 59000 will be :

[tex]z=\dfrac{59000-62000}{\dfrac{6200}{\sqrt{32}}}\approx\dfrac{-3000}{\dfrac{6200}{5.6568}}=-2.73716129032\approx-2.78[/tex]

Now, by using the standard normal z-table , the probability that the mean salary of the sample is less than ​$59,000 :-

[tex]P(z<-2.78)=0.002718[/tex]

Hence, the probability that the mean salary of the sample is less than ​$59,000= 0.002718