Round the number 45 to one significant figure

Answer:

If the sales 2 years ago were x, the sales last year were 1.12x and this year's sales were 1.12 * (1.12x). We can write 1.12 * (1.12x) = 81427 so that means x = $64913.

The critical value for the r-test is 1.796.

Using the t-distribution table, we need to find the p-value interval for a two-tailed test with n=13 and α = 0.0095.

In the t-distribution table with degrees of freedom (df) = n - 1 = 13 - 1 = 12 and level of significance α = 0.0095, we find that the t-value is approximately equal to ±2.718 (rounded to three decimal places).

Therefore, the P-value interval for a two-tailed test with n=13 and α = 0.0095 is:0.0095 < P-value < 0.9905

To find the critical value(s) for the r test for a right-tailed test with α = 0.05 and df = n - 2, we use the t-distribution table.

For a right-tailed test with α = 0.05 and df = n - 2 = 13 - 2 = 11, the critical t-value is approximately equal to 1.796 (rounded to three decimal places).

Hence, the critical value for the r test is 1.796.

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Answer and Step-by-step explanation:

The answer is D.

The rate of the function will be 3.5, while the initial amount is 15, so you do y - 3.5x + 15.

#teamtrees #PAW (Plant And Water)

Therefore , the solution of the given problem of linear equation comes out to be the linear equation of the image is y = 3x + 22.

The formula y = mx+b is used to produce a simple regression curve. The slope is B, and the y-intercept is m. The preceding line indicates independent components, yet is commonly referred to as a "math equation combining many variables". In bivariate linear equations, there are just two variables. Application problems involving linear equations have no known solutions. Y=mx+b.

Here,

Let's first find the coordinates of two points on the original line. We can choose any two points, but for simplicity, let's choose (0,4) and (1,7), which lie on the line y=3x+4.

The distance between these two points is:

d = √((1-0)² + (7-4)²) = √(10)

To find the new coordinates of these points after dilation, we multiply the distance by 4 and use the center of dilation (1,1) as the reference point:

New coordinates of (0,4):

x' = 1 + 4(0-1) = -3

y' = 1 + 4(4-1) = 13

New coordinates of (1,7):

x' = 1 + 4(1-1) = 1

y' = 1 + 4(7-1) = 25

So the image of the line y=3x+4 after dilation by scale factor 4 centered at (1,1) is the line passing through the points (-3,13) and (1,25). We can find the equation of this line by finding the slope and the y-intercept.

Slope:

m = (y2-y1)/(x2-x1) = (25-13)/(1-(-3)) = 12/4 = 3

Y-intercept:

y = mx + b

13 = 3(-3) + b

b = 22

Therefore, the linear equation of the image is y = 3x + 22.

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What? What do you mean?

Answer:

\( = { \tt{ \frac{120}{10} }} \\ \\ = { \tt{12 \: steps \: per \: minute}}\)

Answer:

120/10 = 12 steps per minute

Step-by-step explanation:

(thats 3 steps every 15 seconds....i think he walked instead of running)

Answer:

True

Step-by-step explanation:

When displaying any sort of data, it is important to make the table or chart as easy to understand and read as possible without compromising the data. In this case, it is simpler to understand the pie chart if we use as few slices as possible that still makes sense for displaying the data set.

Answer:

The cost of 100 cans of tennis balls is $250

The cost of 120 cans of tennis balls is $300.

The cost of 250 cans of tennis balls is $625

Step-by-step explanation:

0/12 = 2.5 per can

100*2.5 = 250

120*2.5 = 300

250*2.5= 625

Answer:

b because y moves same amount per x

The table that represent a linear function is option c, as it showing constant progress.

Linear Functions. In Mathematics, a linear function is defined as a function that has either one or two variables without exponents. It is a function that graphs to the straight line.

here, we have,

It is appealing because it is straightforward and manageable mathematically. It has a lot of crucial applications.

The graph of a linear function is a straight line.

The following is the format of a linear function.

a + bx = y = f(x)

One independent variable and one dependent variable make up a linear function. X and Y are the independent and dependent variables, respectively.

The constant term a, also known as the y intercept. When x = 0, it represents the value of the dependent variable.

B is the independent variable's coefficient. It provides the dependent variable's rate of change and is also referred to as the slope.

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Answer:.

Step-by-step explanation:

The approximate percentage of 1-mile long roadways with potholes numbering between 41 and 61, using the empirical rule, is 81.5%.

The empirical rule, also known as the 68-95-99.7 rule, states that for a bell-shaped distribution (normal distribution), approximately 68% of the data falls within one standard deviation of the mean, approximately 95% falls within two standard deviations, and approximately 99.7% falls within three standard deviations.

In this case, the mean of the distribution is 49, and the standard deviation is 4. To find the percentage of 1-mile long roadways with potholes numbering between 41 and 61, we need to calculate the percentage of data within one standard deviation of the mean.

Since the range from 41 to 61 is within one standard deviation of the mean (49 ± 4), we can apply the empirical rule to estimate the percentage. According to the rule, approximately 68% of the data falls within this range.

Therefore, the approximate percentage of 1-mile long roadways with potholes numbering between 41 and 61 is 68%.

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Complete Question:

online homework manager Course Messages Forums Calendar Gradebook Home > MAT120 43550 Spring2022 > Assessment Homework Week 7 Score: 14/32 11/16 answered X Question 6 < > Score on last try: 0 of 2 pts. See Details for more. > Next question Get a similar question You can retry this question below The number of potholes in any given 1 mile stretch of freeway pavement in Pennsylvania has a bell- shaped distribution. This distribution has a mean of 49 and a standard deviation of 4. Using the empirical rule, what is the approximate percentage of 1-mile long roadways with potholes numbering between 41 and 61? Do not enter the percent symbol. ans = 81.5 % Question Help: Post to forum Calculator Submit Question ! % & 5 B tab caps Hock A N 2 W S X #3 E D C $ 54 R T G 6 Y H 7 00 * 8 J K 1

Answer:

13 cm

Step-by-step explanation:

Add up all of the sides and you should get 13.

3 + 3 + 3.5 + 3.5 = 13

Answer:

71

Step-by-step explanation:

2n-1=141

2n=142

n=71

u in the plane in set of real numbers rℝcubed3 spanned by the columns, then the one solution matrix is,   \(\left[\begin{array}{ccc}x_{1} &\\x_{2} &\\\end{array}\right]\) = \(\left[\begin{array}{ccc}2&\\6.66&\\\end{array}\right]\)

Given data,

Let u [-16 18 10] and A [2 -3 1 -4 5 1]

u in the plane in set of real numbers rℝcubed3 spanned by the columns of A,

So,

We can write,

u = \(\left[\begin{array}{ccc}-16&\\18&\\10&\end{array}\right]\) and A = \(\left[\begin{array}{ccc}2&-3\\1&-4\\5&1\end{array}\right]\)

Then,

AX = b

w = \(\left[\begin{array}{ccc}2&-3&-16\\1&-4&18\\5&1&10\end{array}\right]\)

A = \(\left[\begin{array}{ccc}2&-3\\1&-4\\5&1\end{array}\right]\)

null A = \(\left[\begin{array}{ccc}x_{1} &\\x_{2} &\\\end{array}\right]\)

\(\left[\begin{array}{ccc}2&-3\\1&-4\\5&1\end{array}\right]\)    \(\left[\begin{array}{ccc}x_{1} &\\x_{2} &\\\end{array}\right]\)= \(\left[\begin{array}{ccc}0&\\0&\\0&\end{array}\right]\)

2\(x_{1}\) - 3\(x_{2}\) = 0

\(x_{1}\) - 4\(x_{2}\) = 0

5\(x_{1}\) + 1\(x_{2}\) = 0

We can solve the equations,

2\(x_{1}\) - 3\(x_{2}\) = 0

\(x_{1}\) - 4\(x_{2}\) = 0

-----------------------

\(x_{1}\) - 7\(x_{2}\) = 0

\(x_{1}\) = 7\(x_{2}\)

We can substitute \(x_{1}\) values,

2\(x_{1}\) - 3\(x_{2}\) = 0

2*7\(x_{2}\) - 3\(x_{2}\) = 0

\(14x_{2}\) - 3\(x_{2}\) = 0

11\(x_{2}\) = 0

\(x_{2}\) = 0

\(\left[\begin{array}{ccc}2&-3\\1&-4\\5&1\end{array}\right]\)    \(\left[\begin{array}{ccc}x_{1} &\\x_{2} &\\\end{array}\right]\)= = \(\left[\begin{array}{ccc}-16&\\18&\\10&\end{array}\right]\)

2\(x_{1}\) - 3\(x_{2}\) = -16

\(x_{1}\) - 4\(x_{2}\) = 18

5\(x_{1}\) + 1\(x_{2}\) = 10

We can solve the matrix equations,

2\(x_{1}\) - 3\(x_{2}\) = -16

\(x_{1}\) - 4\(x_{2}\) = 18

--------------------------

\(x_{1}\) - 7\(x_{2}\) = 2

\(x_{1}\) = 2 + 7\(x_{2}\)

\(x_{1}\) = 2 + 7(0)

\(x_{1}\) = 2

2\(x_{1}\) - 3\(x_{2}\) = -16

2*2 - 3\(x_{2}\) = -16

4 - 3\(x_{2}\) = -16

- 3\(x_{2}\) = -16 - 4

- 3\(x_{2}\) = -20

3\(x_{2}\) = 20

\(x_{2}\) = 20/3

\(x_{2}\) = 6.66

Therefore,

u in the plane in set of real numbers rℝcubed3 spanned by the columns, then the one solution matrix is,   \(\left[\begin{array}{ccc}x_{1} &\\x_{2} &\\\end{array}\right]\) = \(\left[\begin{array}{ccc}2&\\6.66&\\\end{array}\right]\)

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Answer:

x>355

Step-by-step explanation:

Answer:

x>355

...

...

...

...

...

1. Supplementary

2. Complementary

3. Adjacent angles

4. Vertically opposite angles

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The bottom is moving at a rate of \(4\sqrt{3}\) ft/s along the ground when the botton of the ladder is at a distance of 6 ft from the wall.

Here we have been mentioned that a 12 feet ladder is leaning against the wall which is represened by AC in the below given triangle.

We have AB = y = distance of top of the ladder from the wall

BC = x = distance of bottom of the ladder from the wall

AC = z = height of the ladder = 12 ft (given)

The top of the ladder AB is sliding down at a rate of 4 \(ft/s\)

∴ \(\frac{dy}{dt}\) = - 4 ft/s (negative sign indicates that the top is sliding downwards)

Let the rate at which the bottom of the ladder is sliding be \(\frac{dx}{dt}\)

Now in the right angled triangle \(ABC\) by using Pythagoras theorem we have,

\(AB^{2} + BC^{2} =\)\(AC^{2}\)

\(y^{2} +x^{2} = z^{2}\)    (equation 1)

When x = 6ft and z = 12 ft,

\(y^{2} +6^{2} = 12^{2}\)

⇒ \(y^{2} = 144 - 36\)

⇒ \(y^{2} = 108\)

⇒ y = √108

⇒ y = 6 √3 ft

Differentiating equation 1 with respect to \(time\) we get,

\(\frac{d}{dt} (y^{2} +x^{2} ) = \frac{d}{dt} (z^{2})\)

⇒ 2y\(\frac{dy}{dt}\) + 2x\(\frac{dx}{dt}\) = 2z\(\frac{dz}{dt}\)

Putting the values of dz/dt = 0 (as the height is constant) , dy/dt = - 4 ft/s , y = 6 √3 ft, x = 6ft and z = 12ft we have,

∴ 2(6√3) (-4) + 2(6)\(\frac{dx}{dt}\) = 2(12)(0)

⇒ -48√3 + 12\(\frac{dx}{dt}\) = 0

⇒ 12\(\frac{dx}{dt}\) = 48√3

⇒ \(\frac{dx}{dt} = \frac{48\sqrt{3} }{12}\)

⇒ \(\frac{dx}{dt} = 4\sqrt{3}\) ft/s

Hence the rate at which the bottom of the ladder is sliding is \(\frac{dx}{dt} = 4\sqrt{3}\) ft\s

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Answer:

8

Step-by-step explanation:

\( \frac{80}{100} \times 10 = 8\)

The concept of consecutive integers is explained as follows:

Consecutive numbers, or consecutive integers, are integers that follow each other in order. The difference between any two consecutive numbers is always 1. For example, the consecutive numbers starting from 1 would be 1, 2, 3, 4, 5, and so on. Similarly, the consecutive numbers starting from -3 would be -3, -2, -1, 0, 1, 2, and so on.

It is important to note that if we have a consecutive sequence of integers, one number will be even, and the next number will be odd. This is because the parity (evenness or oddness) alternates as we move through consecutive integers.

Furthermore, the sum of two consecutive numbers (and, in fact, the sum of any number of consecutive numbers) is always an odd number. This is because when we add an even number to an odd number, the result is always an odd number.

To generate a sequence of consecutive integers, we can start with any integer n and then use n, n+1, n+2, and so on to obtain consecutive integers. For example, if n is an integer, then n, n+1, and n+2 will be consecutive integers.

Here are some examples of consecutive integers:

- Starting from 1: 1, 2, 3, 4, 5, ...

- Starting from -3: -3, -2, -1, 0, 1, 2, ...

- Starting from 1004: 1004, 1005, 1006, 1007, ...

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Answer:

s<30

Step-by-step explanation:

s is the number of students. the staement says the number of students is less than 30

The expression for \(v_0(t)\) is \(v_0(t) = 4 + 2e^{(-t)}\). In the voltage output \(v_0(t)\) in the circuit is given by \(v_0(t) = 4 + 2e^{(-t)}\) by using Laplace Transform.

The voltage output \(v_0(t)\) in the circuit can be found using the Laplace Transform method. To apply the Laplace Transform, we need to convert the circuit into the Laplace domain by representing the elements in terms of their Laplace domain equivalents.

Given:

\(vs(t) = 4e^{(-2tu(t))\) - The input voltage

i(0) = 1 - Initial current through the inductor

\(v_0(0) = 2\) - Initial voltage across the capacitor

R = 2Ω - Resistance in the circuit

The Laplace Transform of the input voltage vs(t) is \(V_s(s)\), the Laplace Transform of the output voltage v0(t) is \(V_0(s)\), and the Laplace Transform of the current through the inductor i(t) is I(s).

To solve for v0(t), we can apply Kirchhoff's voltage law (KVL) to the circuit in the Laplace domain. The equation is as follows:

\(V_s(s) = I(s)R + sL*I(s) + V_0(s)\)

Substituting the given values, we have:

\(4/s + 2I(s) + V_0(s) = I(s)2 + s1/s*I(s) + 2/s\)

Rearranging the equation to solve for V_0(s):

\(V_0(s) = 4/s + 2I(s) - 2I(s) - s*I(s)/s + 2/s\\= 4/s + 2/s + 2I(s)/s - sI(s)/s\\= (6 + 2I(s) - sI(s))/s\)

To obtain v0(t), we need to take the inverse Laplace Transform of \(V_0(s)\) However, we don't have the expression for I(s). To find I(s), we can apply the initial conditions given:

Applying the initial condition for the current through the inductor, we have:

\(I(s) = sLi(0) + V_0(s)\\= 2s + V_0(s)\)

Substituting this back into the equation for  \(V_0(s)\):

\(V_0(s) = (6 + 2(2s + V_0(s)) - s(2s + V_0(s)))/s\)

Simplifying further:

\(V_0(s) = (6 + 4s + 2V_0(s) - 2s^2 - sV_0(s))/s\)

Rearranging the equation to solve for \(V_0(s)\):

\(V_0(s) + sV_0(s) = 6 + 4s - 2s^2\\V_0(s)(1 + s) = 6 + 4s - 2s^2\\V_0(s) = (6 + 4s - 2s^2)/(1 + s)\)\(i(0) = 1v_0(0) = 2\)

Now, we can take the inverse Laplace Transform of \(V_0\)(s) to obtain \(v_0(t)\):

\(v_0(t)\)  = Inverse Laplace Transform{\((6 + 4s - 2s^2)/(1 + s)\)}

The expression for \(v_0(t)\) is the inverse Laplace Transform of \((6 + 4s - 2s^2)/(1 + s)\). To find the inverse Laplace Transform of this expression, we need to decompose it into partial fractions.

The numerator of the expression is a quadratic polynomial, while the denominator is a linear polynomial. We can start by factoring the denominator:

1 + s = (1)(1 + s)

Now, we can express the expression as:

\((6 + 4s - 2s^2)/(1 + s) = A/(1) + B/(1 + s)\)

To determine the values of A and B, we can multiply both sides by the denominator and equate the coefficients of the like terms on both sides. After performing the algebraic manipulation, we get:

\(6 + 4s - 2s^2 = A(1 + s) + B(1)\)

Simplifying further:

\(6 + 4s - 2s^2 = A + As + B\)

Comparing the coefficients of the like terms, we have the following equations:

\(-2s^2: -2 = 0\)

4s: 4 = A

6: 6 = A + B

From the equation \(-2s^2 = 0\), we can determine that A = 4.

Substituting A = 4 into the equation 6 = A + B, we can solve for B:

6 = 4 + B

B = 2

Now that we have the values of A and B, we can express the expression as:

\((6 + 4s - 2s^2)/(1 + s) = 4/(1) + 2/(1 + s)\)

Taking the inverse Laplace Transform of each term separately, we get:

Inverse Laplace Transform(4/(1)) = 4

Inverse Laplace Transform\((2/(1 + s)) = 2e^{(-t)}\)

Therefore, the expression for \(v_0(t)\) is \(v_0(t) = 4 + 2e^{(-t)}\).

The voltage output \(v_0(t)\) in the circuit is given by \(v_0(t) = 4 + 2e^{(-t)}\).

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Question:Using Laplace Transform find \(v_o(t)\) in the circuit below

\(vs(t) = 4e^{(-2tu(t))\),\(i(0)=1,v_0(0)=2V.\)

The maximum value in the feasible region is P = 45.

We have,

To solve this problem, we need to graph the system of constraints and find the feasible region.

Then, we evaluate the objective function P = 4x + 3y at the vertices of the feasible region to determine the maximum value.

Let's start by graphing the constraints.

The constraint 6 ≤ x + y can be rewritten as y ≥ -x + 6.

We'll graph the line y = -x + 6 and shade the region above it.

The constraint x ≥ 3 represents a vertical line passing through x = 3. We'll shade the region to the right of this line.

The constraint y ≥ 1 represents a horizontal line passing through y = 1. We'll shade the region above this line.

Combining all the shaded regions will give us a feasible region.

Now, we need to evaluate the objective function P = 4x + 3y at the vertices of the feasible region to find the maximum value.

The vertices of the feasible region are the points where the shaded regions intersect.

By observing the graph, we can identify three vertices: (3, 1), (6, 7), and (13, -6).

Now, we substitute these vertices into the objective function to find the maximum value:

P(3, 1) = 4(3) + 3(1) = 12 + 3 = 15

P(6, 7) = 4(6) + 3(7) = 24 + 21 = 45

P(13, -6) = 4(13) + 3(-6) = 52 - 18 = 34

Among these values, the maximum value is P = 45.

Therefore,

The maximum value in the feasible region is P = 45.

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Answer:

Step-by-step explanation:

The probability that the coin is fair given that it lands heads is 0.3571.

Given that a coin is picked from the box and flipped, the probability of the coin being fair is 1/3.

The probability of the coin being biased towards heads and the coin being biased towards tails is 1/3.

Therefore, the probability that the coin is fair and lands heads is (1/3) x 0.5

= 0.1667.

The probability that the coin is biased towards heads and lands heads is (1/3) x 0.8

= 0.2667.

The probability that the coin is biased towards tails and lands heads is (1/3) x 0.1

= 0.0333.

Therefore, the total probability that the coin lands heads is 0.1667 + 0.2667 + 0.0333

= 0.4667.

Using Bayes' Theorem, the probability of the coin being fair given that it lands heads is:

P(fair|heads)

= P(heads|fair) * P(fair) / P(heads)

= 0.5 * 1/3 / 0.4667

= 0.3571.

Thus, the probability that the coin is fair given that it lands heads is 0.3571.

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a. Mason's error is that he flipped the inequality sign. The correct solution is x > 6.

b. This indicates that any value of x greater than 6 (but not equal to 6) will satisfy the inequality x > 6.

An inequality is a mathematical statement that compares two values or expressions using inequality symbols such as "<" (less than), ">" (greater than), "<=" (less than or equal to), ">=" (greater than or equal to), or "≠" (not equal to). Inequalities can be used to describe a range of values that a variable can take, rather than just a single value.

a. Mason's error is that he flipped the inequality sign. The correct solution is x > 6.

b. To show the correct solution on the number line, we start by marking 6 on the number line with an open circle (since x is not equal to 6). Then, we shade the area to the right of 6, because x is greater than 6. The graph should look like this:

0 1 2 3 4 5 6o 7 8 9

--------

shaded

This indicates that any value of x greater than 6 (but not equal to 6) will satisfy the inequality x > 6.

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Answer:

1/2 2/4 4/8

Step-by-step explanation:

The sum of angles in a hexagon is 720⁰. The missing interior angles in Noah's list is 110⁰  

The sum of interior angles of a triangle is equal to 180⁰. The sum of interior angles of a polygon is equal to:

sum of interior angles of a polygon = number of triangles x 180⁰

A hexagon consists of 4 triangles, hence, the sum of interior angles in a hexagon = 4 x 180⁰ = 720⁰

In general, we can find the sum of interior angles in a polygon using the formula:

(n - 2) x 180⁰

Where n = number of sides

In the given problem, let's denote the missing interior angle as q. The sum of interior angles are:

80⁰ + 85⁰ + 95⁰ + 105⁰ + 120⁰ + 125⁰ + q = 720⁰

610⁰ + q = 720⁰

q = 720⁰ - 610⁰ =  110⁰  

The picture in your question is missing. Most likely it was like on the attached picture.

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Answer:

a) 0 seconds.

b) The stunt diver is in the air for 2.81 seconds.

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

\(ax^{2} + bx + c, a\neq0\).

This polynomial has roots \(x_{1}, x_{2}\) such that \(ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})\), given by the following formulas:

\(x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}\)

\(x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}\)

\(\Delta = b^{2} - 4ac\)

Height of the diver after t seconds:

\(h(t) = -4.9t^2 + 12t + 5\)

a) How long is the stunt diver above 15 m?

Quadratic equation with \(a < 0\), so the parabola is concave down, and it will be above 15m between the two roots that we found for \(h(t) = 15\). So

\(h(t) = -4.9t^2 + 12t + 5\)

\(15 = -4.9t^2 + 12t + 5\)

\(-4.9t^2 + 12t - 10 = 0\)

Quadratic equation with \(a = -4.9, b = 12, c = -10\). Then

\(\Delta = 12^{2} - 4(-4.9)(-10) = -52\)

Negative \(\Delta\), which means that the stunt diver is never above 15m, so 0 seconds.

b) How long is the stunt diver in the air?

We have to find how long it takes for the diver to hit the ground, that is, t for which \(h(t) = 0\). So

\(h(t) = -4.9t^2 + 12t + 5\)

\(0 = -4.9t^2 + 12t + 5\)

\(-4.9t^2 + 12t + 5 = 0\)

Quadratic equation with \(a = -4.9, b = 12, c = 5\). Then

\(\Delta = 12^{2} - 4(-4.9)(5) = 242\)

\(x_{1} = \frac{-12 + \sqrt{242}}{2*(-4.9)} = -0.36\)

\(x_{2} = \frac{-12 - \sqrt{242}}{2*(4.9)} = 2.81\)

Time is a positive measure, so we take 2.81.

The stunt diver is in the air for 2.81 seconds.