Elly buys 75 shares of stock in a mutual fund for a total investment of $450. She sells 50 shares of her stock for total proceeds of $400. Determine if Elly made a profit or loss, and how much per share.
a.
loss of $0.67 per share
b.
loss of $2 per share
c.
profit of $0.67 per share
d.
profit of $2 per share

The B. distance of a javelin throw is an example of a continuous random variable.

A continuous random variable is a variable that can take any value within a specified range, often associated with measurements and quantities. Unlike discrete variables, it can assume an infinite number of values, typically represented by intervals on the real number line.

Out of the given options, the distance of a javelin throw is an example of a continuous random variable. A continuous random variable is one that can take on any value within a given range. In the case of a javelin throw, the distance can be any real number value within a certain range, such as 0 to infinity. This is in contrast to the other options, which are discrete random variables that can only take on specific whole number values.

Answer:

The other factors are (x-1) and (x+2).

The perimeters of two rhombi with equal areas are sometimes equal, but not always.

Let's consider the formula for the area of a rhombus, which is given by [tex]\( A = \frac{1}{2} \times d_1 \times d_2 \), where \( d_1 \) and \( d_2 \)[/tex] are the lengths of the diagonals. For the perimeter P, we have [tex]\( P = 4 \times s \)[/tex], where s is the length of one side of the rhombus.

For two rhombi to have equal areas, the product of their diagonals must be the same. That is, if we have two rhombi with diagonals [tex]\( (d_1, d_2) \) and \( (d_1', d_2') \), then \( d_1 \times d_2 = d_1' \times d_2' \)[/tex].

However, the perimeter depends only on the length of one side, s, and not on the diagonals. Therefore, two rhombi with the same area can have different side lengths and thus different perimeters.

Let's consider an example:

Rhombus 1:

[tex]- Diagonals \( d_1 = 8 \) units and \( d_2 = 4 \) units[/tex]

[tex]- Area \( A = \frac{1}{2} \times 8 \times 4 = 16 \) square units[/tex]

[tex]- Side length \( s \), using the Pythagorean theorem (since the diagonals bisect each other at right angles), is \( s = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} \) units[/tex]

[tex]- Perimeter \( P = 4 \times s = 4 \times \sqrt{20} \) units[/tex]

Rhombus 2:

[tex]- Diagonals \( d_1' = 10 \) units and \( d_2' = 3.2 \) units[/tex]

[tex]- Area \( A' = \frac{1}{2} \times 10 \times 3.2 = 16 \) square units (equal to the area of Rhombus 1)[/tex]

[tex]- Side length \( s' \), using the Pythagorean theorem, is \( s' = \sqrt{1.6^2 + 2.56^2} = \sqrt{2.56 + 6.5536} = \sqrt{9.1136} \) units[/tex]

[tex]- Perimeter \( P' = 4 \times s' = 4 \times \sqrt{9.1136} \) units[/tex]

Comparing the perimeters:

[tex]- \( P = 4 \times \sqrt{20} \approx 4 \times 4.472 = 17.888 \) units[/tex]

[tex]- \( P' = 4 \times \sqrt{9.1136} \approx 4 \times 3.0188 = 12.0752 \) units[/tex]

Clearly, the perimeters are not equal, even though the areas are the same.

However, it is possible for two rhombi with equal areas to have equal perimeters if their side lengths are the same. For instance, if Rhombus 1 and Rhombus 2 both had side lengths of s = s' = 4 units, then their perimeters would both be P = P' = [tex]4 \times 4 = 16 \)[/tex] units, regardless of the lengths of their diagonals, as long as the diagonals satisfy the area condition [tex]\( d_1 \times d_2 = d_1' \times d_2' \)[/tex].

In conclusion, the perimeters of two rhombi with equal areas can sometimes be equal, specifically when the side lengths are the same, but they are not always equal, as demonstrated by the example above.

Answer:

ANSWER IS

Step-by-step explanation:

Answer:

I think it's b, please correct me if I'm wrong.

Step-by-step explanation:

I'm taking the test rn so I hope its right.

Answer:

its b i just did it

Step-by-step explanation:

Group terms that contain the same variable, and move the constant to the opposite side of the equation

Rewrite as perfect squares

Answer:  The required values are

[tex]f^{-1}(x)=x-4~~~~~~~~\textup{and}~~~~~~~~~f^{-1}(4)=0.[/tex]

Step-by-step explanation:  We are given the following function f(x) :

[tex]f(x)=x+4~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

We are to find the values of [tex]f^{-1}(x)[/tex] and [tex]f^{-1}(4).[/tex]

Let us consider that

[tex]y=f(x)~~~~~~~~\Rightarrow x=f^{-1}(y).[/tex]

So, from equation (i), we get

[tex]f(x)=x+4\\\\\Rightarrow y=f^{-1}(y)+4\\\\\Rightarrow f^{-1}(y)=y-4\\\\\Rightarrow f^{-1}(x)=x-4.[/tex]

Substituting x = 4 in the above equation, we get

[tex]f^{-1}(4)=4-4=0\\\\\Rightarrow f^{-1}(4)=0.[/tex]

Thus, the required values are

[tex]f^{-1}(x)=x-4~~~~~~~~\textup{and}~~~~~~~~~f^{-1}(4)=0.[/tex]

The algebraic expression is D = 50t where D is the distance covered in miles and t is the time taken in hours.

Distance is defined as the product of speed and time.

Arithmetic operations can also be specified by adding, subtracting, dividing, and multiplying built-in functions.

The operator that performs the arithmetic operations is called arithmetic operator.

* Multiplication operation: Multiplies values on either side of the operator

For example 4*2 = 8

/ Division operation: Divides left-hand operand by right-hand operand

For example 4/2 = 2

Given that the distance a car travels in t hours at a rate of 50 miles per hour

Let the distance covered in miles = D

And t is the time taken in hours

Since distance is defined as the multiplication of speed and time.

So an algebraic expression for the verbal description as

⇒ D = 50t

Hence, the algebraic expression is D = 50t where D is the distance covered in miles and t is the time taken in hours.

Learn more about Arithmetic operations here:

brainly.com/question/25834626

#SPJ2

To find the value of 'b' that gives the maximum value of 86 for the function f(x) = -x² + bx - 14, substitute the x-coordinate of the vertex (-b/2a) into the function, equate it with 86 and solve for 'b'.

To find the values of b for which the quadratic function f(x) = -x² + bx - 14; Maximum value: 86 holds true, we need to utilize the properties of quadratic functions. In a quadratic function in the form f(x) = ax² + bx + c, the maximum value occurs at the vertex of the parabola. The x-coordinate of the vertex is given by -b/2a.

Given that our quadratic is a maximum (opens downwards since a=-1), the y-coordinate of the vertex, which is our maximum value, is 86. Substituting these into our function, we get f(-b/2a) = 86, replacing a=-1, this reduces to -b²/4 -14 = 86. Simplifying this equation will give you the value of b.

https://brainly.com/question/35505962

#SPJ2

The actual distance between the houses is 18.9 miles.

The unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value.

Multiplication is the mathematical operation that is used to determine the product of two or more numbers.

Given that On a​ map, 1 inch equals 9.2 miles. Two houses are 3.5 inches apart on the map.

To find the actual distance between the​ houses.

Given that,

1 inch = 9.2 miles

Therefore,

∴ we get

3.5 inches = 3.5 × 9.2 miles

                = 13.8 miles

The actual distance is 13.8 miles.

Learn more about the unitary method, please visit the link given below;

https://brainly.com/question/23423168

#SPJ5

Final answer:

To find the actual distance between the houses, we can set up a proportion using the given map scale. The actual distance between the houses is 32.2 miles. Using the ratio 1 inch/9.2 miles = 3.5 inches/x is more helpful in solving the problem.

Explanation:

To find the actual distance between the houses, we need to use the scale of 1 inch equals 9.2 miles. We are given that the two houses are 3.5 inches apart on the map.

First, we can set up a proportion to find the actual distance:

1 inch / 9.2 miles = 3.5 inches / x

Cross-multiplying, we have:

1x = 3.5 * 9.2 miles

Simplifying, we get:

x = 32.2 miles

The actual distance between the houses is 32.2 miles.

To represent the scale with two different ratios, we can use the proportion:

1 inch / 9.2 miles = 3.5 inches / x

This ratio is more helpful for solving the problem as it allows us to directly calculate the actual distance using the given map distance and the scale ratio.

Answer:

Rewrite the inequality so there is a single rational expression on one side, and 0 on the other.

Combine under a common denominator.

Test points in the critical regions.

Construct the solution.

Step-by-step explanation:

Answer Above

a² + b² = c²

c² - a² = b²

17² - 7² = x²   ↓

289 - 49 = √240

√240 = 15.49193338482967 (rounded 15.49 or 15.5)

Hope this helps,

♥A.W.E.S.W.A.N.♥

Length is [tex]\boldsymbol{4.4}[/tex] feets and width is equal to [tex]\boldsymbol{8.8}[/tex] feets.

The area of a two-dimensional region, form, or planar lamina in the plane is the quantity that expresses its extent.

A quadrilateral having four right angles is known as a rectangle.

Let [tex]\boldsymbol{w}[/tex] feets denotes width of a rectangle.

Length of a rectangle [tex]=\boldsymbol{w-4}[/tex] feets

Area of a rectangle [tex]=\boldsymbol{42}[/tex] square feet

Length [tex]\times[/tex] Width [tex]=42[/tex]

      [tex]w(w-4)=42[/tex]

[tex]w^2-4w-42=0[/tex]

[tex]w=\frac{4\pm \sqrt{16+168}}{2}[/tex]

   [tex]=2\pm \sqrt{46}[/tex]

As dimension can not be negative, [tex]2-\sqrt{46}[/tex] is rejected.

So,

[tex]w=2+ \sqrt{46}[/tex]

   [tex]=\boldsymbol{8.8}[/tex] feets

Length [tex]=8.8-4[/tex]

            [tex]=\boldsymbol{4.4}[/tex] feets

So, length is [tex]4.4[/tex] feets and width is equal to [tex]8.8[/tex] feets.

Find out more information about rectangle here:

https://brainly.com/question/10046743?referrer=searchResults