Joseph has a bag filled with 2 red, 6 green, 15 yellow, and 7 purple marbles. Determine P(not green) when choosing one marble from the bag.

90%
80%
60%
20%

Answer:

Step-by-step explanation:

first, we convert the minutes into an hour

there are 60 minutes in an hour

so then we do 60 divided by 12

that equals to 5

so 5 miles is the answer

To solve this problem, we need to understand the conditions under which the inequality $x^2 nx 15 < 0$ has no real solutions. This inequality can be rewritten as $nx(x-15/n) < 0$, which tells us that either $n>0$ and $x<0$ or $x>15/n$, or $n<0$ and $x>0$ or $x<15/n$.

In either case, we have a product of two factors that must be negative, which means that either both factors are negative or both factors are positive.
Since we are looking for the number of different possible values of $n$, we need to consider all possible combinations of signs for $n$ and $x-15/n$. If $n>0$, then $x-15/n$ must also be positive, which means that $x>15/n$. If $n<0$, then $x-15/n$ must be negative, which means that $x<15/n$. In either case, we can see that $n$ must be either positive or negative, and that there is only one possible value of $n$ that satisfies the given condition: $n = \frac{15}{x^2}$.

To find the number of different possible values of $n$, we need to consider all possible values of $x$. If $x=0$, then the inequality is trivially true for any value of $n$. If $x\neq 0$, then we can see that $n$ can take any value in the interval $(0,\infty)$ or $(-\infty,0)$, which means that there are infinitely many possible values of $n$ that satisfy the given condition.

Therefore, the answer to the question is that there are infinitely many different possible values of $n$.
To find the number of different possible values of $n$ for which the inequality $x^2 + nx + 15 < 0$ has no real solutions in $x$, we first need to analyze the inequality.

Step 1: Find the discriminant of the quadratic inequality.
The discriminant, $D$, is given by the formula $D = b^2 - 4ac$, where $a$, $b$, and $c$ are the coefficients of the quadratic expression. In this case, $a = 1$, $b = n$, and $c = 15$.

So, $D = n^2 - 4(1)(15) = n^2 - 60$.

Step 2: Determine the condition for the inequality to have no real solutions.
For a quadratic inequality to have no real solutions, the parabola must not intersect the x-axis, which means the discriminant must be less than 0.

So, we need to solve the inequality $D < 0$:

$n^2 - 60 < 0$

Step 3: Solve the inequality for $n$.
To solve the inequality, find the range of values of $n$ that satisfy the inequality.

$(n - \sqrt{60})(n + \sqrt{60}) < 0$

Since $\sqrt{60}$ is between 7 and 8, we can rewrite the inequality as:

$-8 < n < 8$

Step 4: Count the number of possible integer values of $n$.
The inequality indicates that $n$ must be an integer between -8 and 8 (not inclusive). Therefore, the possible values of $n$ are -7, -6, -5, -4, -3, -2, -1, 1, 2, 3, 4, 5, 6, and 7.

There are 14 different possible values of $n$ for which the inequality $x^2 + nx + 15 < 0$ has no real solutions in $x$.

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(a) The probability of selecting a senior on any given selection is 0.65. The probability of selecting a non-senior (either a junior or a sophomore) is 0.35. To select exactly three award-winners until a senior is selected, we need to select two non-seniors followed by a senior. The probability of this sequence is:

0.35 * 0.35 * 0.65 = 0.080

So the probability of selecting exactly three award-winners until a senior is selected is 0.080.

(b) The probability of selecting a junior on any given selection is 0.24. The probability of selecting a non-junior (either a senior or a sophomore) is 0.76. To select at least three award-winners until a junior is selected, we need to select two or more non-juniors followed by a junior. The probability of this sequence is:

(0.76 * 0.76 * 0.24) + (0.76 * 0.24) + (0.24) = 0.334

So the probability of selecting at least three award-winners until a junior is selected is 0.334.

(c) The probability of selecting a sophomore on any given selection is 0.11. The probability of selecting a non-sophomore (either a senior or a junior) is 0.89. To select 2 or fewer award-winners until a sophomore is selected, we need to select 1 or 2 non-sophomores followed by a sophomore. The probability of these sequences is:

(0.89 * 0.89 * 0.11) + (0.89 * 0.11) + (0.11) = 0.214

So the probability of selecting 2 or fewer award-winners, until a sophomore is selected, is 0.214.
(a) To find the probability that we will select exactly three award-winners, and the third one is a SENIOR, we need to consider the following probabilities:

1. First student is not a senior (i.e., a junior or a sophomore)
2. Second student is not a senior (i.e., a junior or a sophomore)
3. Third student is a senior

Probability (not a senior) = 1 - Probability (senior) = 1 - 0.65 = 0.35
Probability (exactly 3 award-winners with the third being a senior) = 0.35 * 0.35 * 0.65 ≈ 0.079625

(b) To find the probability that we will select at least three award-winners until a JUNIOR is selected, we can find the probability of selecting 1 or 2 award-winners and subtract it from 1.

Probability (1 award-winner and it's a junior) = 0.24
Probability (2 award-winners, first is not a junior and second is a junior) = (1-0.24) * 0.24 = 0.1816

Total probability (1 or 2 award-winners) = 0.24 + 0.1816 = 0.4216
Probability (at least 3 award-winners) = 1 - 0.4216 ≈ 0.5784

(c) To find the probability that we will select 2 or fewer award-winners until a SOPHOMORE is selected:

Probability (1 award-winner and it's a sophomore) = 0.11
Probability (2 award-winners, first is not a sophomore and second is a sophomore) = (1-0.11) * 0.11 ≈ 0.0989

Total probability (2 or fewer award-winners) = 0.11 + 0.0989 ≈ 0.2089

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Five nickels plus two dimes totaling

combining for a sum of $0.45.

three nickels and four dimes respectively

combining for a sum of $0.55.

two nickels and five dimes denoting x = 2, y = 5

combining for a sum of $0.60.

3 ways to achieve a total of 7 coins with nickels (N) and dimes (D), and their corresponding values are

Five nickels plus two dimes totaling x = 5, y = 2 respectively. The value of five nickels = $0.25

two dimes = $0.20

combining for a sum of $0.45.

three nickels and four dimes respectively depositing x = 3, y = 4

3 nickels = $0.15

4 dimes = $0.40

combining for a sum of $0.55.

two nickels and five dimes denoting x = 2, y = 5

2 nickels = $0.10

5 dimes = $0.50

combining for a sum of $0.60.

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To prove that df ^ dg = |df/dx df/dy| dxdy.|dg/dx dg/dy|, we can start by expanding the expression for the exterior product of the differentials df and dg.

df ^ dg = (df/dx dx + df/dy dy) ^ (dg/dx dx + dg/dy dy)

Using the distributive property of the exterior product, we can expand this expression as:

df ^ dg = (df/dx dx) ^ (dg/dx dx) + (df/dx dx) ^ (dg/dy dy) + (df/dy dy) ^ (dg/dx dx) + (df/dy dy) ^ (dg/dy dy)

Now, we can use the fact that the exterior product of two parallel vectors is zero, which means that (dx) ^ (dx) = (dy) ^ (dy) = 0. This allows us to simplify the expression as:

df ^ dg = (df/dx df/dy dy ^ dx) ^ (dg/dx dg/dy dy ^ dx)

Since dy ^ dx = -dx ^ dy, we can further simplify the expression as:

df ^ dg = -|df/dx df/dy| dx ^ dy ^ (dg/dx dg/dy) dx ^ dy

Now, we can use the fact that dx ^ dy = -dy ^ dx, which means that (dx ^ dy) ^ (dx ^ dy) = 0. This allows us to simplify the expression as:

df ^ dg = -|df/dx df/dy dg/dx dg/dy| (dx ^ dy) ^ (dx ^ dy)

Since (dx ^ dy) ^ (dx ^ dy) = 0, we can conclude that:

df ^ dg = 0

Therefore, we have proven that df ^ dg = |df/dx df/dy| dxdy.|dg/dx dg/dy|.

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Daniel's balance before he wrote the check on Friday was $248.

To solve this problem, we can start by subtracting the $60 deposit from the $158 check, which gives us a balance of $98 before the check was cashed.

Since the account was overdrawn by $8 on Wednesday, we can subtract $8 from the balance to get $90.

Finally, we must add back the $158 check that was cashed which will give a balance of:

= $158 + $90

= $248

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To find the number of integers x such that x^2 < 10x – 21, follow these steps:

1. Rearrange the inequality to have all terms on one side:
x^2 - 10x + 21 < 0

2. Factor the quadratic expression:
(x - 7)(x - 3) < 0

3. Determine the critical points by finding the zeros of the factors:
x - 7 = 0 => x = 7
x - 3 = 0 => x = 3

4. Create intervals based on the critical points:
(-∞, 3), (3, 7), and (7, ∞)

5. Test a number from each interval in the inequality (x - 7)(x - 3) < 0:

- Interval (-∞, 3): Choose x = 2, (2 - 7)(2 - 3) = 5 * -1 < 0, interval is valid
- Interval (3, 7): Choose x = 4, (4 - 7)(4 - 3) = -3 * 1 > 0, interval is not valid
- Interval (7, ∞): Choose x = 8, (8 - 7)(8 - 3) = 1 * 5 > 0, interval is not valid

6. Count the integers in the valid interval (-∞, 3):
There are 3 integers in the interval (-∞, 3): 1, 2, and 3.

Therefore, there are 3 integers x such that x^2 < 10x - 21.

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Hi! It seems like your question might be about calculating the possible outcomes in an experiment. Based on the terms provided, I'll try my best to help you.

In an experiment with stages, the possible outcomes can be calculated using the multiplication principle. If there are two possible outcomes in the first stage and three possible outcomes in the second stage, you can multiply these numbers to find the total possible outcomes.

Total outcomes = (Outcomes in stage 1) x (Outcomes in stage 2)

Total outcomes = 2 x 3 = 6

Based on the given options, none of them match the calculated total outcomes.

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The perimeter of the square is greater than its area.

We have,

The area of a square is given by the formula A = s²,

where s is the length of one side of the square.

The perimeter is given by the formula P = 4s,

where s is the length of one side of the square.

In our case,

The length of one side of the square is 10 units.

So,

Area = s² = 10² = 100 square units

Perimeter = 4s = 4(10) = 40 units

We can see that the perimeter of the square (40 units) is greater than the area of the square (100 square units).

This makes sense because the perimeter is measuring the total distance around the square, while the area is measuring the amount of space inside the square.

To explain why the perimeter is greater than the area, we can imagine that we are trying to measure the perimeter of the square by walking around its edge, while we are trying to measure the area of the square by filling it with small square tiles.

We can see that we would need more tiles to fill the space inside the square than we would need to walk around its edge, which explains why the area is smaller than the perimeter.

Therefore,

The perimeter of the square is greater than its area.

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In exercise 9.38(b), we are given a random sample Y1, Y2, ..., Yn from a normal distribution with mean μ and unknown variance σ^2. The likelihood function for this sample is:

L(μ, σ^2) = (2πσ^2)^(-n/2) exp[-∑(Yi-μ)^2/(2σ^2)]

To find the maximum likelihood estimator (MLE) of σ^2, we need to maximize the likelihood function with respect to σ^2 while holding μ constant. Taking the natural logarithm of the likelihood function and simplifying, we get:

ln L(μ, σ^2) = -n/2 ln(2π) - n/2 ln(σ^2) - ∑(Yi-μ)^2/(2σ^2)

Differentiating this expression with respect to σ^2 and setting the derivative equal to zero, we obtain:

d/dσ^2 ln L(μ, σ^2) = -n/(2σ^2) + ∑(Yi-μ)^2/(2σ^4) = 0

Solving for σ^2, we get:

σ^2 = ∑(Yi-μ)^2/n

Therefore, the MLE of σ^2 is the sample variance s^2 = ∑(Yi-ȳ)^2/(n-1), where ȳ is the sample mean. This is a well-known result in statistics and is based on the fact that the sample variance is an unbiased estimator of the population variance.

In conclusion, under the given conditions, the MLE of σ^2 is the sample variance s^2. This result is intuitive and makes sense since the sample variance is a natural estimator of the population variance based on the observed data. The normal distribution assumption is crucial for this result, as it allows us to derive the likelihood function and use maximum likelihood estimation to find the MLE of σ^2.

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To answer the question about whether the process standard deviation of the cylindrical metal bar production should be adjusted to (I) a higher level than 0.25 cm or (II) a lower level than 0.25 cm to minimize the chance of production costs exceeding $1000, the suggestion is to adjust the standard deviation to (II) a lower level than 0.25 cm.

By reducing the standard deviation, the variation in the lengths of the produced metal bars will decrease, resulting in more consistent and controlled production. This will ultimately help minimize the chances of the production cost of a metal bar exceeding the $1000 threshold. A lower standard deviation ensures that the production process has fewer outliers and deviations from the mean length of 11 cm, leading to cost efficiency and reduction of waste or rework due to bars not meeting the desired specifications.

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Answer:

Alot

Step-by-step explanation:

Think abt it

From the trigonometric ratios, the value of sine trigonometric ratio for angle C, i.e., sin(C) in above right angled triangle CDE, is equals to the 0.55.

Trigonometry is a branch of mathematics. The trigonometric ratios are special measurements of a right triangle the right angle trigonometric ratios, these ratios describe the relationship between the sides and angles in a right triangle. The six trigonometric ratios in a right angled triangle are defined as sine, cosine, tangent, cosecant, secant, and cotangent. The symbols used for them are sin, cos, sec, tan, csc, cot. The three main ratios are defined as below

We have a right angled triangle CDE with 90° measure of angle D present in above figure. We have to determine the value of sine angle of C. Consider angle C priority,

Height or opposite of triangle = 11

Length of hypotenuse of triangle = 20

Using the above formula for sine trigonometric ratio, [tex]sin \: C = \frac{opposite}{hypotenuse}[/tex]

Substitute all known values in above formula, [tex]= \frac{11}{20}[/tex]

= 0.55

Hence, required value is 0.55.

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Complete question :

The above figure complete the question.

Find the value of sin C rounded to the nearest hundredth, if necessary.

20

E

11

D

Help PLEASE

The lengths of the diagonals in the isosceles trapezoid are 11.045 units and 7.2 units.

From the given figure, the vertices of the quadrilateral are (1, 6), (3, 0), (-5, 0) and (-1, 6).

From lower left to upper right: (-5, 0) and (1, 6)

Here, length = √(6+5)²+(1-0)²

= √122

= 11.045 units

From lower right to upper left: (3, 0) and (-1, 6)

Here, length = √(-1-3)²+(6-0)²

= √52

= 7.2 units

Therefore, the lengths of the diagonals in the isosceles trapezoid are 11.045 units and 7.2 units.

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The height of the second cylinder should be 56.25% greater than the height of the first cylinder. (Option 1)

Let's assume the radius of the first cylinder to be 'r' and its height to be 'h'. So, its volume can be represented as V1 = πr^2h.

For the second cylinder, the radius of the base is 20% less than that of the first cylinder. So, the radius of the second cylinder can be represented as 0.8r. Let the height of the second cylinder be represented as 'H'. So, its volume can be represented as V2 = π(0.8r)²H.

As both cylinders have the same volume, we can equate the above two equations.

πr²h = π(0.8r)²H

h = (0.8)²H

H = (1/(0.8)²)h

H = (1.5625)h

Therefore, the height of the second cylinder should be 56.25% greater than the height of the first cylinder.

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Complete Question:

Two cylinders have the same volume, but the radius of the base of the second cylinder is 20% less than the radius of the base of the first. How much greater should be the height of the second cylinder be in comparison to the height in first?

Options:

The inequalities are x ≤ -2 and x <-5.

We have,

x is less than or equal to -2 and x is less than -5.

writing the inequality in mathematical expression as

x is less than or equal to -2  = x ≤ -2.

and, x is less than -5 = x <-5.

Now, for x ≤ -2  the number line start from -2 with closed dot and move towards -3, -4, -5, ....

and, for x < -5 the number line start from -5 with open dot move towards -6, -7, -8, ....

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The volume of the satellite is,

⇒ V = 12π m³

Given that;

A satellite is in the shape of a cylinder with two hemispheres fitted snugly on either end.

And, The diameter of the cylinder is 2 m and its length is 12 m.

Now, We know that;

Volume of cylinder = πr²h

Hence, We get;

The volume of the satellite is,

⇒ V = π × (2/2)² × 12

⇒ V = 12π m³

Thus, The volume of the satellite is,

⇒ V = 12π m³

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The experimental probability that the next shirt sold will be purple is [tex]2/5[/tex].

The experimental probability means ratio of the number of times the event occurs to the total number of trials or observations.

In this case, the event is the sale of a purple shirt and the trials are the total number of shirts sold.

So, total number of shirts sold is:

= 6 purple shirts + 9 other color shirts

= 15 shirts

The number of purple shirts sold is 6.

The experimental probability of selling a purple shirt on the next sale will be:

= Number of purple shirts sold / Total number of shirts sold

= 6 / 15

= 2 / 5.

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The event that will have a sample space of S = {h, t} is (a) Flipping a fair, two-sided coin

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

Sample space of S = {h, t}

The sample size of the above is

Size = 2

Analyzing the options, we have

Hence, the event is (a)

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The needed, following the property of intersecting chords, length of segment b is 2 cm,

To find the length of segment b, we need to use the property of intersecting chords inside or outside a circle, which states that the product of the two segments of each chord is equal.

Given that:

ab = 6 cm

cd = 4 cm

ac = 3 cm

bd = b cm (length of segment b, to be found)

The property states:

ab * bd = cd * ac

Substitute the given values:

6 cm * b cm = 4 cm * 3 cm

Now, solve for b:

6b = 12

Divide both sides by 6 to isolate variable b:

b = 12 / 6

b = 2 cm

So, the length of segment b is 2 cm.

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The probability that at most 2 columns will fail is 0.98.

This is a binomial distribution problem, where the number of trials n = 16, the probability of success (a column failing) p = 0.05, and we want to find the probability of at most 2 columns failing.

To solve this, we need to calculate the probability of 0, 1, or 2 columns failing and add them up.

P(at most 2 columns failing) = P(0 columns failing) + P(1 column failing) + P(2 columns failing)

P(0 columns failing) = (n choose 0) * p^0 * (1-p)^(n-0) = (16 choose 0) * 0.05^0 * 0.95^16 = 0.45

P(1 column failing) = (n choose 1) * p^1 * (1-p)^(n-1) = (16 choose 1) * 0.05^1 * 0.95^15 = 0.38

P(2 columns failing) = (n choose 2) * p^2 * (1-p)^(n-2) = (16 choose 2) * 0.05^2 * 0.95^14 = 0.15

P(at most 2 columns failing) = 0.45 + 0.38 + 0.15 = 0.98

Therefore, the probability that at most 2 columns will fail is 0.98.

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Writing the total surface area of North America, which is approximately 9,540,000 square miles in Scientific Notation, is 9.54 x 10^6.

Scientific notation is shorthand way of writing very large or very small numbers in a standard form.

A number is written in scientific notation when a number between 1 and 10 is multiplied by a power of 10.

For instance, 9,540,000 square miles can be written in scientific notation as 9.54 x 10^6 square miles.

Thus, we can state that, in scientific notation, 9,540,000 square miles equal 9.54 x 10^6 square miles.

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The population proportion of members who favor holding the charity event in the civil club is approximately 94.12%. To calculate the population proportion of members in the civil club who favor holding the charity event, follow these steps:

The population proportion of members who favor holding the charity event can be calculated by dividing the number of members who said yes by the total number of members in the club.Proportion = Number of members who said yes / Total number of members in the club
Step:1. Identify the total number of members in the civil club: 85 members.
Step:2. Identify the number of members who said yes to holding the charity event: 80 members.
Step:3. Divide the number of members who said yes by the total number of members: 80 / 85.
Step:4. Convert the result to a percentage by multiplying by 100: (80 / 85) x 100.
So, the population proportion of members who favor holding the charity event in the civil club is approximately (80 / 85) x 100 = 94.12%.

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a) The total number of capsules needed for 4 weeks is C, 112 capsules.

b) The total milliliters needed for the month of October is C, 465 ml.

a) To determine the number of Rimactane 150mg capsules needed for a 10-year-old boy weighing 30 kg, who has been prescribed a dose of 10 mg/kg twice daily for 4 weeks, follow these steps:

1. Calculate the total daily dose: 30 kg * 10 mg/kg = 300 mg/day
2. Determine the number of daily doses: 300 mg/day / 150 mg/capsule = 2 capsules/day
3. Calculate the total number of capsules needed for 4 weeks: 2 capsules/day * 7 days/week * 4 weeks = 112 capsules

The answer is C, 112 capsules.

b) To calculate the number of milliliters of Epanutin suspension (phenytoin 30 mg/5 ml) that a 5-year-old boy weighing 18 kg with epilepsy will take at a dose of 5 mg/kg twice daily during the month of October, follow these steps:

1. Calculate the total daily dose: 18 kg * 5 mg/kg = 90 mg/day
2. Determine the number of milliliters needed for each daily dose: 90 mg/day * (5 ml/30 mg) = 15 ml/day
3. Calculate the total milliliters needed for the month of October: 15 ml/day * 31 days = 465 ml

The answer is C, 465 ml.

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The height of the cylinder is 7/2 inches.

Remember that for a cylinder of radius R and height H, the volume is:

V = pi*R²*H

Where pi = 22/7

We know that:

R = (1/3) in

V = (1 + 2/9) in³ = 11/9 in³

Replacing these values we will get:

11/9  = (22/7)*(1/3)²*H

11/9 = (22/7)*(1/9)*H

11 =(22/7)*H

11*(7/22) = H

7/2 = H

The answer is 7 halves inches.

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Examples of Algebraic Multigrid Method (AMG) applied to linear systems include solving partial differential equations (PDEs) such as Poisson's equation and the Helmholtz equation, as well as computational fluid dynamics (CFD) problems.

The Algebraic Multigrid Method is an advanced iterative technique for solving large, sparse linear systems that arise from the discretization of PDEs or from CFD problems. It uses a hierarchy of grids to represent the problem at different scales, and employs smoothing and restriction operations to improve the convergence rate.

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Billy's health score go down  by 75% if his sugar resources are quadrupled

Let the amount of sugar in Billy's kitchen be denoted by S and the number of cookies he can bake be denoted by C. Let his health score be denoted by H. Then we have the following relationships:

C ∝ S (directly proportional)

C ∝ 1/H (inversely proportional)

Combining these two relationships, we get:

C ∝ S/H

If S is quadrupled, then C will also quadruple according to the first relationship. However, H will decrease by some percentage x according to the second relationship. To find x, we can use the fact that C is proportional to S/H:

C = k*S/H

where k is a constant of proportionality. If S is quadrupled, then C will also quadruple, so we have:

4C = k4S/H

C = kS/(H/4)

This tells us that if S is quadrupled, then C will be divided by H/4. In other words, C/H will be divided by 4. So, the percentage decrease in H can be found as follows:

C/H → (C/H)/4 = (S/H)/(4/k) → x = 100%*(1 - 1/4) = 75%

Therefore, if Billy's sugar resources are quadrupled, his health score will go down by 75%.

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Answer:

$71.75

Rounded : $72

Step-by-step explanation:

To budget for gasoline this month, you can calculate the average amount spent on gasoline over the past four months:

Average = (67 + 78 + 53 + 89) / 4 = amount you should budget (x)

Average = 287 / 4 = x

71.75 = x

(Answer Rounded if that’s what you need but you didn’t ask: $72)

Therefore, you should budget around $71.75 or $ 72 for gasoline this month, assuming your driving habits and gas prices remain relatively constant. However, keep in mind that unexpected changes in gas prices or driving habits may affect your actual spending.

[tex]\textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ h=96 \end{cases}\implies V=\cfrac{\pi r^2 (96)}{3} \\\\[-0.35em] ~\dotfill\\\\ \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~\hspace{9em}\stackrel{\textit{since we know both Volumes are equal}}{\cfrac{4\pi r^3}{3}~~ = ~~\cfrac{\pi r^2 (96)}{3}}[/tex]

[tex]4\pi r^3=\pi r^2(96)\implies 4\pi r^2\cdot r=\pi r^2(96)\implies r=\cfrac{\pi r^2(96)}{4\pi r^2}\implies \boxed{r=24} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{ \textit{\LARGE cone} }{\cfrac{\pi (24)^2(96)}{3}}\implies \stackrel{ \textit{\LARGE sphere} }{\cfrac{4\pi (24)^3}{3}}\implies18432\pi ~~ \approx ~~ \text{\LARGE 57905.84}~units^3[/tex]