You are offered two jobs selling college textbooks. One company offers an annual salary of $30,000 plus a year-end bonus of 1% of your total sales. The other company offers an annual salary of $25,000 plus a year-end bonus of 2% of your total sales. How much would you need to sell in a year to make the second offer the better offer?

The simplified form of the given expression as a fraction is 7x/x-7

Fractions  are written as a ratio of two integers. For instance a/b is a fraction where a and b are integers.

Given the expression below;

(1/7+1/x)/(1/49+1/x²)

Find the LCM to have:

(x+7/7x)/(x²-49)/49x²

Divide to have:

x+7/7x * 49x²/(x²-49)

(x+7) * 7x/(x+7)(x-7)

Cancel out the like terms to have:

1 * 7x/x-7

= 7x/x-7

Hence the simplified form of the given expression as a fraction is 7x/x-7

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Answer: C: 43

Step-by-step explanation:

As we are calling the function with 2 for x, we can substitute 2 for every x we see in the function and solve.

[tex]f(2)=2(2)^3-19(2)^2+28(2)+47\\=2(8)-19(4)+28(2)+47\\=16-76+56+47\\=43[/tex]

Hence, f(2) is 43.

Answer:

None of these

Step-by-step explanation:

[tex]\frac{360}{n}=24 \implies n=15[/tex]

This is called a pentadecagon.

The expression involving exponents to represent the shaded area in square inches of the diagram is: 6²- (3² + 2²). The shaded area in squares inches of the diagram is: 23 square inches.

The expression is:

6²- (3² + 2²)

The shaded area:

Shaded area=6²- (3² + 2²)

Shaded area=36-(9+4)

Shaded area=36-13

Shaded area=23 square inches

Therefore the expression involving exponents to represent the shaded area in square inches of the diagram is: 6²- (3² + 2²). The shaded area in squares inches of the diagram is: 23 square inches.

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Using the given definition and [tex]\Delta x=\frac{0-(-2)}n=\frac2n[/tex], we have

[tex]\displaystyle \int_{-2}^0 (7x^2+7x) \,dx = \lim_{n\to\infty} \sum_{i=1}^n \left(7\left(-2+\frac{2i}n\right)^2 + 7\left(-2+\frac{2i}n\right)\right) \frac2n \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac2n \sum_{i=1}^n \left(14 - \frac{42i}n + \frac{28i^2}{n^2}\right)[/tex]

Recall the well-known power sum formulas,

[tex]\displaystyle \sum_{i=1}^n 1 = \underbrace{1 + 1 + 1 + \cdots + 1}_{n\,\rm times} = n[/tex]

[tex]\displaystyle \sum_{i=1}^n i = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2[/tex]

[tex]\displaystyle \sum_{i=1}^n i^2 = 1 + 4 + 9 + \cdots + n^2 = \frac{n(n+1)(2n+1)}6[/tex]

Reducing our sum leads to

[tex]\displaystyle \int_{-2}^0 (7x^2+7x) \,dx = \lim_{n\to\infty} \frac2n \left(\frac{7n}3 - 7 + \frac{14}{3n}\right) = \lim_{n\to\infty} \left(\frac{14}3 - \frac{14}n + \frac{28}{3n^2}\right)[/tex]

As [tex]n[/tex] goes to ∞, the rational terms containing [tex]n[/tex] will converge to 0, and the definite integral converges to

[tex]\displaystyle \int_{-2}^0 (7x^2+7x) \,dx = \boxed{\frac{14}3}[/tex]

Answer:

C

Step-by-step explanation:

[tex]\cos E=\frac{11}{18} \\ \\ E=\cos^{-1} \left(\frac{11}{18} \right) \\ \\ E \approx 52.33^{\circ}[/tex]

If the sum of the length and width of rectangle is 60 and rectangle is having maximum area then the dimensions are 30 units each.

Given that the sum of length and breadth of rectangle is 60.

We are required to find the dimensions of the rectangle that will have the maximum area. Area is basically how much part of surface is being covered by that particular shape or substance.

Let the length of rectangle be x.

According to question the breadth will be (60-x).----2

Area of rectangle=Length *Breadth

A=x(60-x)

A=60x-[tex]x^{2}[/tex]

Differentiate A with respect to x.

dA/dx=60-2x

Again differentiate with respect to x.

[tex]d^{2} A/dA^{2}[/tex]=-2x

-2x<0

So the area is maximum because x cannot be less than or equal to 0.

Put dA/dx=0

60-2x=0

60=2x

x=30

Put the value of x in 2 to get the breadth.

Breadth=60-x

=60-30

=30

Hence if the sum of the length and width of rectangle is 60 and rectangle is having maximum area then the dimensions are 30 units each.

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The conclusion of the Hypothesis Conclusion is; that there is sufficient evidence to support the claim that the fund has moderate risk.

We are given;

Sample size; n = 36

Population standard deviation; σ₀ = 10

Sample standard deviation; s = 4.95

Significance level; α = 0.05

Claim: Standard deviation less than 10

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis needs to contain an equality and the value mentioned in the claim. If the claim is the null hypothesis, then the alternative hypothesis states the opposite of each other. Thus;

Null Hypothesis; H₀: σ = 10

Alternative Hypothesis; H₁: σ < 10

Compute the value of the test statistic:

χ2 = [(n - 1)/(σ²)] * s²

χ2 = [(36 - 1)/(10²)] * 4.95²

χ2 = 8.576

The critical value of the left-tailed test is given in the row with df = n - 1 = 36 - 1 = 35 and in the column with 1 − α = 0.95 of the chi-square distribution table online, we have;

χ2_{1 - α} = 21.77

The rejection region then contains all values smaller than 21.77

If the test statistic is in the rejection region, then reject the null hypothesis:

8.576 < 13.848

Thus, we will reject H₀ and conclude that there is sufficient evidence to support the claim that the fund has moderate risk.

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Answer:

Step-by-step explanation:

(1). A = π r²

A = 4 π

Area of shaded region is [tex]\frac{4\pi *260}{360}[/tex] = [tex]\frac{26}{9} \pi[/tex] or [tex]\frac{26\pi }{9}[/tex]

(2). C = 4 π

In this case, the length of the bigger arc and the area of shaded region happen to be the same.

The length of the arc ADB is  [tex]\frac{26}{9} \pi[/tex] or [tex]\frac{26\pi }{9}[/tex]

Answer:

A convenient formula to use is

S = ((1 + i)^n - 1) / i where S is the value of 1$ deposited for n periods at an interest rate of i

in this case n = 12 * 28 = 336 periods of deposit at an interest rate of

.0021 / 12 = .00175 = i

S = (1.00175^336 - 1) / .00175 = 456.8338  the value of 1$ after 336 periods

350 * 456.8338 = 159891.81   the value of 350 deposited monthly

Note that 350 * 336 would be 117,600

One must be careful to distinguish the above formula from

(1 - (1 + i)^-n) / i which gives the value of 1$ when the borrower is "paying" an interest rate of i - this would be the case for a mortgage - or what is  the value of 1$ paid for n periods when paying an interest rate of i

Step-by-step explanation:

volume = area of circle × hight of cylinder ....

V = πr² × h

so V = πr²h

Answer:

  10831 mm³

Step-by-step explanation:

The volume of any solid figure with a uniform cross section parallel to the base can be found using the volume formula ...

  V = Bh

where B is the area of the base, and h is the height perpendicular to the base.

The base of this cylindrical stack of pennies is a circle of diameter 19.05 mm. The area formula for a circle in terms of diameter is ...

  A = (π/4)d²

The stack of pennies has a base area of ...

  A = (π/4)(19.05 mm)² ≈ 285.023 mm²

The volume formula tells us the stack of pennies has a volume of ...

  V = Bh

  V = (285.023 mm²)(38 mm) ≈ 10830.9 mm³

The volume of the stack of pennies is approximately 10831 mm³.

The probability that it will rain on exactly one of the three days they are there is 0.375.

Given that Hiking Club plans to go camping in a State park where the probability of rain on any given day is 50%.

The binomial distribution is used when there are exactly two outcomes of a trial that are mutually exclusive.

Use binomial probability:

P = ⁿCʳ pʳ (1−p)ⁿ⁻ʳ

where n is the number of trials,

r is the number of successes,

and p is the probability of success.

probability of rain on any given day =p =0.50

total number of ways,n=3

let X is number of days with rain out of 3 days

So, [tex]X\sim Bin(3,0.50)[/tex]

P(x=X)=ⁿCₓpˣ(1-p)ⁿ⁻ˣ

P(x=1)=³C₁(0.50)¹(1-0.50)³⁻¹

P(x=1)=3×0.50×(0.5)²

P(x=1)=1.5×0.25

P(x=1)=0.375

Hence, the probability that it will rain on exactly one of the three days they are there where the probability of rain on any given day is 50% is 0.375.

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Answer:

23

Step-by-step explanation:

if you have 8 apples in a box, and then you put 15 additional apples into the box - how many apples are in the box ?

you see that is the same problem as the symbolized scenario with 2 lines being connected.

you have a string (e.g. for a kite) of 8 meters. and then you tie another string of 15 meters to it - how long is the new combined string ?

you really need help with that ?

8 + 15 = 23

Answer:

I dont know if this is right but i got -7/3.

Sorry if its wrong

Answer:

10x

Step-by-step explanation:

g(x)=5x

f(x)=2x

f(g(x))=f(5x)

f(g(x))=2*5x=10x

Answer:

7

Step-by-step explanation:

→ Find the scale factor

30 ÷ 25 = 1.2

→ Multiply answer by 20

20 × 1.2 = 24

→ Equate equation to 24

4x - 4 = 24

→ Add 4 to both sides

4x = 28

→ Divide both sides by 4

x = 7

Angles 2 and 4 are corresponding angles. They are congruent, not supplementary because they have the same measure and do not add up to 180 degrees. Therefore, the answer is the third option. Corresponding angles are congruent.

Answer:

None of these answers are correct.

Step-by-step explanation:

[tex]QR=\frac{25+45}{2}=35[/tex]

[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]

By Trapezoid mid - segment theorem :

[tex] \qquad❖ \: \sf \:QR = \cfrac{MN+ OP}{2} [/tex]

[tex] \qquad❖ \: \sf \:QR= \cfrac{25+ 45}{2} [/tex]

[tex] \qquad❖ \: \sf \:QR = \cfrac{70}{2} [/tex]

[tex] \qquad❖ \: \sf \:QR= 35 \: \: units[/tex]

[tex] \qquad \large \sf {Conclusion} : [/tex]

The required rewritten equation in terms of variable 'r' is r = -u/s²t. By applying simple arithmetic operations, the required equation is obtained.

Consider an equation c = ax + by. Solving for variable 'a'.  

Step1: Write the required variable terms on one side

Step2: Add/Subtract like terms if any or take common if any

Step3: Divide/Multiply the coefficient of the required variable

Step4: simplify the obtained terms for the required variable

As follows:

c = ax + by

⇒ ax = by - c

⇒ ax/x = (by - c)/x

∴ a = (by - c)/x

The given equation is u = -s²rt

dividing by 't' on both sides:

⇒ u/t = -s²rt/t

⇒ u/t = -s²r

dividing by 's²' on both sides:

⇒ u/s²t = -s²r/s²

⇒ u/s²t = -r

∴ r = -u/s²t

Thus, the required equation for variable 'r' is r = -u/s²t.

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Step-by-step explanation:

a)

z = (140 - 118)/16 = 22/16 = 11/8 = 1.375 ≈ 1.38

in the z-table this gives us the p-value : 0.91621

that is the probability of values 140 and below.

for above 140 we need to calculate the value of the other side of the bell-curve :

1 - 0.91621 = 0.08379

b)

z = (90 - 118)/16 = -28/16 = -7/4 = -1.75

in the z-table this gives us the p-value : 0.04006

that is the probability of values of 90 and below.

Step-by-step explanation:

second one

explanation: trust me bro

The computation shows that:

G n M = Anael and Max

G u S = Acel, Action, Anael, Max, Carl, Dario, Catlin, Kai, and Barek.

From the information, ten students from a school appear in one or more subjects for an inter school quiz competition as shown in the table.

The subjects include general Knowledge Math Science.

Therefore, G represents the set of students appearing for General Knowledge, M represents the set of students appearing for Math, and S represents the set of students appearing for Science. The intersection is illustrated above.

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Answer + Step-by-step explanation:

the correct question:

For r ≠ 1 ,Prove using the mathematical induction method that :

[tex]1+\cdots+r^n =\frac{1-r^{n+1}}{1-r}[/tex]

………………………………………………………………………………………………………………

for n = 0 :

1⁰ = 1  and  (1 - r⁰⁺¹)/(1 - r) = (1 - r)/(1 - r) = 1

Then the property is true for n = 0.

For n ≥ 0 :

Suppose

[tex]1+\cdots+r^n =\frac{1-r^{n+1}}{1-r}[/tex]

And prove that

[tex]1+\cdots+r^{n+1} =\frac{1-r^{n+2}}{1-r}[/tex]

Since :

[tex]1+\cdots+r^{n+1} =(1+\cdots+r^n)+r^{n+1}[/tex]

Then

[tex]1+\cdots+r^n+r^{n+1} =\frac{1-r^{n+1}}{1-r}+r^{n+1}[/tex]

[tex]= \frac{1-r^{n+1}+r^{n+1}(1-r)}{1-r}[/tex]

[tex]= \frac{1-r^{n+2}}{1-r}[/tex]

Then according to the mathematical induction method

[tex]1+\cdots+r^n =\frac{1-r^{n+1}}{1-r}[/tex]

Where n is a natural number and r ≠ 1.

The given number of balloons, green = 10, purple = 4, red = 5, tie-dye = 2, black = 3, gives;

a. 5/138

b. 25/276

c. 5/1012

d. 35/46

Given parameters;

Number of balls;

Green = 10

Purple = 4

Red = 5

Tie-dye = 2

Black = 3

Mode of selection = Without replacement

Number of balloons = 10+4+5+2+3 = 24

a. Probability of popping a red balloon = 5/24

Probability of popping a second red balloon = 4/23

Therefore;

Probability of popping two reds consecutively = 5/24 × 4/23 = 5/138

b. Probability of popping a red balloon = 5/24

Probability of popping a green balloon next = 10/23

Therefore;

Probability of popping a red and then a green balloon = 5/24 × 10/23 = 25/276

c. Probability that the first balloon that pops is a red = 5/24

Next balloon is a black = 3/23

Third balloon is red = 4/22

The probability, P, is therefore;

d. The probability that the first balloon is a tie-dye = 2/24 = 1/12

Therefore;

Probability that the first balloon is not a tie-dye = 1 - 1/12 = 11/12

Probability that the second balloon is not a tie-dye = 21/23

Similarly;

Probability that the third balloon is not a tie-dye = 20/22 = 10/11

Which gives;

The probability, P, of popping anything but a tie-dye on three consecutive throws is therefore;

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[tex]~~~~~~ \textit{Annual Percent Yield Formula} \\\\ ~~~~~~~~~~~~ APY=\left(1+\frac{r}{n}\right)^{n}-1 ~\hfill \begin{cases} r=rate\to 10\%\to \frac{10}{100}\dotfill &0.1\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12 \end{cases} \\\\\\ APY=\left(1+\frac{0.1}{12}\right)^{12}-1\implies APY=\left( \frac{121}{120} \right)^{12}-1\implies APY\approx 0.1047[/tex]

The given height of the shape is calculated to be 1,895.04 inches.

We have the following details required to solve this problem

The distance b1 = 800 m

The angle of elevation β1 = 50°

distance b2 = 1450 m

- The new angle of elevation  α = 33°

We first have to find the height of h1

800 x tan 50 degrees

= 953.40

The height of h2

1450 x tan 33

= 941.641

The height of the shape is the sum of the heights of h1 and h2

953.40 + 941.641

= 1,895.04

Hence the height of the shape is given as 1,895.04 inches.

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The true statement about the series [tex]\sum\limits^{\infty}_{n=1} \frac{9}{n(n +3)}[/tex] is that (a) the series diverges

The series is given as:

[tex]\sum\limits^{\infty}_{n=1} \frac{9}{n(n +3)}[/tex]

Take the limit of the function to infinity

[tex]\lim_{n \to \infty} \frac{9}{n(n +3)}[/tex]

This gives

[tex]\lim_{n \to \infty} \frac{9}{n(n +3)} = \frac{9}{\infty * (\infty +3)}[/tex]

Evaluate the sum

[tex]\lim_{n \to \infty} \frac{9}{n(n +3)} = \frac{9}{\infty * \infty}[/tex]

Evaluate the product

[tex]\lim_{n \to \infty} \frac{9}{n(n +3)} = \frac{9}{\infty}[/tex]

Evaluate the quotient

[tex]\lim_{n \to \infty} \frac{9}{n(n +3)} = 0[/tex]

Since the limit is 0, then it means that the series diverges

Hence, the true statement about the series [tex]\sum\limits^{\infty}_{n=1} \frac{9}{n(n +3)}[/tex] is that (a) the series diverges

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Answer:

B. The series converges to [tex]\displaystyle{\frac{11}{2}}[/tex].

Step-by-step explanation:

Before evaluating the infinite series, the expression can be decomposed as the sum of two fractions (partial fraction decomposition) as follows.

Let [tex]\textit{A}[/tex] and [tex]\textit{B}[/tex] be constants such that

                                       [tex]{\displaystyle{\frac{9}{n\left(n+3\right)}}}} \ \ = \ \ \displaystyle{\frac{A}{n} \ \ \ + \ \ \frac{B}{n+3}}[/tex]

Multiply both sides of the equation by the denominator of the left fraction,

[tex]n\left(n+3\right)[/tex], yielding

                                              [tex]9 \ \ = \ \ A\left(n+3\right) \ \ + \ \ B \-\hspace{0.045cm} n[/tex]

Now, let [tex]n \ = \ 0[/tex], thus

                                             [tex]\-\hspace{0.2cm} 9 \ \ = \ \ A\left(0 + 3\right) \ + \ B\left(0\right) \\ \\ 3 \-\hspace{0.035cm} A \ = \ \ 9 \\ \\ \-\hspace{0.11cm} A \ \ = \ \ 3[/tex].

Likewise, let [tex]n \ = \ -3[/tex], then

                                            [tex]\-\hspace{0.5cm} 9 \ \ = \ \ A\left(-3 + 3\right) \ + \ B\left(-3\right) \\ \\ -3 \-\hspace{0.035cm} B \ = \ \ 9 \\ \\ \-\hspace{0.44cm} B \ \ = \ \ -3[/tex]

Hence,

                                       [tex]\displaystyle{\sum_{n=1}^{\infty} {\frac{9}{n\left(n+3\right)}}} \ = \ \displaystyle\sum_{n=1}^{\infty} \left(\frac{3}{n} \ - \ \frac{3}{n+3}\right)[/tex].

First and foremost, write the nth partial sum (first nth terms) of the series,

          [tex]\displaystyle\sum_{n=1}^{n} \left(\frac{3}{n} \ - \ \frac{3}{n+3}\right) \ \ = \ \-\hspace{0.33cm} \displaytstyle{\frac{3}{1} \ - \frac{3}{4} + \ \frac{3}{2} \ - \frac{3}{5} \ + \frac{3}{3} \ - \frac{3}{6} \ + \frac{3}{4} \ - \frac{3}{7}} \\ \\ \\ \-\hspace{3.58cm} + \ \displaystyle{\frac{3}{5} \ - \ \frac{3}{8} \ + \ \frac{3}{6} \ - \ \frac{3}{9} \ + \ \frac{3}{7} \ - \ \frac{3}{10}} \\ \\ \\ \-\hspace{3.58cm} + \ \ \dots[/tex]

                                                [tex]+ \ \ \displaystyle{\frac{3}{n-3} \ - \ \frac{3}{n} \ + \ \frac{3}{n-2} \ - \ \frac{3}{n+1}} \\ \\ \\ \ + \ \frac{3}{n-1} \ - \ \frac{3}{n+2} \ + \ \frac{3}{n} - \ \frac{3}{n+3}}[/tex].

Notice that the expression forms a telescoping sum where subsequent terms cancel each other, leaving only

        [tex]\displaystyle\sum_{n=1}^{n} \left(\frac{3}{n} \ - \ \frac{3}{n+3}\right) \ \ = \ \-\hspace{0.33cm} \displaytstyle{\frac{3}{1} \ + \ \frac{3}{2} \ + \frac{3}{3} \ - \ \frac{3}{n+1}} - \ \frac{3}{n+2} \ - \ \frac{3}{n+3}}}[/tex].

To determine if this infinite series converges or diverges, evaluate the limit of the nth partial sum as [tex]n \ \rightarrow \ \infty[/tex],

         [tex]\displaystyle\sum_{n=1}^{\infty} \left(\frac{3}{n} \ - \ \frac{3}{n+3}\right) \ \ = \ \-\hspace{0.33cm} \lim_{n \to \infty} \left(\displaytstyle{\frac{11}{2} \ - \ \frac{3}{n+1} \ - \ \frac{3}{n+2} \ + \ - \ \frac{3}{n+3}\right) \\ \\ \\ \-\hspace{3.25cm} = \ \ \ \displaystyle{\frac{11}{2} \ - \ 0 \ - \ 0 \ - \ 0} \\ \\ \\ \-\hspace{3.25cm} = \ \ \ \displaystyle{\frac{11}{2}[/tex]

The upper quartile for the given data is 51 .

Given data: {51,49,52,46,50,38,38,45,34,52,46}

Data in arranged form: {34,38,38,45,46,46,49,50,51,52,52}

Find the median first, then the upper quartile. There are eleven data points, thus look at term six to determine the median as there are five data points on any side. The median is 46 because the sixth term is 46.

The upper extreme, 52, and the median are then used to determine the upper quartile; make sure not to include the median data point when dividing the groups. The upper quartile is the third term (between the median and upper extreme) because there would be two data points on each side if there were only five data points from the median to the upper extreme. The third term, which is 51, falls into the upper quartile.

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