A dealer made a profit of 22.5% by selling a car for GH¢5,880. Find the percentage profit he would have realized if he had sold it for GH¢6,120.​

Answer:

3.7517%

Step-by-step explanation:

take 3.7 and divide it by 4 to get 0.925

then apply 0.925% to x 4 times to find out the compounded rate

Answer:

$3:00

Step-by-step explanation:

let 'x' be the money that the parents gave to Jane and her sister each

Jane will have 7+x dollars

Jane's sister will have 2 + x dollars

Jane had twice as much money as her sister thus

7 + x = 2(2 + x)

7 + x = 4 + 2x

4 + 2x = 7 + x

2x - x = 7 - 4

x = $3

the parents gave $3 to each.

Answer:

Step-by-step explanation:

[tex]\displaystyle\\ x^8-\frac{1}{81} \\\\x^8-\frac{1}{3^4} =\\\\x^8-(\frac{1}{3})^4 =\\\\x^{4*2}-(\frac{1}{3})^{2*2}= \\\\(x^4)^2-((\frac{1}{3} )^2)^2=\\\\(x^4-(\frac{1}{3})^2)(x^4+(\frac{1}{3} ) ^2)=\\\\(x^{2*2}-(\frac{1}{3})^2)(x^4+\frac{1^2}{3^2} )=\\\\((x^2)^2-(\frac{1}{3})^2)(x^4+\frac{1}{9})=\\\\(x^2-\frac{1}{3} )(x^2+\frac{1}{3})(x^4+\frac{1}{9})[/tex]

Using z-scores, it is found that due to the higher z-score, Luna did better on the exam.

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

From the two bullet points, it is found that the student who did "better" is the one whose grade has the higher z-score.

For Hazel, the parameters are given as follows:

[tex]X = 93, \mu = 80, \sigma = 8[/tex]

Hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

Z = (93 - 80)/8

Z = 1.625.

For Luna, the parameters are given as follows:

[tex]\mu = 68, X = 40, \sigma = 15[/tex]

Hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

Z = (68 - 40)/15

Z = 1.87.

Due to the higher z-score, Luna did better on the exam.

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The required probability for a number of problems given is as follows, 1.) 1 / 3 2.) 1 / 5, and  3.) 1 / 4.

Here,
A jar consists of 4 red marbles, 5 yellow marbles, 3 green marbles, and 3 blue marbles.
Total number of marbles = 15

1) probability that the outcome contains green and/or blue marbles only,
= 2 / 6 = 1 / 3

2.)  probability that the outcome contains only green marbles,
= 3 / 15 = 1 / 5

3.) the probability that the outcome contains no green marbles,
  = 3 / 12 = 1 / 4

Thus, the required probability for a number of problems given is as follows, 1.) 1 / 3 2.) 1 / 5, and  3.) 1 / 4.


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The dimensions of the rectangle are; length 63 inches and width is 20 inches.

Let 'L' be the length of the rectangle.

Let 'W' be the breadth of the rectangle.

As, the length of a rectangle is three more than triple the width.

Thus,

L = 3 + 3W

The formula for the perimeter is;

Perimeter = 2 (L+W)  

The value of perimeter is 166 inches.

Thus,

166 = 2(3 + 3W +W)

166 = 2(3 + 4W)

166 = 6 + 8W

166 - 6 = 8W

W = 20

W = width = 20 inches

L = 3 + 3W

L = 3 + 3(20)

L = 3 + 60

L = 63 inches.

Thus, the dimensions of the rectangle are; length 63 inches and width is 20 inches.

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Answer: 2x²-4x-16

Step-by-step explanation:

[tex](2x+4)(x-4)=\\2x(x)+2x(-4)+4(x)+4(-4)=\\2x^2-8x+4x-16=\\2x^2-4x-16[/tex]

There are 11 scores, (N)  in the population.

The given parameters are:

Variance, s^2 = 5

Sum of squares, SS = 50

The variance of the population is calculated using

s^2 = SS/[N - 1]

So, we have

5 = 50/[N - 1}

Cross multiply

So, we have

5(N - 1) = 50

This gives

N - 1 = 10

Add 1 to both sides

N = 11

Hence, there are 11 scores, (N)  in the population.

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1. 1/2

2. 3/10

3. 2 and 3/5

4. 1 and 7/12

5. 1 and 1/3

[tex]f(x)=3x^2+3\hspace{5em}\displaystyle \lim_{h\to 0}~\cfrac{f(2+h)~~ - ~~f(2)}{h} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\cfrac{f(2+h)~~ - ~~f(2)}{h}\implies \cfrac{[3(2+h)^2+3]~~ - ~~[3(2)^2+3]}{h} \\\\\\ \cfrac{[3(4+4h+h^2)+3]~~ - ~~[3(4)+3]}{h}\implies \cfrac{[(12+12h+3h^2)+3]~~ - ~~[15]}{h} \\\\\\ \cfrac{[12h+3h^2+15]~ - ~~[15]}{h}\implies \cfrac{12h+3h^2}{h}\implies \cfrac{h(12+3h)}{h}\implies 12+3h \\\\[-0.35em] ~\dotfill\\\\ \displaystyle \lim_{h\to 0}~\cfrac{f(2+h)~~ - ~~f(2)}{h}\implies \lim_{h\to 0}~12 + 3h\implies 12+3(0)\implies \text{\LARGE 12}[/tex]

Answer:

$600

Step-by-step explanation:

450 = 3/4 (because 75%=3/4)

450/3=150. 150 = 1/4, or 25%. Now it's just as simple as multiplying 150 4 times to get 100%, or the original cost, 150 times 4 equals 600. Therefore the notebooks original price is $600

Clare will have traveled 9 miles in 45 minutes.

So, Clare is riding her bike at a speed of 12 miles per hour.

Distance traveled in 45 minutes or 45/60 hours at the same speed will be = Speed Time

Therefore, Clare will have traveled 9 miles in 45 minutes.

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The domain and range of the function is -3 ≤ x < 7 and -7 ≤ x < 6 respectively.

The domain of a function is the set of values that we are allowed to plug into our function. This set is the x values in a function such as f(x). The range of a function is the set of values that the function assumes. This set is the values that the function shoots out after we plug an x value in.

For domain :-

Since, the domain is the set of all allowed x inputs of a function.

The smallest x can get is x = -3 as shown by the left-most point.

The largest x can get is x = 7, though we can't actually reach this value due to the open hole at the right-most point.

The interval of all possible x values is -3 ≤ x < 7  which converts to the interval notation [-3, 7)

We use a square bracket to include the endpoint and a parenthesis to exclude the endpoint.

We can say "x is any number between -3 and 2, including -3 but excluding 2".

For range :-

Since, the range is the set of all allowed y inputs of a function.

The smallest x can get is x = -7 as shown by the left-most point.

The largest x can get is x = 6, though we can't actually reach this value due to the open hole at the right-most point.

The interval of all possible x values is -7 ≤ x < 6  which converts to the interval notation [-7, 6)

We use a square bracket to include the endpoint and a parenthesis to exclude the endpoint.

Hence, the domain and range of the function is -3 ≤ x < 7 and -7 ≤ x < 6 respectively.

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The yield on the corporate bond of a face value of $1000 is 7.77%.

What is the percentage discount?

The percentage discount is the discount given on a product as compared to the given discount on 100 rupees.

Given, the face value of the bond is $1000.

Discounted price of the bond is $900.

Therefore, the fixed interest on the bond for that period will be

= $1000 × 7/100 = $70.

Now, the yield on that corporate  bond = 70 × 100/900 % = 7.77% .

Hence, the yield on the corporate bond of a face value of $1000 is 7.77%.

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Answer:

x=2

Step-by-step explanation:

Because these angles are the same you can do: 10x-14=7x-8. By doing simple math you'll get x=3 as you answer

Answer:

x = -18/7 y simplified as -2.57143y

Step-by-step explanation:

to figure this out and find the excluded value, first set the denominator equal to 0, because fractions represent division and dividing by 0 isn't defined. that will give us 7x+18y=0.. move the variable to the right hand side and change it sign.. which will be 7x=-18y, now divide both side by 7, = x=-18/7y / -2.57143y

The amount when 500 invested at 9% interest compounded monthly be worth after 4 years will be $715.7.

Compound interest is the interest on a loan or deposit calculated based on the initial principal and the accumulated interest from the previous period.

We know that the compound interest is given as

[tex]\rm A = P \left ( 1 + \dfrac{r}{n} \right ) ^{nt}[/tex]

Where A is the amount, P is the initial amount, r is the rate of interest, and t is the number of years.

For interest compounded monthly, the value of n is 12.

The amount when 500 invested at 9% interest compounded monthly be worth after 4 years will be

[tex]\rm A = 500 \left ( 1 + \dfrac{0.09}{12} \right ) ^{12 \times 4}[/tex]

Simplify the equation, then we have

A = 500 (1.0075)⁴⁸

A = 500 x 1.4314

A = $715.70

The amount when 500 invested at 9% interest compounded monthly be worth after 4 years will be $715.7.

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The width of the rectangle = 15 yd

The length of the rectangle = 60 yd.

Perimeter of a rectangle = 2(length + width).

Given the following:

Width of the rectangle = w

Length of the rectangle = 4w

Perimeter of the rectangle = 150 yd.

Therefore:

2(4w + w) = 150

2(5w) = 150

10w = 150

10w/10 = 150/10

w = 15

Width of the rectangle = w = 15 yd

Length of the rectangle = 4w = 4(15) = 60 yd.

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Answer:

x = 0

Step-by-step explanation:

Let's solve the problem,

→ (x/9) + 8 = 8

→ x/9 = 8 - 8

→ x = 0 × 9

→ [ x = 0 ]

Thus, the value of x is 0.

Answer:

97

Step-by-step explanation:

To find x² + y², use the identity (x - y)² = x² + y² - 2xy.

(x - y)² = x² + y² - 2xy

Substitute xy = 36 and x - y = 5 in the above formula,

   5²    =  x² + y² - 2*36

 25     = x² + y² - 72

Add 72 to both sides,

25 + 72 = x² + y²

x² + y² = 97

Based on the packaging of the cookies, the number of ways you can buy 36 cookies is 12 ways.

The ways you can buy the cookies include:

In conclusion, there are 12 ways to buy the cookies.

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The distance between the school and the hospital on the map  is 8 cm

Here, we want to find the distance between the school and the hospital on the map

The scale is that 4cm to 5km

So ;

if 4 cm = 5km

Then x cm = 10km

x * 5 = 4 * 10

5x = 40

x = 40/5

x = 8 cm

Therefore the distance between the school and the hospital on the map  is 8 cm

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The value of integral ₀∫¹ ( eˣ - 2 )² dx is 8 + ( e² - 1) / 2 - 4e.

When infinitesimal data are combined, notions such as displacement, area, and volume can be created. An integral assigns numbers to functions in a way that describes these concepts. Integration is the action of locating integrals.

Consider the integral expression,

₀∫¹ ( eˣ - 2 )² dx

Using ( a + b )² = a² + b² + 2ab

[tex]\int_0^{1}( e^{x} - 2 )^{2} dx = \int_0^{1}(e^{2x}+4-2e^{x})dx[/tex]

[tex]\int_0^{1}( e^{x} - 2 )^{2} dx = \int^0_{1}e^{2x} dx +\int_0^{1}4 dx - \int_0^{1}4e^{x} dx[/tex]

Now, integration of eˣ is eˣ + c where c is constant.

Therefore,

[tex]\int_{0}^{1} ( e^{x} - 2 )^{2} dx = [ \frac{e^{2x}}{2} ]_0^{1} + [ 4x ]_0^{1} - [ 4e^{x}]_0^{1}[/tex]

₀∫¹ ( eˣ - 2 )² dx = ( e² / 2 - e⁰/2 ) + ( 4 - 0 ) - ( 4e¹ - 4e⁰ )

₀∫¹ ( eˣ - 2 )² dx =  e² / 2 - 1/2 + 4 - 4e + 4

₀∫¹ ( eˣ - 2 )² dx = e² / 2 - 4e + 8 - 1/2

₀∫¹ ( eˣ - 2 )² dx =  = 8 + ( e² - 1) / 2 - 4e

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The value of x in the equation x = [a^2 + a^2]/[ab + 1] is [2a^2]/[ab + 1]

Equations are mathematical expressions that are represented by the equal to or equality sign i.e =

Examples of equations are:

a + b = c

y = mx + c

Ax + By = C

y = ax^2 + bx + c

y =  x

The equation is given as

x = [a^2 + a^2]/[ab + 1]

To start with, we evaluate the like terms i.e. we add a^2 and a^2

This is represented as

a^2 + a^2 = 2a^2

Substitute 2a^2 for a^2 + a^2 in the given equation

So, we have

x = [2a^2]/[ab + 1]

The above equation cannot be further simplified

Hence, the value of x in the equation x = [a^2 + a^2]/[ab + 1] is [2a^2]/[ab + 1]

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Answer:

the answer would be x=8

Step-by-step explanation:

6×4=24

24/3=8

Based on the information given, it can be deduced that the chart that is being described is the OHLC chart.

A statistical graph or chart is defined as the pictorial representation of statistical data in graphical form. The statistical graphs are used to represent a set of data to make it easier to understand and interpret statistical information.

It should be noted that an open-high-low-close chart (OHLC) chart simply means a type of chart that's used to illustrate the movements in the price of a financial instrument for a particular period of time.

The vertical line on the chart illustrates the price range over a unit of time. It's useful in assessing volatility as it shows the full price range of the period.

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The domain of the function is all the real numbers and the range is from -2 to 2.

The range of values that we are permitted to enter into our function is known as the domain of a function. The x values for a function like f make up this set (x).

A function's range is the collection of values it can take as input. After we enter an x value, the function outputs this sequence of values.

In the image, we can see that the value of the function varies from -2 to 2 at all the possible values of the real numbers.

Therefore, the domain of the function is all the real numbers and the range is from -2 to 2.

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From the graph of the fifth root function, we have that:

In a graph:

Hence, since the given graph has no holes, and keeps going to infinity, we have that both the domain and the range are all real values.

Since the function has no holes, it is also continuous, and has no vertical asymptotes.

For this problem, we have that the function is moving right and up over it's entire domain, hence it is increasing for all real values, and never decreasing nor constant.

Since the graph passes through the origin, we have that:

For this problem, looking at the graph, we have the the second statement holds true for all values of x, hence the function is odd.

It is given by the values of y when x -> ∞.

Hence:

For the end behavior, y -> - ∞ when x -> -∞ and y -> +∞ when x -> +∞.

Since f(x) -> ∞ when x -> ∞, the function has no asymptotes.

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