What is 0.232323... written as a
fraction?

A dependent variable is the variable that being check in a scientific experiment.

The dependent variable is "dependent" on the other independent variable. As the experimenter alter the independent variable, the change in the dependent variable is observed and recorded. When you take data in an experiment, the dependent variable is the one being advised.

A scientist is testing the effect of light and dark on the efforts of moths by turning a light on and off. The independent variable is the amount of light and the moth's reply is the dependent variable.

A change in the independent variable (total amount of light) straight source a change in the dependent variable (moth behavior).

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Equivalent expressions are expressions of equal values

The equivalent expressions are 4x+ (y - 8y) + (2z-5z) +6   and 6x-3x-6x + (2y - 10y) + (4 - 8) + (z - 88z)

The first expression has been solved.

So, we have the following expressions

4x−7y−5z+6 and -3x−8y−4−87z

4x−7y−5z+6

We have:

4x-7y-5z+6

Rewrite as:

4x+ (y - 8y) + (2z-5z) +6

-3x−8y−4−87z

We have:

-3x−8y−4−87z

Rewrite as:

3x-6x + (2y - 10y) + (4 - 8) + (z - 88z)

Hence, the equivalent expressions are 4x+ (y - 8y) + (2z-5z) +6   and 6x-3x-6x + (2y - 10y) + (4 - 8) + (z - 88z)

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9514 1404 393

Answer:

  5x +20y = T

Step-by-step explanation:

There are 5 envelopes in each of x small packets, for a total of 5x envelopes in small packets.

There are 20 envelopes in each of y large packets, for a total of 20y envelopes in large packets.

The total number envelopes is then ...

  T = 5x +20y

The sum of 200000000000 + 123456789012345678912345678 is 123456789012365678912345678.

Here we have implemented the mathematical operation called addition.

A mathematical procedure called addition combines two or more values to create a new value. One of the fundamental operations in mathematics, "+" stands in for it.

Addition can be applied to other mathematical objects like vectors or matrices as well as numbers of any sort, including integers, decimals, and fractions.

The outcome is unaffected by the order in which the values are added because the addition is commutative.

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Answer:

it is A Scalene Triangle

Answer:

c

Step-by-step explanation:

In 60 minutes, minute hand completes 360

0

 i.e., one rotation.

⇒ in 1 minute, minute hands completes  

60

360

∘

​

=6

∘

Hence, in 5 minutes the angle covered is is 6×5=30

o

.

The angle of sector formed is 30

o

.

The area swept by the minute hand in 5 minutes=

360

30

​

×π×14×14

Using π=

7

22

​

, we get

The area swept by the minute hand in 5 minutes =

3

154

​

 sq. cm

Answer:

6 Apples

Step-by-step explanation:

There would be six apples and 2 oranges left over.

Hope this Helps!

:D

Answer:

I was kind of confused on this one but is it 15 apples?

Step-by-step explanation:

3, 6, 9, 12, 15

4, 8, 12, 16, 20

I think the answer is 15 apples.

The mean of a set of log-normally distributed random variables is equal to the mean of their logs.

Therefore, if we simulate a lot of lognormal(5,1) random variables, take their logs and then take the mean, it will be closest to the mean of the logs, which is 5. This is because lognormal(5,1) tells us that the mean of the original variables is exp(5+1^2/2), which is equal to exp(5.5), or 148.413. Taking the log of this number returns 5.

To calculate this more precisely, let's assume we simulate n lognormal(5,1) random variables and denote them by x_1, x_2, ..., x_n. We take the logs of each variable, producing the values y_1, y_2, ..., y_n. The mean of the logs is then calculated as (y_1+y_2+...+y_n)/n. Since each y_i is equal to the log of one of the x_i's, which is equal to 5, the mean of the logs is 5. Therefore, if we simulate a lot of lognormal(5,1) random variables, take their logs and then take the mean, it will be closest to 5.

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The dimension of the box will be (40 inch × 40 inch × 10 inch).

How to determine the maximum and minimum values of an expression using differentiation?
Any given expression can have both minimum and maximum outcomes for any number in a real world scenario, which can be determined using differentiation; first order differentiation when equated to zero reveals the value that will solute to produce the maximum or minimum and then the upon differentiating the obtained expression again and then substituting the variable in acceptable positions, if the outcome is a positive number then the obtained number in first differentiation suggests a minimum value of the given expression.

Given, the volume of the box described = 4000 inch³
Let the dimension of the base of the box be = 'a' inch × 'a' inch = a² inch²
Therefore, height of the box = (4000 inch³)/(a² inch²) = (4000/a²) inch
Total surface area of the described box = S = a² + [4a×(4000/a²)]
= [a² + 16000/a] inch
For the limiting values of S, dS/da = 0 ⇒ d[a² + 16000/a]/da = 0
⇒ 2a - [16000/a²] = 0 ⇒ 2a = 16000/a² ⇒ a³ = 8000 ⇒ a = ∛8000 = 20
Therefore, the value of a is 20 inch.
Thus, height of the box = 4000/a² = 4000/400 = 10 inch
For S to have the minimum limited value, d²S/da² > 0 for a = 20 inch, which is satisfied by the available value of S.
∴ The minimal surface area = a² + 16000/a = 400 + (16000/400)
= 440 inch²
We are aware that the least amount of material will be needed to build the box when its overall surface area is the smallest.
Thus, to minimize the amount of material in consideration, the dimension of the box will be (40 inch × 40 inch × 10 inch).

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Answer:

2,3 bro bro and also my leg

Step-by-step explanation:

Given that,

The area of a blackboard is \(1\dfrac{1}{3}=\dfrac{4}{3}\ \text{yards}^2\)

The area of poster is \(\dfrac{8}{9}\ \text{yards}^2\)

We need to find the unit rate of the blackboard's area to the poster's area. So, it can be calcualted by dividing blackboard's area to the poster's area.

So,

\(\dfrac{\dfrac{4}{3}}{\dfrac{8}{9}}=\dfrac{4}{3}\times \dfrac{9}{8}\\\\=1.5\)

So, the rate of \(1.5\ \text{yard}^2\).

This is about understanding sets.

Option A is correct.

Thus, the subsets would be;

{a}, {b}, {c}, {a,b}, {a, b, c}, {a, c}, {b,c}

Applying the same principle used in the example above, we can write the subsets as;

{2};{-2};{-2,2}

Thus, option A is the correct answer.

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Answer:

a

Step-by-step explanation:

The probability of observing any particular value for a four-degrees-of-freedom t-distribution is nearly zero, and the likelihood that a population item will have a value for this distribution is equal to 1.

When estimating the mean of a normal t-distribution from a small sample in statistics, the t-distribution with four degrees of freedom emerges as a continuous probability distribution.

Given that a t-value t-distribution can be any real number, the likelihood that a population item will have a value for a t-distribution with four degrees of freedom is equal to 1.

In other words, since the distribution is continuous and has an infinite number of possible values, the likelihood of detecting any specific value for a t-distribution with four degrees of freedom is infinitesimally small.

Therefore, The probability of observing any particular value for a four-degrees-of-freedom t-distribution is nearly zero, and the likelihood that a population item will have a value for this distribution is equal to 1.

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The matrices A and B that were defined are the integer matrices:

A = [1 0 0; 0 1 0]

B = [1 1 0; 0 0 0]

Both A and B are integer matrices, and they have the same null space and column space but are not multiples of each other. Therefore, they satisfy the requirements " Integer matrices A,B not multiples of each other such that Nul(A)=Nul(B) and Col(A)=Col(B)."

Let A and B be 3x3 matrices given by:

A = [1 0 0; 0 1 0]

B = [1 1 0; 0 0 0]

We can see that A and B have the same null space and column space, but they are not multiples of each other. To show that A and B have the same null space, we need to find the null space of both matrices.

For A, we need to solve the equation Ax = 0:

[1 0 0; 0 1 0] [x1; x2; x3] = [0; 0]

This Equations has the unique solution x = [0; 0; 0], so null space of A is the trivial subspace {0}.

For B, first solve the equation Bx = 0:

[1 1 0; 0 0 0] [x1; x2; x3] = [0; 0]

This equations has the general solution x = [-x2; x2; x3], where x2 and x3 are arbitrary integers, so null space of B is the subspace spanned by the vector [-1; 1; 0].

Show that A and B have the same column space, we have to show the columns of A and B span the same subspace. The columns of A are [1; 0] and [0; 1], which span the entire 2-dimensional space. The columns of B are [1; 0] and [1; 0], which span the 1-dimensional subspace { [x; x] : x is an integer }. But, the column space of A is also the subspace { [x; y] : x and y are integers }, which is the same as the column space of B.

Therefore, A and B have the same column space.

Since A and B have the same null space and column space but are not multiples of each other, they satisfy the conditions " integer matrices A,B not multiples of each other such that Nul(A)=Nul(B) and Col(A)=Col(B)."

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db/dt = 40/12 = 10/3 Feet per second, or approximately 3.33 feet per second. So the foot of the ladder is being pulled away from the wall at a rate of about 3.33 feet per second when it is 8 feet away from the wall and the top of the ladder is sliding down at 4 feet per second.

We can use the Pythagorean theorem to relate the length of the ladder, the distance of its foot from the wall, and the height it reaches on the wall:

\(a^2 + b^2 = c^2\)

where c is the length of the ladder, a is the distance of its foot from the wall, and b is the height it reaches on the wall. Differentiating with respect to time, we get:

2a da/dt + 2b db/dt = 2c dc/dt

We are interested in finding db/dt when a = 8 feet and dc/dt = -4 feet per second (negative because the top of the ladder is sliding down). We also know that c = 10 feet, so we can plug in these values and solve for db/dt:

2(8) da/dt + 2b db/dt = 2(10) (-4)

Simplifying:

16 da/dt + b db/dt = -40

We also know that when a = 8 feet and b = 6 feet (from the Pythagorean theorem), the ladder is at a height of 6 feet on the wall. Therefore, we can plug in these values and solve for da/dt:

8 da/dt + 6 db/dt = 0

Simplifying:

da/dt = -(3/4) db/dt

Now we can substitute this expression for da/dt in the first equation, and solve for db/dt:

2(8) (-(3/4) db/dt) + 2(6) db/dt = -40

Simplifying:

-12 db/dt = -40

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Answer:

(-65/23, 72/23)

Step-by-step explanation:

Easiest and fastest way to do this is to graph the systems of equation on a graphing calculator, then analyzing the graph to see where they intersect. The intersection point is your solution to the systems of equations.

The estimated probability that all four randomly selected young American men are 68 inches or shorter is approximately 0.004 or 0.4%.

To calculate the probability that all four randomly selected young American men are 68 inches or shorter, we need to consider the probability for each individual man and multiply them together.

Let's assume that the probability of an individual young American man being 68 inches or shorter is p. Since we are selecting four men at random, the probability of each man being 68 inches or shorter is the same, and we can multiply their probabilities together.

The probability of one man being 68 inches or shorter is p. Therefore, the probability of all four men being 68 inches or shorter is p × p × p × p = p^4.

However, we are not given the specific value of p in the problem statement. If we assume that the height of young American men follows a normal distribution, we can look up the corresponding z-score for a height of 68 inches or shorter and use the standard normal distribution to estimate the probability.

For example, if we find that a height of 68 inches corresponds to a z-score of -1.0, we can use a standard normal distribution table or a calculator to determine the probability of a z-score less than or equal to -1.0. Let's say this probability is approximately 0.1587.

Therefore, the estimated probability that all four randomly selected young American men are 68 inches or shorter would be (0.1587)^4 = 0.004.

Thus, the probability is approximately 0.004 or 0.4% rounded to three decimal places.

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The table of values that represents exponential decay is (c)

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

The table of values

An exponential function is represented as

y = abˣ

Where

Rate = b

When the rate is less than 1, then the table represents a decay

i.e when y reduces as x increases

The table that has the above features is the table (c)

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Answer:

Step-by-step explanation:

Answer:

C

Step-by-step explanation:

step by step fr

Answer:

x=26

m∠HJN=53°

Step-by-step explanation:

The two angles share a right angle, meaning they are complementary and have a sum of 90°.

\((2x+1)+37=90\)

\(2x+38=90\)

\(2x=52\)

\(x=26\)

If x=26,

m∠HJN

\(=2(26)+1\)

\(=52+1\)

\(=53\)

Answer:

x = 1

Step-by-step explanation:

Using the rules of exponents

\((a^m)^{n}\) = \(a^{mn}\)

\(a^{m}\) × \(a^{n}\) ⇔ \(a^{(m+n)}\)

Note that 9 = 3² and 27 = 3³ , thus

\(9^{x-2}\) × \(27^{x}\)

= \((3^2)^{x-2}\) × \((3^3)^{x}\)

= \(3^{2x-4}\) × \(3^{3x}\)

= \(3^{(2x-4+3x)}\)

= \(3^{5x-4}\) , then

\(3^{5x-4}\) = \(3^{1}\)

Since the bases on both sides are equal, equate the exponents

5x - 4 = 1 ( add 4 to both sides )

5x = 5 ( divide both sides by 5 )

x = 1

The y-intercept of the graph is -3

The y-intercept is the point where the line intersects the y-axis. This is the value of y when x is equals to zero.

Using slope intercept formula,

y = mx + b

where

Now, let's trace the value of y when x = 0 in the graph.

Therefore,

when x = 0, y = -3

Hence, the y-intercept of the graph is -3

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The AM signal plot on the given graph paper will show the message signal with a Root Mean Square (RMS) of 2√2, along with the carrier signal and the modulated signal, denoted by V and Vm respectively. The modulation index is 40%.

Step 1: Determine the peak voltage of the message signal.

Given that the message signal's RMS voltage is 2√2, we can find the peak voltage (Vm) using the formula:

Vm = RMS × √2

Vm = 2√2 × √2

Vm = 2 × 2

Vm = 4

Step 2: Calculate the modulation index (m).

The modulation index (m) is given as 40%, which can be written as 0.4.

m = 0.4

Step 3: Determine the amplitude of the carrier signal.

The carrier signal's amplitude (V) can be calculated by dividing the peak voltage of the modulated signal by the modulation index:

V = Vm / m

V = 4 / 0.4

V = 10

Step 4: Plot the signals on graph paper.

Using an appropriate scale, plot the message signal, carrier signal, and modulated signal on the graph paper.

Label the x-axis as time.

Label the y-axis as voltage.

Mark the values for time and voltage on the axes.

Draw the message signal, which has an RMS of 2√2, as a sine wave with an amplitude of 2√2.

Draw the carrier signal, which has an amplitude of 10, as a horizontal line at a fixed voltage of 10.

Draw the modulated signal, denoted as Vm, which is obtained by multiplying the message signal with the carrier signal, as a sine wave with an amplitude of 4.

Mark the values for Vm, V, and other related voltages on the plot accordingly.

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The AM signal can be plotted on the graph paper with appropriate scaling. The message Root Mean Square (RMS) is 2√2, and the modulation index is 40%.

To plot the AM signal, we first need to understand the concept of modulation index. Modulation index (m) is a measure of the extent of modulation imposed on the carrier signal by the message signal. In this case, the modulation index is 40%, which means that the amplitude of the carrier signal varies by 40% of the peak amplitude due to modulation.

The message Root Mean Square (RMS) value represents the amplitude of the message signal. Given that the RMS is 2√2, we can calculate the peak voltage (Vm) of the message signal using the formula Vm = √2 * RMS. Therefore, Vm = √2 * 2√2 = 4V.

Next, we need to determine the carrier signal amplitude (V). The carrier signal remains constant in amplitude but varies in frequency. Since the modulation index is 40%, the carrier signal will have a peak-to-peak variation of 40% * Vm = 0.4 * 4V = 1.6V.

Now, we can plot the AM signal on the graph paper. The x-axis represents time, and the y-axis represents voltage. The carrier signal will have a constant amplitude of V, while the message signal will vary between -Vm and +Vm.

On the plot, we can mark the values of Vm and V to indicate the amplitudes of the message and carrier signals, respectively. Additionally, we can mark the related voltages, such as -0.4Vm, 0.4Vm, -Vm, Vm, etc., to represent different points on the AM signal.

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Answer:

5

Step-by-step explanation:

y= 4x+17

73=4x+17 (subtract 17 by both sides)

20=4x (divide by both sides)

5= x, 5 days

Answer:

B

Step-by-step explanation:

B is the only one that has only 2 sides of the same length and one that doesn't match the other 2.

A doesn't have any of the lengths being the same

C. is a right triangle, although there can be isosceles right triangles they still need to have only 2 lengths that are the same and in this one, all the values are different

D. This is an equilateral triangle, this has more than 2 lengths that are the same

The given expressions, when A=1, B=0, and C=1, X evaluates to 1.001; when A=B=C=1, X evaluates to 1.111; and when A=B=C=0, X evaluates to 0.000. These evaluations are based on the given values of A, B, and C, and the notation ¯¯¯¯¯¯BC represents the complement of BC.

To evaluate the expression X = A.¯¯¯¯¯¯BC, we substitute the given values of A, B, and C into the expression.

(a) For A = 1, B = 0, and C = 1:

X = 1.¯¯¯¯¯¯01

To find the complement of BC, we replace B = 0 and C = 1 with their complements:

X = 1.¯¯¯¯¯¯01 = 1.¯¯¯¯¯¯00 = 1.001

(b) For A = B = C = 1:

X = 1.¯¯¯¯¯¯11

Similarly, we find the complement of BC by replacing B = 1 and C = 1 with their complements:

X = 1.¯¯¯¯¯¯11 = 1.¯¯¯¯¯¯00 = 1.111

(c) For A = B = C = 0:

X = 0.¯¯¯¯¯¯00

Again, we find the complement of BC by replacing B = 0 and C = 0 with their complements:

X = 0.¯¯¯¯¯¯00 = 0.¯¯¯¯¯¯11 = 0.000

In conclusion, when A = 1, B = 0, and C = 1, X evaluates to 1.001. When A = B = C = 1, X evaluates to 1.111. And when A = B = C = 0, X evaluates to 0.000. The evaluation of X is based on substituting the given values into the expression A.¯¯¯¯¯¯BC and finding the complement of BC in each case.

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The band needs to sell at least 175 CDs to make a profit of at least $500.

The inequality that represents the number of CDs, n, the band needs to sell to make a profit of at least $500 is:

5n - 1.25(300) ≥ 500

Explanation:

- The band sells each CD for $5, so the revenue from selling n CDs is 5n.
- The band ordered 300 CDs at a cost of $1.25 each, so the total cost of the CDs is 1.25(300) = $375.
- To make a profit of at least $500, the revenue from selling the CDs (5n) must be greater than or equal to the total cost of the CDs plus $500 (375 + 500 = 875).
- Putting it all together, we get the inequality:

5n - 1.25(300) ≥ 500

Simplifying this inequality gives:

5n - 375 ≥ 500

5n ≥ 875

n ≥ 175

Therefore, the band needs to sell at least 175 CDs to make a profit of at least $500.

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