Use the fact 0.09=1/11 to write 0.27 as a fraction

Answer:

A:The Reqd. No. of Permutations=3360

Step-by-step explanation:

I couldn’t find the answer to B..sorry

The value of the expression when g = -2 is -1

Given the expression;

(5+2g)exp5

(5+2g)^5

For g = -2

Let's substitute the value of g in the expression

= ( 5 + 2 ( -2) ) ^5

Expand the bracket

= ( 5 - 4) ^ 5

Find the difference

= (-1) ^5

= -1

Thus, the value of the expression when g = -2 is -1

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

[tex]\textsf{ \underline{\underline{Steps to solve the problem} }:}[/tex]

Take HJ = a, GH = b and GJ = c

put the value of a from equation 1 in equation 2

now, put the value of a and c in equation 3

Now, we need to find HJ (a)

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

Answer:

Step-by-step explanation:

I am going to be honest here. I know the answer is 22 but I cant really explain it you kinda just have to trust I'm right.

The solutions to the quadratic equation of the sprinkler in the exact form are x = 7 + 2√110 and x = 7 - 2√110

Quadratic equations are second-order polynomial equations and they have the form y = ax^2 + bx + c or y = a(x - h)^2 + k

A quadratic equations can be split to several equations and it can be solved as a whole

In this case, the quadratic equation is given as

-x^2 + 14x + 61 = 0

Using the form of the quadratic equation y = ax^2 + bx + c, we have

a = -1, b = 14 and c = 61

The quadratic equation can be solved using the following formula

x = (-b ± √(b^2 - 4ac))/2a

Substitute the known values of a, b and c in the above equation

x = (-14 ± √(14^2 - 4*-1*61))/2*-1

Evaluate the exponent

x = (-14 ± √(196 - 4*-1*61))/2*-1

Evaluate the products

x = (-14 ± √(196 + 244))/-2

Evaluate the sum

x = (-14 ± √(440))/-2

Express 440 as 4 * 110

x = (-14 ± √(4 * 110))/-2

Take the square root of 4

x = (-14 ± 2√110)/-2

Evaluate the quotient

x = 7 ± 2√110

Split the equation

x = 7 + 2√110 and x = 7 - 2√110

Hence, the solutions to the quadratic equation of the sprinkler in the exact form are x = 7 + 2√110 and x = 7 - 2√110

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The Gauss-Jordan elimination method different from the Gaussian elimination method in that unlike the Gauss-Jordan approach, which reduces the matrix to a diagonal matrix, the Gauss elimination method reduces the matrix to an upper-triangular matrix.

Gauss-Jordan Elimination is a technique that may be used to discover the inverse of any invertible matrix as well as to resolve systems of linear equations.

It is based on the following three basic row operations that one may apply to a matrix: Two of the rows should be switched around. Multiply a nonzero scalar by one of the rows.

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Part (a)

[tex]z=\frac{85-75}{11.4} \approx \boxed{0.88}[/tex]

By similar logic, the answers to the other parts are

(b) 1.75

(c) -2.19

(d) 0

A null hypothesis (H₀) can be defined the opposite of an alternate hypothesis (H₁) and it asserts that two (2) possibilities are the same.

The test statistic can be calculated by using this formula:

[tex]t=\frac{x\;-\;u}{\frac{\delta}{\sqrt{n} } }[/tex]

Where:

For this clinical trial (study), we should use a t-test and the null and alternative hypotheses would be given by:

H₀: μ = 7

H₀: μ ≠ 7

Next, we would calculate the t-test as follows:

[tex]t=\frac{7.08\;-\;7}{\frac{0.25}{\sqrt{14} } }\\\\t=\frac{0.08}{\frac{0.25}{3.7417 } }[/tex]

t = 0.08/0.0668

t = 1.198.

For the p-value, we have:

P-value = P(t < 1.198)

P-value = 0.1262.

Therefore, the p-value (0.1262) is greater than α = 0.10. Based on this, we should fail to reject the null hypothesis.

In conclusion, yes the mean discharge differs from 7 fluid ounces.

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She paid a simple interest of $3,125.00 on the borrowing of $25,000

What is simple interest?

The simple interest on a loan or an investment can be determined as the principal multiplied by interest multiplied by the number of periods.

This is quite different from compound interest where the interest earned previously would earn interest in the future alongside the principal

I=PRT

I=interest on loan=unknown

P=amount borrowed=$25,000

R=interest rate=1% per month

T=12.5 months( from December 2 2011 to December 1 2012 makes one year and from December 1-16 gives 15 days, which is 0.5 of one month)

I=$25,000*1%*12.5

I=$3,125.00

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The equation of the parabola with the given vertex and directrix in vertex form is y = (1/16)( x + 5 )² - 9.

Hence, option C is the correct answer.

Given the data in the question;

To find the equation, we use the equation of the parabola that opens up or down since the directrix ( y = -13 ) is vertical.

The equation is expressed as;

( x - h )² = 4p( y - k )

First, we find the distance from the focus to the vertex.

|p| is the distance rom the focus to the vertex and from the vertex to the directrix.

p = -9 + 13

p = 4

We substitute the values into the equation;

( x - h )² = 4p( y - k )

( x - (-5) )² = 4(4)( y - (-9) )

( x + 5 )² = 16( y + 9 )

Multiply both side by 1/16

(1/16)( x + 5 )² =  y + 9

Make y the subject of the formula

(1/16)( x + 5 )² - 9 =  y

y = (1/16)( x + 5 )² - 9

The equation of the parabola with the given vertex and directrix in vertex form is y = (1/16)( x + 5 )² - 9.

Hence, option C is the correct answer.

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Using translation concepts, it is found that pi is the phase shift of the graph, and since it is negative, the graph is shifted right pi units.

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction either in it’s definition or in it’s domain. Examples are shift left/right or bottom/up, vertical or horizontal stretching or compression, and reflections over the x-axis or the y-axis.

In this problem, the change is given as follows:

x -> x - pi

It means that the change is in the domain, in which pi is the phase shift of the graph, and since it is negative, the graph is shifted right pi units.

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Answer: [tex]-3x^2+6y^3-9z[/tex]

Step-by-step explanation:

[tex](2x^2-y^3+5z)-(5x^2-7y^3+14z)[/tex]    given

[tex]2x^2-y^3+5z-5x^2+7y^3-14z[/tex]  use distributive property and distribute the -

[tex]-3x^2+6y^3-9z[/tex]          final answer

Answer:

-3x^2 + 6y^3 - 9z

Step-by-step explanation:

(2x^2 - y^3 + 5z) - (5x^2 - 7y^3 + 14z)

open the brackets:

2x^2 - y^3 + 5z - 5x^2 + 7y^3 - 14z

simplify:

=> -3x^2 - y^3 +5z + 7y^3 - 14z               [2x^2 - 5x^2 = -3x^2]

=> -3x^2 + 6y^3 + 5z - 14z                       [ -y^3 + 7y^3 = 6y^3]

=>  -3x^2 + 6y^3 - 9z                               [  5z - 14z = -9z]

Answer:

5,-1  

Step-by-step explanation:

edited.... I gave 1/3   AC  earlier (misread the Q)

   This means AC is divided into fourths

x    from  6 to 2 is -4     1/4th of this added to 6 is

    6 - 1/4 * 4 =   5

y    from 1 to -7  is -8    1/4th of this added to 1 is

                   1 - 1/4 * 8 = -1  

Using proportions, it is found that out of the next 75 couples asked, 49 will have at least one spouse that prefers action movies.

A proportion is a fraction of a total amount, and the measures are related using a rule of three. Due to this, relations between variables, either direct or inverse proportional, can be built to find the desired measures in the problem.

Researching this problem on the internet, it is found that out of 150 couples initially surveyed, 26 + 25 + 12 + 22 + 13 = 98 couples had at least one spouse preferring action movies, hence the proportion is:

p = 98/150 = 0.6533.

Out of the next 75 couples, the amount is:

A = 0.6533 x 75 = 49

Hence, out of the next 75 couples asked, 49 will have at least one spouse that prefers action movies.

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The total amount in the transaction account is

= $625.

The amount in the transaction account is

=75

Your salary and any other payments that you get may be deposited into an account known as a transaction account, which is an account that you utilize on a day-to-day basis. When connected to a Visa card, your transaction account may also be used to pay bills, go shopping, and make other types of payments that are commonplace. Balances in transaction accounts do not accrue any interest for the account holder.

b) Now if the reserve requirement changes to 5%, we need to reconstruct the balance sheet of the total banking system.

Calculate the initial total reserve using the new reserve requirement as follows:

= 4% x500

= 0.04×500

=20

Calculate the excess reserve as follows:

=30-20

= $10

Calculate the money multiplier as follows:

=1/4%

=1/0.04

=25

Calculate the increase in transaction account as follows:

=25x $5

= $125

Hence, the total amount in the transaction account is

$500+ $125

= $625.

c)

Therefore, the amount in the transaction account is

=600-525

=75

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Step-by-step explanation:

x×1/2×1/4×3/4

x×3/32

3/32x

Answer:

4 hours

Step-by-step explanation:

Hour1-

Bicycle = 24.50

ElectricScooter = 32

Hour2-

Bicycle = 34

ElectricScooter = 39

Hour3-

Bicycle = 43.5

ElectricScooter = 46

Hour4 -

Bicycle = 53

ElectricScooter = 53

9. The curve passes through the point (-1, -3), which means

[tex]-3 = a(-1) + \dfrac b{-1} \implies a + b = 3[/tex]

Compute the derivative.

[tex]y = ax + \dfrac bx \implies \dfrac{dy}{dx} = a - \dfrac b{x^2}[/tex]

At the given point, the gradient is -7 so that

[tex]-7 = a - \dfrac b{(-1)^2} \implies a-b = -7[/tex]

Eliminating [tex]b[/tex], we find

[tex](a+b) + (a-b) = 3+(-7) \implies 2a = -4 \implies \boxed{a=-2}[/tex]

Solve for [tex]b[/tex].

[tex]a+b=3 \implies b=3-a \implies \boxed{b = 5}[/tex]

10. Compute the derivative.

[tex]y = \dfrac{x^3}3 - \dfrac{5x^2}2 + 6x - 1 \implies \dfrac{dy}{dx} = x^2 - 5x + 6[/tex]

Solve for [tex]x[/tex] when the gradient is 2.

[tex]x^2 - 5x + 6 = 2[/tex]

[tex]x^2 - 5x + 4 = 0[/tex]

[tex](x - 1) (x - 4) = 0[/tex]

[tex]\implies x=1 \text{ or } x=4[/tex]

Evaluate [tex]y[/tex] at each of these.

[tex]\boxed{x=1} \implies y = \dfrac{1^3}3 - \dfrac{5\cdot1^2}2 + 6\cdot1 - 1 = \boxed{y = \dfrac{17}6}[/tex]

[tex]\boxed{x = 4} \implies y = \dfrac{4^3}3 - \dfrac{5\cdot4^2}2 + 6\cdot4 - 1 \implies \boxed{y = \dfrac{13}3}[/tex]

11. a. Solve for [tex]x[/tex] where both curves meet.

[tex]\dfrac{x^3}3 - 2x^2 - 8x + 5 = x + 5[/tex]

[tex]\dfrac{x^3}3 - 2x^2 - 9x = 0[/tex]

[tex]\dfrac x3 (x^2 - 6x - 27) = 0[/tex]

[tex]\dfrac x3 (x - 9) (x + 3) = 0[/tex]

[tex]\implies x = 0 \text{ or }x = 9 \text{ or } x = -3[/tex]

Evaluate [tex]y[/tex] at each of these.

[tex]A:~~~~ \boxed{x=0} \implies y=0+5 \implies \boxed{y=5}[/tex]

[tex]B:~~~~ \boxed{x=9} \implies y=9+5 \implies \boxed{y=14}[/tex]

[tex]C:~~~~ \boxed{x=-3} \implies y=-3+5 \implies \boxed{y=2}[/tex]

11. b. Compute the derivative for the curve.

[tex]y = \dfrac{x^3}3 - 2x^2 - 8x + 5 \implies \dfrac{dy}{dx} = x^2 - 4x - 8[/tex]

Evaluate the derivative at the [tex]x[/tex]-coordinates of A, B, and C.

[tex]A: ~~~~ x=0 \implies \dfrac{dy}{dx} = 0^2-4\cdot0-8 \implies \boxed{\dfrac{dy}{dx} = -8}[/tex]

[tex]B:~~~~ x=9 \implies \dfrac{dy}{dx} = 9^2-4\cdot9-8 \implies \boxed{\dfrac{dy}{dx} = 37}[/tex]

[tex]C:~~~~ x=-3 \implies \dfrac{dy}{dx} = (-3)^2-4\cdot(-3)-8 \implies \boxed{\dfrac{dy}{dx} = 13}[/tex]

12. a. Compute the derivative.

[tex]y = 4x^3 + 3x^2 - 6x - 1 \implies \boxed{\dfrac{dy}{dx} = 12x^2 + 6x - 6}[/tex]

12. b. By completing the square, we have

[tex]12x^2 + 6x - 6 = 12 \left(x^2 + \dfrac x2\right) - 6 \\\\ ~~~~~~~~ = 12 \left(x^2 + \dfrac x2 + \dfrac1{4^2}\right) - 6 - \dfrac{12}{4^2} \\\\ ~~~~~~~~ = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4[/tex]

so that

[tex]\dfrac{dy}{dx} = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4 \ge 0 \\\\ ~~~~ \implies 12 \left(x + \dfrac14\right)^2 \ge \dfrac{27}4 \\\\ ~~~~ \implies \left(x + \dfrac14\right)^2 \ge \dfrac{27}{48} = \dfrac9{16} \\\\ ~~~~ \implies \left|x + \dfrac14\right| \ge \sqrt{\dfrac9{16}} = \dfrac34 \\\\ ~~~~ \implies x+\dfrac14 \ge \dfrac34 \text{ or } -\left(x+\dfrac14\right) \ge \dfrac34 \\\\ ~~~~ \implies \boxed{x \ge \dfrac12 \text{ or } x \le -1}[/tex]

13. a. Compute the derivative.

[tex]y = x^3 + x^2 - 16x - 16 \implies \boxed{\dfrac{dy}{dx} = 3x^2 - 2x - 16}[/tex]

13. b. Complete the square.

[tex]3x^2 - 2x - 16 = 3 \left(x^2 - \dfrac{2x}3\right) - 16 \\\\ ~~~~~~~~ = 3 \left(x^2 - \dfrac{2x}3 + \dfrac1{3^2}\right) - 16 - \dfrac13 \\\\ ~~~~~~~~ = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3[/tex]

Then

[tex]\dfrac{dy}{dx} = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3 \le 0 \\\\ ~~~~ \implies 3 \left(x - \dfrac13\right)^2 \le \dfrac{49}3 \\\\ ~~~~ \implies \left(x - \dfrac13\right)^2 \le \dfrac{49}9 \\\\ ~~~~ \implies \left|x - \dfrac13\right| \le \sqrt{\dfrac{49}9} = \dfrac73 \\\\ ~~~~ \implies x - \dfrac13 \le \dfrac73 \text{ or } -\left(x-\dfrac13\right) \le \dfrac73 \\\\ ~~~~ \implies \boxed{x \le 2 \text{ or } x \ge \dfrac83}[/tex]

The order of the probabilities from the highest to the least would be

4 > 2 > 3 > 5 >1. This tells us tgat the highest probability is number 4 while the lowest is 1.

In mathematics, this is the concept that is used to talk about the likelihood of having an event occurring.

The probability of these items have been calculated below

1. Winning the Lottery

probability = 1/49P6

= 0.00000000009932

2. Rolling a 7:

we have to find the group of 7's

= (2,5) (1,6), (3,4), (4, 3), (6,1), (5,2)

= 6/36 = 0.1666

3. Drawing the Ace of Spade

1/52 = 0.019

4. 2 Heads in a Row:

1/2 x 1/2

= 0.25

5. Marbles:

3 + 4+ 5 = 12

3/12 * 3/12 * 3/12

= 0.01562

Hence the order of the probabilities from the highest to the least would be

4 > 2 > 3 > 5 >1

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I'm not sure if the last two apostrophes are part of the quote - "Solve ... " - or if you mean the second derivative [tex]x''[/tex]. I think you mean the first interpretation, but I'll include both cases since they are both solvable.

If the former is correct, separate variables to solve.

[tex]x' = -9tx \implies \dfrac{dx}{dt} = -9tx \implies \dfrac{dx}x = -9t\,dt[/tex]

Integrate both sides to get

[tex]\ln|x| = -\dfrac92 t^2 + C[/tex]

Solve for [tex]x[/tex].

[tex]e^{\ln|x|} = e^{-9/2\,t^2 + C} \implies \boxed{x = Ce^{-9/2\,t^2}}[/tex]

If you meant the latter, then the ODE can be rewritten as

[tex]9t x'' + x' = 0[/tex]

Reduce the order of the equation by substituting [tex]y(t) = x'(t)[/tex] and [tex]y'(t) = x''(t)[/tex].

[tex]9t y' + y = 0[/tex]

Solve for [tex]y'[/tex] and separate variables.

[tex]y' = -\dfrac y{9t} \implies \dfrac{dy}{dt} = -\dfrac y{9t} \implies \dfrac{dy}y = -\dfrac{dt}{9t}[/tex]

Integrate.

[tex]\ln|y| = -\dfrac19 \ln|t| + C[/tex]

Solve for [tex]y[/tex].

[tex]e^{\ln|y|} = e^{-1/9 \,\ln|t| + C} \implies y = Ct^{-1/9}[/tex]

Solve for [tex]x[/tex] by integrating.

[tex]x' = Ct^{-1/9} \implies x = C_1 t^{8/9} + C_2[/tex]

The missing value in the logarithm are as follows:

Using logarithm rule,

logₐ b - logₐ c = logₐ (b / c)

logₐ b + logₐ c = logₐ (b × c)

Therefore,

log₃ 7 - log₃ 2 = log₃ (7 / 2)

log₉ 7 + log₉ 4 = log₉ (7 × 4) =  log₉ 28

log₆ 1 / 81 = log₆ 81⁻¹ = log₆ 3⁻⁴ =  - 4 log₆ 3

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Using a system of equations, it is found that Allison bought 2 bottles of juice and 9 bottles of soda.

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

For this problem, the variables are:

Allison purchased a total of 11 bottles of juice, hence:

x + y = 11 -> x = 11 - y.

These 11 bottles contain 445 grams of sugar, hence, considering the amounts of each bottle, we have that:

20x + 45y = 445

Since x = 11 - y:

20(11 - y) + 45y = 445

25y = 225

y = 225/25

y = 9.

x = 11 - y = 11 - 9 = 2.

She bought 2 bottles of juice and 9 bottles of soda.

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Answer:

I think its A

Step-by-step explanation:

since there are 365 days in a year, so 180 days is really just 180/365 of a year, so

[tex]~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$1295\\ r=rate\to 7\%\to \frac{7}{100}\dotfill &0.07\\ t=years\to \frac{180}{365}\dotfill &\frac{36}{73} \end{cases} \\\\\\ I = (1295)(0.07)(\frac{36}{73})\implies I\approx 44.7[/tex]

The nth term of the sequence is 2n - 8

The nth term of an arithmetic progression is expressed as;

Tn = a + (n - 1)d

where

a is the first term

d is the common difference

n is the number of terms

Given the following parameters

a = f(1)=−6

f(2) = −4

Determine the common difference

d = f(2) - f(1)

d = -4 - (-6)
d = -4 + 6

d = 2

Determine the nth term of the sequence

Tn = -6 + (n -1)(2)

Tn = -6+2n-2
Tn = 2n - 8

Hence the nth term of the sequence is 2n - 8

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By definition, we have

[tex]f(n) = f(n - 1) + f(n - 2)[/tex]

so that by substitution,

[tex]f(n-1) = f(n-2) + f(n-3) \implies f(n) = 2f(n-2) + f(n-3)[/tex]

[tex]f(n-2) = f(n-3) + f(n-4) \implies f(n) = 3f(n-3) + 2f(n-4)[/tex]

[tex]f(n-3) = f(n-4) + f(n-5) \implies f(n) = 5f(n-4) + 3f(n-5)[/tex]

[tex]f(n-4) = f(n-5) + f(n-6) \implies f(n) = 8f(n-5) + 5f(n-6)[/tex]

and so on.

Recall the Fibonacci sequence [tex]F(n)[/tex], whose first several terms for [tex]n\ge1[/tex] are

[tex]\{1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots\}[/tex]

Let [tex]F_n[/tex] denote the [tex]n[/tex]-th Fibonacci number. Notice that the coefficients in each successive equation form at least a part of this sequence.

[tex]f(n) = f(n-1) + f(n-2) = F_2f(n-1) + F_1 f(n-2)[/tex]

[tex]f(n) = 2f(n-2) + f(n-3) = F_3 f(n-2) + F_2 f(n-3)[/tex]

[tex]f(n) = 3f(n-3) + 2f(n-4) = F_4 f(n-3) + F_3 f(n-4)[/tex]

[tex]f(n) = 5f(n-4) + 3f(n-5) = F_5 f(n-4) + F_4 f(n-5)[/tex]

[tex]f(n) = 8f(n-5) + 5f(n-6) = F_6 f(n-5) + F_5 f(n-6)[/tex]

and so on. After [tex]k[/tex] iterations of substituting, we would end up with

[tex]f(n) = F_{k+1} f(n - k) + F_k f(n - (k+1))[/tex]

so that after [tex]k=n-2[/tex] iterations,

[tex]f(n) = F_{(n-2)+1} f(n - (n-2)) + F_{n-2} f(n - ((n-2)+1)) \\\\ f(n) = f(2) F_{n-1} + f(1) F_{n-2} \\\\ \boxed{f(n) = -4 F_{n-1} - 6 F_{n-2}}[/tex]

Secured loans exist as loans for which the borrower posts some collateral. This exists in contrast to loans that exist created only on the ground of creditworthiness and trust in the borrower, for which no collateral exists pledged (unsecured loans).

The difference between a secured loan and an unsecured loan exists a secured loan needs collateral and an unsecured loan does not.

Secured loans exist backed by an asset, like a house in the case of a mortgage loan or a car with an auto loan. An unsecured loan on the other hand lives not connected to any of your assets and the lender can't automatically seize your property as payment for the loan.

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The number of strings that can be formed by ordering the letters SCHOOL using some or all of the letters is 1440 strings

Permutation can be defined as number of ways in which things can arranged.

We were given to find how many strings that can be formed by ordering the letters SCHOOL using some or all of the letters.

First of all, How many distinct letters are in the word SCHOOL ?

The distinct letters are 5 in numbers.

What is the total number of letters in the word SCHOOL ?

The total number of letters is 6.

Then

6! + 5[tex]P_{5}[/tex]

That is, 6 factorial + 6 permutation 5

( 6 x 5 x 4 x 3 x 2 x 1 ) + 6!/( 6 - 5)!

720 + 720

1440 strings

Therefore, the number of strings that can be formed by ordering the letters SCHOOL using some or all of the letters is 1440 strings

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Answer:

Below

Step-by-step explanation:

Using the given law of cosines    c = 26   a = 50   b = 29

      and cos c   is  cos x  

26^2 = 50^2 + 29^2 - 2(50)(29) cos x

-2665 = -2900 cos x

cos x = 2665/2900

x = arc cos ( 2665/2900) = 23.2°

Answer:

1.60

Step-by-step explanation: