Ron deposits $2,000 into an account that receives 3.1% interest compounded continuously. How much money is in the account after 9 years?
A.) $2,623.70
B.) $2,632.44
C.) $2,643.61
D.) $2,605.83

The final amount is $7,615.27

A = P(1 + r/n)^t

Where,

A = Final amount

P = principal = $7, 200

r = interest rate = 2.5% = 0.025

n = number of periods = 4

t = time = 9 years

A = P(1 + r/n)^t

= 7,200(1 + 0.025/4)^9

= 7,200(1 + 0.00625)^9

= 7,200(1.00625)^9

= 7,200(1.0576769512798)

= 7,615.2740492152

Approximately,

A = $7,615.27

https://brainly.com/question/14003110

Answer:

you will add the numerator and the denominator and or you look for lowest common factor

9514 1404 393

Answer:

  asymptotes: x = ±6

  zero: x = 0

Step-by-step explanation:

The vertical asymptotes of the function will be at the values of x where the denominator is zero. The denominator is x^2 -36, so has zeros for values of x that satisfy ...

  x^2 -36 = 0

  x^2 = 36

  x = ±√36 = ±6

The vertical asymptotes of the function are x = -6 and x = +6.

__

The zero of the function is at the value of x that makes the numerator zero. This will be the value of x that satisfies ...

  6x = 0

  x = 0 . . . . . divide by 6

The zero of the function is x=0.

__

As a check on this work, we have had a graphing calculator graph the function and identify the zero.

Answer:

No

Step-by-step explanation:

4:1

Multiply each side by 3

4*3 : 1*3

12: 3

This is not equal to 12:9

Answer:

no

Step-by-step explanation:

12 : 9

4 : 3 ≠ 4 : 1

Answer:

5765865876+5737555586=11503421462

What do you personaly feel like is most dificult.

For me its rembering minus signs

Answer:

€36 = 30.62 pounds sterling

Answer:

? = 13.6

Step-by-step explanation:

Let the unknown angle be y

so

tan y= p/b

tan y =8/33

y = tan‐¹(8/33)

y = 13.62699486

y = 13.6

Answer:

Step-by-step explanation:

[tex](a^2-1)x^2+(a-1)x+a^2-4a+3=0\\\\Calculate\ and\ identify\ the\ polynomials\\\\\Longleftrightarrow\ a^2x^2-x^2+ax-x+a^2-4a+3=0\\\\\Longleftrightarrow\ a^2x^2+ax+a^2-4a+3=x^2+x+0\\\\\Longleftrightarrow\ \left\{\begin{array}{ccc}a^2&=&1\\a&=&1\\a^2-4a+3&=&0\\\end{array} \right.\\\\\Longleftrightarrow\ \left\{\begin{array}{ccc}(a-1)(a+1)&=&0\\a-1&=&0\\(a-1)(a-3)&=&0\\\end{array} \right.\\\\\\We\ must\ exclude\ a=-1\ and\ a=3\ (not\ solution)\\\Longrightarrow\ a=1\\[/tex]

answer is 6

Step-by-step explanation:

3x+11=y

y=29

3x+11=29

3x=29-11

3x=18

x=18÷3

x=6

Answer:6

Step-by-step explanation:

3x+11=29

3x=29-11

3x=18

X=18/3

X=6

Answer:

682

Step-by-step explanation:

191,919 + 262,626

454545 ÷ 666 = 682.5

Thus meaning 682 bags will have 666 berries and one bag will have 333 berries.

Answer:

Substitute the functions and the value of the functions.

Step-by-step explanation:

Doing all will be long, so i'll present a and d

Here,(no a)

f(x)=3x-1, g(x)=x^2+2

Now,

f(g(x))=f(x^2+2)=3(x^2+2)-1=3x^2+6-1=3x^2+5

g(f(x))=g(3x-1)=(3x-1)^2+2=9x^2-6x+1+2=9x^2-6x+3

Here, (no d)

f(x)=x^2-9, g(x)=√(x+4)

Now,

f(g(x))=f(√(x+4))=(√(x+4))^2-9=x+4-9=x-5

g(f(x))=g(x^2-9)=√(x^2-9+4)=√(x^2-5)

Answer:

210 miles in 1 hour

Step-by-step explanation:

steps are in picture

We can seperate (3b³) into two different parts, the constant and the variable.

The constant (3) and the variable (b) can both be squared and multiplied to get the correct answer, so:

3² = 9

(b³)² = [tex]b^{6}[/tex]

So, [tex](3b^{3})^{2} = 9b^{6}[/tex]

Complete Question

Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and inexpensive to manufacture. However, their nondegradable nature has prompted development of environmentally friendly composites using natural materials. An article reported that for a sample of 10 specimens with 2% fiber content, the sample mean tensile strength (MPa) was 51.1 and the sample standard deviation was 1.2. Suppose the true average strength for 0% fibers (pure cellulose) is known to be 48 MPa. Does the data provide compelling evidence for concluding that true average strength for the WSF/cellulose composite exceeds this value? (Use α = 0.05.)  

t=8.169  

P-value= ?

Answer:

a)  [tex]P-value=0[/tex]

b)  Hence,We FAil to reject the alternative hypothesis and accept that the true average strength for the WSF/  cellulose composite exceeds 48 MPa.

Step-by-step explanation:

From the question we are told that:

Sample size [tex]n=10[/tex]

Mean [tex]\=x= 51.3[/tex]

Standard deviation [tex]\sigma=1.2[/tex]

Significance level is taken as [tex]\alpha=0.05[/tex]

t test statistics

[tex]t=8.169[/tex]

Therefore

[tex]P-Value=P(t>8.169)[/tex]

Critical point

[tex]t_{\alpha,df}[/tex]

[tex]\alpha=0.05[/tex]

[tex]df=10-1=>9[/tex]

Therefore

P-value from T distribution table

[tex]P-value=0[/tex]

Conclusion

[tex]P-value (0)< \alpha(0.05)[/tex]

We Reject the Null Hypothesis [tex]H_0[/tex]

Hence,We FAil to reject the alternative hypothesis and accept that the true average strength for the WSF/  cellulose composite exceeds 48 MPa.

Answer:

Tandin has 16 children.

Step-by-step explanation:

Total of children:

3+5 = 8(first woman)

2(youngest wife)

1 + 1 = 2(two of the  remaining wives)

So

8 + 2 + 2 = 12

If the ratio of children of  5th wife was 1:3 with the children of other wives.

Thus the 5th wife has 12/3 = 4 children.

How many  children does Tandin have?

12 + 4 = 16

Tandin has 16 children.

Answer:

D

Step-by-step explanation:

Surface area is area of base plus all 4 sides.  Base is a square with s = 8 so area is 64sq yd.  Sides are triangles, with A = 1/2 bh, so 1/2(8)(5) = 20.  Four sides would be 80sq yds.  Add base and sum is 144 sq yds.

first we shall learn the rules.when numbers with same sign are divided it gives pisitive sign but, when numbers of different signs are divided it gives negetive sign.

here,

7. (-154) ➗ (-14) =11

11. (-40) ➗10=-4

15. 90 ➗ (-15)=-6

16. 108 ➗ (-9)=-12

17. (-48) ➗ (-6)=8

18. (-105) ➗ 7=-15

hope it helps you......

Answer:

The probability tree is;

                            0.95        [tex](+)[/tex]

                   [tex](S)[/tex]

          0.1              0.05        [tex](-)[/tex]

[  P  ]

          0.9             0.15          [tex](+)[/tex]                                                        

                  [tex](S_{no})[/tex]    

                             0.85         [tex](-)[/tex]        

Step-by-step explanation:

Given the data in the question;

10% of the rugby team members use steroids

so Probability of using steroid; P( use steroid ) = 10% = 0.10

Probability of not using steroid; P( no steroid use ) = 1 - 0.10 = 0.90

Since the test show positive for an athlete who uses steroids, 95% of the time.

Probability of using steroids and testing positive = 95% = 0.95

Probability of using steroids and testing Negative = 1 - 0.95 = 0.05

Also from the test, 15% of all steroid-free individuals also test positive.

so

Probability of not using steroids and testing positive = 15% = 0.15

Probability of not using steroids and testing negative = 1 - 0.15 = 0.85

To set up the probability tree, Let;

[tex](S)[/tex] represent steroid use

[tex](S_{no})[/tex] represent no steroid use

[tex](+)[/tex] represent test positive

[tex](-)[/tex] represent test negative

so we have;

                            0.95        [tex](+)[/tex]

                   [tex](S)[/tex]

          0.1              0.05        [tex](-)[/tex]

[  P  ]

          0.9             0.15          [tex](+)[/tex]                                                        

                  [tex](S_{no})[/tex]    

                             0.85         [tex](-)[/tex]        

Answer: 18π

okokok gg

Step-by-step explanation:

Here angle is given in degree.We have convert it into radian.

[tex] {1}^{\circ} =( { \frac{\pi}{180} } )^{c} \\ \therefore \: {80}^{\circ} = ( \frac{80\pi}{180} ) ^{c} = {( \frac{4\pi}{9} })^{c} \: = \theta ^{c} [/tex]

Area of green shaded regions = A

[tex] \sf \: A = \frac{1}{2} { {r}^{2} }{ { \theta}^{ c} } \\ = \frac{1}{2} \times {9}^{2} \times \frac{4\pi}{9} \\ = 18\pi \: {cm}^{2} [/tex]

Let the number be x

ATQ

[tex]\\ \sf\twoheadrightarrow 6x-18=96[/tex]

[tex]\\ \sf\twoheadrightarrow 6x=96+18[/tex]

[tex]\\ \sf\twoheadrightarrow 6x=112[/tex]

[tex]\\ \sf\twoheadrightarrow x=\dfrac{112}{6}[/tex]

[tex]\\ \sf\twoheadrightarrow x=7[/tex]

Consecutive terms in this sequence are separated by a constant c, so if the 4th term is 15, then the next terms would be

5th: 15 + c

6th: (15 + c) + c = 15 + 2c

7th: (15 + 2c) + c = 15 + 3c

and so on. More generally, since any given number in the sequence depends on the number that came before it, we can write the n-th term in terms of the 4th term,

n-th: 15 + (n - 4) c

Then the 9th term in the sequence is

15 + (9 - 4) c = 35

and solving for c gives

15 + 5c = 35   ==>   5c = 20   ==>   c = 4

Then the 15th term would be

15 + (15 - 4)×4 = 15 + 11×4 = 15 + 44 = 59

Answer:

vertical asymptotes

x=6, x=-6

horizontal asymptotes

y=0

zeros (0,0)

Step-by-step explanation:

f(x) = 6x / ( x^2 - 36)

First factor

f(x) = 6x / ( x-6)(x+6)

Since nothing cancels

The vertical asymptotes are when the denominator goes to zero

x-6 = 0   x+6=0

x=6          x= -6

Since the numerator has a smaller power than the denominator (1 < 2), there is an asymptote at y = 0

To find the zeros, we find where the numerator = 0

6x=0

x=0

[tex]\\ \rm\Rrightarrow y=\dfrac{6x}{x^2-36}[/tex]

The h orizontal asymptote

As x has less degree than x²

Vertical asymptote

Answer:

8 cm^2

Step-by-step explanation:

If the triangle is isosceles the sides are the same

Let the sides be x

We know that we can use the Pythagorean theorem

a^2+b^2 = c^2 where a and b are the sides and c is the hypotenuse

x^2+x^2 = (4 sqrt(2))^2

2x^2 =16(2)

2x^2 = 32

Divide by 2

2x^2/ 32/2

x^2 = 16

Taking the square root of each side

sqrt(x^2) = sqrt(16)

x = 4

The area of the triangle is

A =1/2 bh

A = 1/2 (4) (4)

A = 1/2(16)

A = 8

Answer:

8

Step-by-step explanation:

a^2 + b^2 = c^2

c = [tex]4\sqrt{2}[/tex]

[tex]c^{2} = 32[/tex]

a^2 + b^2 = 32

a=b (isosceles triangle)

a=b=4

base = 4

height = 4

area = 1/2 bh = 1/2(4)(4) = 8

It sounds like R is the region (in polar coordinates)

R = {(r, θ) : 2 ≤ r ≤ 3 and 0 ≤ θ ≤ π/2}

Then the integral is

[tex]\displaystyle \iint_R\frac{\mathrm dx\,\mathrm dy}{\sqrt{x^2+y^2}} = \int_0^{\pi/2}\int_2^3 \frac{r\,\mathrm dr\,\mathrm d\theta}{\sqrt{r^2}} \\\\ = \int_0^{\pi/2}\int_2^3 \mathrm dr\,\mathrm d\theta \\\\ = \frac\pi2\int_2^3 \mathrm dr \\\\ = \frac\pi2r\bigg|_2^3 = \frac\pi2 (3-2) = \boxed{\frac\pi2}[/tex]

Answer:

A)

[tex]k=0[/tex]

B)

[tex]\displaystyle \begin{aligned} 2k + 1& = 2\ln 20 + 1 \\ &\approx 2.3863\end{aligned}[/tex]

C)

[tex]\displaystyle \begin{aligned} k - 3&= \ln \frac{1}{2} - 3 \\ &\approx-3.6931 \end{aligned}[/tex]

Step-by-step explanation:

We are given the function:

[tex]\displaystyle h(x) = 20e^{kx} \text{ where } k \in \mathbb{R}[/tex]

A)

Given that h(1) = 20, we want to find k.

h(1) = 20 means that h(x) = 20 when x = 1. Substitute:

[tex]\displaystyle (20) = 20e^{k(1)}[/tex]

Simplify:

[tex]1= e^k[/tex]

Anything raised to zero (except for zero) is one. Therefore:

[tex]k=0[/tex]

B)

Given that h(1) = 40, we want to find 2k + 1.

Likewise, this means that h(x) = 40 when x = 1. Substitute:

[tex]\displaystyle (40) = 20e^{k(1)}[/tex]

Simplify:

[tex]\displaystyle 2 = e^{k}[/tex]

We can take the natural log of both sides:

[tex]\displaystyle \ln 2 = \underbrace{k\ln e}_{\ln a^b = b\ln a}[/tex]

By definition, ln(e) = 1. Hence:

[tex]\displaystyle k = \ln 2[/tex]

Therefore:

[tex]2k+1 = 2\ln 2+ 1 \approx 2.3863[/tex]

C)

Given that h(1) = 10, we want to find k - 3.

Again, this meas that h(x) = 10 when x = 1. Substitute:

[tex]\displaystyle (10) = 20e^{k(1)}[/tex]

Simplfy:

[tex]\displaystyle \frac{1}{2} = e^k[/tex]

Take the natural log of both sides:

[tex]\displaystyle \ln \frac{1}{2} = k\ln e[/tex]

Therefore:

[tex]\displaystyle k = \ln \frac{1}{2}[/tex]

Therefore:

[tex]\displaystyle k - 3 = \ln\frac{1}{2} - 3\approx-3.6931[/tex]

Answer:

z(max) = 16

z(min) = -24

Step-by-step explanation:

2x - y = 12  multiply by 2

4x - 2y = 24  (1)

4x + 2y = 0  add equations

8x = 24

x = 3

4(3) + 2y = 0

y = -6

so (3, -6) is a common point on these two lines

z = 2(3) + 5(-6) = -24

4x - 2y = 24   (1)

x + 2y = 6   add equations

      5x = 30

        x = 6

6 + 2y = 6

         y = 0

so (6, 0) is a common point on these two lines

z = 2(6) + 5(0) = 12

4x + 2y = 0    multiply by -1

-4x - 2y = 0

  x + 2y = 6  add equations

       -3x = 6

         x = -2

-2 + 2y = 6

         y = 4

so (-2, 4) is a common point on these two lines

z = 2(-2) + 5(4) = 16