Suppose a stock drops in value by 60% one week then increases in value the next week by 75% is the value higher or lower then when u started and by how much

The preparation of the journal entry to record the sale and the settlement of the account receivable is as follows:

January 10, 2022:

Debit Accounts Receivable $16,600

Credit Sales Revenue $16,600

(To record the sale of merchandise on account to Robertsen Co, n/30.)

February 9, 2022:

Debit Notes Receivable $16,600

Credit Accounts Receivable $16,600

(To record the receipt of a 11% promissory note in settlement of account receivable.)

Journal entries refer to the initial records maintained about business transactions.

Journal entries show how the transactions will be posted in the general ledger.

January 10, 2022: Accounts Receivable $16,600 Sales Revenue $16,600

Terms n/30

February 9, 2022: Notes Receivable $16,600 Accounts Receivable $16,600

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Prepare the journal entry to record the sale and the settlement of the account receivable.

the limit expression gives the value of the derivative of a function at a specific point. where f'(x_0) denotes the derivative of f(x) at x = x_0.

what is derivative  ?

The derivative of a function is a measure of how the function changes as its input variable changes. It gives the instantaneous rate of change or slope of the tangent line of the function at a specific point.

In the given question,

The expression you provided represents the limit definition of the derivative of a function f(x) at the point x = x_0. The limit evaluates the instantaneous rate of change or slope of the tangent line of the function f(x) at the point x = x_0.

To evaluate the limit, substitute x = x_0 + h in the expression of the function f(x) and simplify:

[tex]lim h - > 0 [f(x_{0} + h) - f(x_{0})] / h = f'(x_{0})[/tex]

where f'(x_0) denotes the derivative of f(x) at x = x_0.

Therefore, the limit expression gives the value of the derivative of a function at a specific point.

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The two roots of the quadratic equation x² - 5x + 3 = 0 are (5 + √13) / 2 and (5 - √13) / 2.

To find the roots of the quadratic equation x² - 5x + 3 = 0, we can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / 2a

where a, b, and c are the coefficients of the quadratic equation.

Substituting the values of a, b, and c from the given equation, we have:

x = (-(-5) ± √((-5)² - 4(1)(3))) / 2(1)

x = (5 ± √(25 - 12)) / 2

x = (5 ± √13) / 2

Therefore, the roots of the equation are (5 + √13) / 2 and (5 - √13) / 2.

To verify that these are indeed the roots of the equation, we can substitute each of these values for x in the original equation and check if the equation is satisfied. For example, if we substitute (5 + √13) / 2 for x, we get:

(5 + √13)² - 5(5 + √13) + 3 = 0

Simplifying this equation, we get:

13 - 5√13 + 8 + 13 - 5√13 + 3 = 0

This equation is true, which means that (5 + √13) / 2 is a root of the equation x² - 5x + 3 = 0. Similarly, we can verify that (5 - √13) / 2 is also a root of the equation.

These roots can be found using the quadratic formula, which is a useful tool for solving quadratic equations.

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The fraction of the square that is red is 1/6.

A fraction is a number that represents a part or parts of a whole. It consists of two numbers separated by a horizontal or diagonal line, with the number above the line (numerator) representing the part and the number below the line (denominator) representing the whole. For example, the fraction 3/4 represents three parts out of a whole that is divided into four equal parts.

A square divided into six equal parts has six equal sixths.

If Soo colors one of these equal parts red, then she has colored one-sixth of the square red.

Therefore, the fraction of the square that is red is 1/6.

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We can see that the percentage is 20%, then 20 percent of the students got a 10.

To find this, we need to use the formula:

Percentage = 100%(number of students with a 10)/(total number of students)

Here we know the that the revelant numbers are:

Number of students with a 10 = 12

total number of students in the class = 60

Replacing that in the formula we will get the percentage:

P  = 100%*(12/60)

P = 20%

So 20% of the total number of students got a 10.

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It will take about 225.25 years of time for 15 grams to be present.

The half-life of a radioactive substance is the time it takes for half of the substance to decay.

For this substance with a half-life of 170 years, we can use the following formula to find the value of k:

[tex]0.5 = e^(^-^1^7^0^k^)[/tex]

Taking the natural logarithm of both sides, we get:

ln(0.5) = -170k

Solving for k, we get:

k = ln(0.5) / (-170)

= 0.004085

Now, we can use the model [tex]y = y_0e^(^-^k^t^)[/tex] to find the time t required for the substance to decay from 40 grams to 15 grams.

We have:

[tex]15 = 40e^(^-^0^.^0^0^4^0^8^5^t^)[/tex]

Dividing both sides by 40 and taking the natural logarithm of both sides, we get:

ln(15/40) = -0.004085t

Solving for t, we get:

t = ln(15/40) / (-0.004085)

= 225.25 years

Therefore, it will take about 225.25 years for 15 grams to be present.

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a) The function of h in terms of r is h=1400/(πr²).

b) The surface area of the can in terms of r is 2800/r+2πr².

c) The corresponding values of r and h that minimize the amount of required material are r≈14.04 cm and h≈2.98 cm.

a) The volume of a cylinder is given by

V = πr²h.

In this case, the volume of the can is

1400 cm³

So we have

1400 = πr²h.

Solving for h, we get

h = 1400/(πr²).

b) The surface area of a cylinder is given by

A = 2πrh+2πr².

Using the expression we obtained for h in part (a), we can substitute it in the equation and get

A = 2πr(1400/(πr²))+2πr²

= 2800/r+2πr².

c) To minimize the amount of required material, we need to find the value of r that minimizes the surface area.
To do this, we take the derivative of the surface area expression with respect to r and set it equal to zero:

dA/dr = -2800/r²+4πr = 0.

Solving for r, we get

r = √(700/π), which is approximately 14.04 cm.

Substituting this value in the expression for h that we obtained in part (a), we get

h = 1400/(π(14.04)²), which is approximately 2.98 cm.

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Answer: [tex]\frac{2}{5}[/tex] x [tex]\frac{4}{5}[/tex] = [tex]\frac{8}{25}[/tex]

Step-by-step explanation:

We can do this by counting the boxes for both the Width and the Length.

If we look at the width of these boxes (horizontally) then we see that there are 5 boxes all together, with 2 shaded green.

this means that 2 out of 5 of the boxes are shaded for the width, giving us the fraction: [tex]\frac{2}{5}[/tex]

Then for the Length (vertically) we see that there are also 5 boxes, but this time 4 are shaded. This gives us the fraction:  [tex]\frac{4}{5}[/tex].

Then, to fill the equation in, we do: [tex]\frac{2}{5}[/tex] x [tex]\frac{4}{5}[/tex] , which gives us          [tex]\frac{8}{25}[/tex]

2 x 4 = 8

5 x 5 = 25

To double check this answer, we can count how many boxes are shaded green.

We see that 8 are shaded green, out of the 25 boxes that are there.

Therefore, giving us a complete answer of:  [tex]\frac{2}{5}[/tex] x [tex]\frac{4}{5}[/tex] = [tex]\frac{8}{25}[/tex]

We can set up the equation :

0.9x = 110 ,

where x is the unknown number. To solve for x, we can divide both sides by 0.9 :

x = 110 ÷ 0.9

x = 122.22 (rounded to two decimal places)

So the unknown number is approximately 122.22.

To find 9% of that number, we can multiply 122.22 by 0.09 :

0.09 × 122.22 = 11.00 (rounded to two decimal places)

Therefore, 9% of the number is approximately 11.

The strength of the beam is proportional to the width and the square of the depth, so we can express it as Strength = k * width * depth²

where k is the proportionality constant.

In the given figures, the beam is cut from a cylindrical log of diameter d = 24 cm, which means the maximum width and depth of the beam are limited by this diameter. We can see from the figures that the width and depth of the beam change depending on the angle at which it is cut from the log.

In Figure 1, the width and depth of the beam are equal, so we can express the strength of the beam as:

Strength = k * width * depth² = k * x² * (d/2 - x)²

where x is the width and depth of the beam.

In Figure 2, the width and depth of the beam are not equal, so we need to express them separately. Let w be the width and d be the depth, then we can express the strength of the beam as:

Strength = k * width * depth² = k * w * d² = k * ((d/2)sinθ)² * ((d/2)cosθ)²

where θ is the angle at which the beam is cut from the log.

In Figure 3, the width and depth of the beam also vary with the angle θ. We can express the width and depth as follows:

w = d sinθ

d = (d/2)cosθ

Then we can express the strength of the beam as:

Strength = k * width * depth² = k * (d sinθ) * [(d/2)cosθ]²

Therefore, depending on the angle at which the beam is cut from the log, the strength of the beam can be expressed using different equations, but in all cases, the strength is proportional to the width and the square of the depth, with a proportionality constant k.

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Complete question is:

The strength of a beam is proportional to the width and the square of the depth. A beam is cut from a cylindrical log of diameter d = 24 cm. The figures show different ways this can be done. Express the strength of the beam as a function of the angle in the figures. (Use k as your proportionality constant.)

a = amount invested at 4%

how much is 4% of "a"? (4/100) * "a", namely 0.04a.

b = amount invested at 5%

how much is 5% of "b"? (5/100) * "b", namely 0.05b.

we know the total amount invested is 5550, so whatever "a" and "b" might be, we know that a + b = 5550.

we also know that the yielded amount in interest is 259, so if we simply add their interest, that'd be 0.04a + 0.05b = 259.

[tex]a+b=5550\implies b=5550-a \\\\[-0.35em] ~\dotfill\\\\ 0.04a+0.05b=259\implies \stackrel{\textit{substituting from above}}{0.04a+0.05(5550-a)~~ = ~~259} \\\\\\ 0.04a+277.5-0.05a=259\implies 277.5-0.01a=259\implies -0.01a=-18.5 \\\\\\ a=\cfrac{-18.5}{-0.01}\implies \boxed{a=1850}\hspace{9em}\stackrel{ 5550~~ - ~~1850 }{\boxed{b=3700}}[/tex]

The solutions  of the system of equations are (0, -6) and (7, 8)

Here we have the system of equations:

y=x^2−5x−6

y=2x−6

To solve this, we need to solve the equation:

x^2 - 5x - 6 = 2x - 6

x^2 - 7x = 0

x*(x - 7) = 0

We can see that the two solutions are:

x = 0

x = 7

Evaluating the line in these values we get:

y = 2*0 - 6 = -6

y = 2*7 - 6 = 8

Then the solutions are (0, -6) and (7, 8)

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The expected value for playing roulette if a person bet $16 on red is -$0.84.

b. The statement that best describes what this value means is option b. Over the long run, the player can expect to lose about $1.05 for each game played.

To be able to calculate the expected value, you need to multiply the probability of winning by the amount won hence it will be:

Expected value = (18/38) x  $16 - (20/38) x  $16

= -$0.84

Based on the fact that the expected value has a negative sign, it implies that, on average, the player tend to lose money over the long run.

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See full question below

one option in a roulette game is to bet $16 on red. (there are 18 red compartments, 18 compartments, and two compartments that are neither red nor black) if the ball lands on red, you get to keep the $16 you paid to play the game and you are awarded $16. If the ball lands elsewhere, you are awarded nothing and the $16 that you bet is collected.

a. what is the expected value for playing roulette if you bet $16 on red?

$__________ (round to the nearest cent)

b. chosen the statment below that best describes what this value means???

Pick One!

a. over the long run, the player can expect to win about $1.05 for each game played

b. over the long run, the player can expect to lost about $1.05 for each game player

c. over the lung run, the player can expect to break even

The temperature at 1pm is -3°C, at 2pm it's -6°C, at 3pm it's -9°C, and at 6pm it's -18°C. The temperature t hours after noon is T = 0 - 3t

If the temperature is dropping at a rate of 3 degrees per hour, then the temperature t hours after noon can be represented by the equation T = 0 - 3t, where T is the temperature in degrees Celsius.

To find the temperature at a specific time after noon, we need to substitute the corresponding value of t into the equation and solve for T. For example:

At 1pm (t = 1), T = 0 - 3(1) = -3°C

At 2pm (t = 2), T = 0 - 3(2) = -6°C

At 3pm (t = 3), T = 0 - 3(3) = -9°C

At 6pm (t = 6), T = 0 - 3(6) = -18°C

To find the temperature t hours after noon, we can use the same equation: T = 0 - 3t. For example, if we want to find the temperature 4 hours after noon, we can substitute t = 4 into the equation: T = 0 - 3(4) = -12°C. So the temperature 4 hours after noon is -12°C.

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The 99% confidence interval of the population mean is (352.32, 371.68).

To find the 99% confidence interval of the population mean, we need to use the following formula:

CI = x; ± zα/2 * (σ/√n)

where:

x' = sample mean (362)

zα/2 = z-score for 99% confidence level (2.576)

σ = population standard deviation (29.6)

n = sample size (40)

Substituting the values, we get:

CI = 362 ± 2.576 * (29.6/√40)

CI = 362 ± 9.68

This means that if we take many random samples of 40 hours from the population, we can be 99% confident that the true population mean of the number of customers per hour falls between 352.32 and 371.68.

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Complete question is:

A large department store found that it averages 362 customers per hour. Assume that the standard deviation is 29.6 and a random sample of 40 hours was used to determine the average. Find the 99% confidence interval of the population mean.

Therefore here, (5⁴)². 5⁶. (5⁸)².(5⁶)². (5⁸)² simplifies to 5⁸².

The integer's basic form is represented by its smallest equivalent fraction. How to determine the simplest form: Look for shared components in the denominator and numerator.

By applying the exponentiation laws, let's us gradually simplify this an expression is:

[tex](5^4)^2 = 5^(4*2) = 5^85^6 * 5^8 = 5^(6+8) = 5^1425^6 = (5^2)^6 = 5^(2*6) = 5^1225^8 = (5^2)^8 = 5^(2*8) = 5^16[/tex]

When all of these simplifications are put back into the original statement, we get the following:

[tex](5^4)^2. 5^6. 5^8. 25^6. 25^8 = 5^8 * 5^14 * 5^12 * 5^16 * 5^16[/tex]

Using the rule that is  [tex]a^b * a^c = a^(b+c)[/tex],

The first three terms can be combined to create:

[tex]5^8 * 5^14 * 5^12 * 5^16 * 5^16 = 5^(8+14+12) * 5^(16+16) = 5^50 * 5^32[/tex]

Using the same rule again, we can combine these two terms:

5⁵⁰* 5³²

= 5⁵⁰⁺³²

= 5⁸²

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The calculated output value for an input value of 5 is (c) 10

From the question, we have the following parameters that can be used in our computation:

Kerrin is making an input-output table for the rule "add 5."

This means that

Output = input + 5

In this case, we have

input = 5

substitute the known values in the above equation, so, we have the following representation

Output = 5 + 5

Evaluate

Output =10

Hence, the output is 10

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Surface area of similar cone is 44π square inches

What does a cone in maths ?

A cone is referred to as a "right cone" when its vertex is higher than the base's centre (i.e., when the angle formed by the vertex, base center, and any base radius is a right angle); otherwise, the word "oblique" is used. A cone is referred to as an elliptic cone when the base is assumed to be an ellipse rather than a circle.

The surface area of a cone is given by:

surface area = πr(r + l)

where r is the radius and l is the slant height.

surface area of original cone = π(3)(3 + 12) = 45π

slant height of similar cone = (6/3)(12) = 24 inches

surface area of similar cone = π(6)(6 + 18) = 144π

ratio of surface areas = surface area of similar cone / surface area of original cone

= (144π) / (45π)

= 3.2

Therefore, The exact surface area of the similar cone is 6 times the surface area of the original cone,

6 × 45π = 144π square inches

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Answer:

All of them have a combined of 2,227 Stamps

Mean: 2014

Option (b), 62% is correct.

By the given condition,

B/A = A/(A + B) ..... (1)

Now to find the percentage of A = B, we consider B = Ax. From (1), we get

Ax/A = A/(A + Ax)

or, x = 1/(1 + x)

or, x² + x - 1 = 0

Using the quadratic formula, we get

x = {- 1 ± √(1 + 4)}/2

= (- 1 ± √5)/2

Since x cannot be negative, we take

x = (- 1 + √5)/2

≈ 0.62

Therefore the required percentage is

= 0.62 × 100%

= 62%

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B as a percentage ofAis equal to Aas a

percentage of(A+B). How much percent

of A is B?

(a) 60%

(b) 62%

(C) 64%

(d) 66%

The probability that an eighth-grade girl will have long hair and be wearing a dress is 3/8 or 0.375, which is the same thing expressed as a decimal.

what is probability?

Probability is a measure of the likelihood of an event occurring. It is a number between 0 and 1, where 0 means the event is impossible and 1 means the event is certain to happen.

To solve the problem, we need to find the intersection of the events "having long hair" and "wearing a dress" and calculate the probability of that intersection.

Let L be the event "having long hair" and D be the event "wearing a dress". Then we know:

P(L) = 3/4, since three-fourths of the girls have long hair.

P(D) = 1/2, since half of the girls are wearing dresses.

To find P(L ∩ D), we need to multiply the probabilities of the two events:

P(L ∩ D) = P(L) × P(D) = (3/4) × (1/2) = 3/8

Therefore, the probability that an eighth grade girl will have long hair and be wearing a dress is 3/8 or 0.375, which is the same thing expressed as a decimal.

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(a) The value of the row operation of -2(row1) + row (2) is [-10     4    -12].

(b) The interchange of row 1 and row 2 is  [0     6    -3]

                                                                       [5    -2     6]

The value of the row operation of -2(row1) + row (2) is calculated as follows;

-2 (row 1) = 2 [5   -2    6]

-2 (row 1) = [-10     4    -12]

The addition of -2(row1) to row (2) is calculated as follows;

[-10     4    -12] + [0    6    -3]

= [-10   10  - 15]

The interchange of row 1 and row 2 is calculated as follows;

R₁ ↔ R₂

= [5    -2     6]    =   [0     6    -3]

  [0     6    -3]         [5    -2     6]

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Answer: 18.75

Step-by-step explanation:

small greeting cards = 1.80

large greeting cards = 3.85

1.80 x 4 = 7.20

3.85 x 3 = 11.55

7.20 + 11.55 = 18.75

The given total area including the frame is y = (8+2x)(10+2x).

As per the question the thickness of the frame is x.

The area of a rectangle is given by A=L*B

For a picture with frame it will be A=(L+2t)*(B+2t)

1. Comparing the above equation with the given equation it can be concluded that the Length is 10 and the width is 8.

2. The given equation is 100=(8+2x)(10+2x)

Upon solving it simplifies to [tex]x^2+9x-5=0[/tex]

x=0.525.

Thus the thickness x=0.525.

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Answer: 15y2 - 19y + 6 = 0

Step-by-step explanation:

The first term is,  15y2  its coefficient is  15 .The middle term is,  -19y  its coefficient is  -19 .The last term, "the constant", is  +6 Step-1 : Multiply the coefficient of the first term by the constant   15 • 6 = 90 Step-2 : Find two factors of  90  whose sum equals the coefficient of the middle term, which is   -19 .

The length of BC, to the nearest tenth of a centimeter, is calculated as: 12.6 centimeter.

Recall that two similar triangles will always have side lengths that are proportional in length. The triangles shown are similar to each other, therefore, the proportion below would be true:

AC/CE = BC/CD

Given the following:

AC = 7.2 units

CE = 2.4 units

CD = 4.2 units

BC = ?

Plug in the values:

7.2/2.4 = BC/4.2

Cross multiply:

BC = 7.2 * 4.2 / 2.4

BC = 12.6 cm

Thus, the length of side BC in the similar triangles shown is approximately 12.6 cm.

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Using the ratios, 1 yd/3 feet, 1 mi/5,280 feet, and 200 yards/1 feet, the number of flags required to mark the racecourse with a flag every 200 yards is 22.

Ratios are relative sizes of one quantity or value compared to another.

Ratios are depicted as fractional values using decimals, percentages, or fractions, including the standard ratio form (:).

The total distance of the racecourse = 2.5 miles

The common distance between each flag = 200 yards

1 mile = 1,760 yards

2.5 miles = 4,400 yards

1 yard = 3 feel

The number of flags required to cover the distance = 22 (4,400 ÷ 200)

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Answer: 14.5 kg

Step-by-step explanation:

The ratio of true positives to false positives, given the accuracy of the test would be 239 : 711 .

Here. we have,

First, find the number of people who have the flu:

= 5. 2 % x 10, 000

= 520 people

Those who don't have it:

= 10, 000 - 520

= 9, 480 people

True positive :

= 92 % x 520

= 478 people

False negatives:

= 8 % x 520

= 42 people

False positives:

= 15 % x 9, 480

= 1, 422 people

The ratio of true positives to false positives is therefore:

478 : 1, 422

239 : 711

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