question is in the picture, please explain how you got the answer with steps.

Answer:

41)  Yes, the relation is a function.

42)  The domain of the function is [-2, 4].

43)  The range of the function is [-1, 3].

Step-by-step explanation:

A relation is a set of ordered pairs where each input (x-value) is associated with one or more outputs (y-values).

A function is a special type of relation where each input (x-value) is associated with exactly one output (y-value).

We can determine if a graphed relation is a function by applying the Vertical Line Test. It states that if a vertical line intersects the graph at more than one point, then the relation does not pass the test and is not a valid function.

As the given graph passes the Vertical Line Test, the relation is a function.

[tex]\hrulefill[/tex]

The domain of a function is the set of all possible input values (x-values).

The range of a function is the set of all possible output values (y-values).

From inspection of the given graph, the continuous curve begins in quadrant II at point (-2, 1) and ends in quadrant IV at point (4, -1).

The endpoints of the graph are represented by closed circles, which means that the corresponding x and y values are included in the domain and range.

Therefore, the domain of the function is the x-values of the endpoints: [-2, 4].

The minimum point of the curve is endpoint (4, -1) and the maximum point is (0, 3). Therefore, the range of the function is the y-values of the minimum and maximum points: [-1, 3].

The box plot D is the correct box plot that displays the given data.

To find the correct box plot that displays the given data, we need to first determine the five-number summary of the data: minimum value, first quartile (Q1), median (Q2), third quartile (Q3), and maximum value.

The minimum value is 8, and the maximum value is 18. To find the quartiles, we need to first determine the median of the data set. Since there are 18 data points, the median is the average of the middle two values:

(11 + 11)/2 = 11

The median is 11.

To find Q1, we look at the median of the lower half of the data set:

(9 + 9)/2 = 9

Q1 is 9.

To find Q3, we look at the median of the upper half of the data set:

(16 + 17)/2 = 16.5

Q3 is 16.5.

Now that we have the five-number summary, we can compare the box plots to see which one is correct.

Box plot A has the correct minimum and maximum values, but the median (Q2) is too high and the whiskers are not the correct length.

Box plot B has the correct median and whisker length, but the minimum value is too low.

Box plot C has the correct minimum, median, and whisker length, but the maximum value is too high.

Box plot D has the correct minimum, median, and maximum values, as well as the correct whisker length. (d)

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The graph of the function f(x)=√(x + 1) is in the first option

A radical graph, also known as a square root graph, represents the graph of a square root function. A square root function is a mathematical function that takes the square root of the input variable.

The general form of a square root function is f(x) = √(ax + b) + c,

where a, b, and c are constants that determine the characteristics of the graph.

In the given function:

a = 1

b = 1

c = 0

The graph is plotted and attached

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Step-by-step explanation:

To find the coordinates of the focus of a parabola in the general form y = ax^2 + bx + c, you can use the formula (h, k) where h = -b/(2a) and k = (4ac - b^2)/(4a).

In the given equation y = -18x^2 - 2x - 4, we can identify that a = -18, b = -2, and c = -4. Plugging these values into the formulas, we get:

h = -(-2)/(2*(-18)) = 1/18

k = (4*(-18)*(-4) - (-2)^2)/(4*(-18)) = -71/9

Therefore, the focus of the parabola is (1/18, -71/9).

None of the given answer options match the coordinates of the focus calculated, so none of the options (A, B, C, D) are correct.

Given that the side of a square field is 52 m so, the area of the square field is 2704 m².

The side of a square field is given as 52 m.

Now, Let’s find the area of the square field using the given information.

As we know, area of a square can be calculated by using the formula:

A = a², where ‘a’ is the side of the square.

Now, by substituting the given value of ‘a’ in the given formula above we will get the area of the square field as,

A = (52)²

A = 2704 m²

Therefore, the area of the square field of given side i.e. 52m is 2704 m².

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Answer:

The values of b that satisfy the equation are:

b = (2√3 - 3) / 2

b = (-2√3 - 3) / 2

In other words, b can take the values (2√3 - 3) / 2 or (-2√3 - 3) / 2.

Step-by-step explanation:

To find the values of b that satisfy the equation 3(2b + 3)² = 36, we can solve for b by following these steps:

1. Divide both sides of the equation by 3:

  (2b + 3)² = 12

2. Take the square root of both sides:

  √[(2b + 3)²] = √12

  Simplifying further:

  2b + 3 = ±√12

3. Subtract 3 from both sides:

  2b = ±√12 - 3

4. Divide both sides by 2:

  b = (±√12 - 3) / 2

  Simplifying further:

  b = (±√4 * √3 - 3) / 2

  b = (±2√3 - 3) / 2

Therefore, the values of b that satisfy the equation are:

b = (2√3 - 3) / 2

b = (-2√3 - 3) / 2

In other words, b can take the values (2√3 - 3) / 2 or (-2√3 - 3) / 2.

Terri have to react 1.42 seconds before the volleyball hits the ground.

Quadratics can be defined as a polynomial equation of a second degree, which implies that it comprises a minimum of one term that is squared. It is also called quadratic equations. The general form of the quadratic equation is:

[tex]\text{ax}^2 + \text{bx} + \text{c} = 0[/tex]

Given data:

The height can be modeled by a quadratic equation.

[tex]h(t)=-16t^2+v_0t+h_0[/tex]

Where h is the height and t is the time.

[tex]h(t)=-16t^2+19.5t+4.5[/tex]

[tex]-16t^2+19.5t+4.5=0[/tex]

[tex]a = -16, b = 19.5, c = 4.5[/tex]

It looks like a quadratic equation. we can solve it by quadratic formula.

[tex]\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]

[tex]\rightarrow t=\dfrac{-19.5\pm\sqrt{(-19.5)^2-4\times(-16)(4.5)} }{2(-16)}[/tex]

[tex]\rightarrow t=\dfrac{-19.5\pm\sqrt{380.25+288} }{-32}[/tex]

[tex]\rightarrow t=\dfrac{-19.5\pm25.851 }{-32}[/tex]

[tex]\rightarrow t=\dfrac{-19.5-25.851 }{-32}, \ t=\dfrac{-19.5+25.851 }{-32}[/tex]

[tex]\rightarrow t=1.42, \ t=-0.20[/tex]

Time cannot be in negative. So neglect t = –0.235.

Hence, Terri have to react 1.42 seconds before the volleyball hits the ground.

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Using the physics concept of projectile motion and inputting the given values into the appropriate equation, we can determine the time it takes for the volleyball to hit the ground after being served

This question is a classic use of physics, more specifically, the concept of projectile motion. Here, the volleyball can be conceived as a projectile. When Patricia serves the ball upward, the ball will first ascend and then descend due to gravity.

Let's use the following equation which is a version of kinematic equations to solve this problem, adjusting for the fact that we're dealing with an initial height of 4.5 ft and an ending height of 0 ft (when the ball hits the ground). The equation y = yo + vot - 0.5gt² , where:

Setting y=0, yo=4.5 feet, vo=19.5 feet/second, and g=32.2 feet/second², and plug these values into the equation, we'll get a quadratic equation in the form of 0 = 4.5 + 19.5t - 16.1t². Solve that equation for t to find the time it takes for the ball to hit the ground.

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In the given question, it is stated that a premium candy manufacturer makes chocolate candies that, when finished, vary in color. The hue value of a randomly selected candy follows an approximately normal distribution with mean 30 and a standard deviation of 5. A quality inspector discards 12% of the candies due to unacceptable hues (which are equally likely to be too small or too large). So, the largest hue value that the inspector would find acceptable is 36.85

We are required to find out the largest hue value that the inspector would find acceptable. We are given that the mean value is 30, standard deviation is 5, and 12% of candies are discarded due to unacceptable hues. Now, we need to find out the largest hue value that the inspector would find acceptable.

To find the largest acceptable hue value we can use the Z score formula. Z = (X - μ) / σ

Now, substituting the values in the formula we have: Z = (X - 30) / 5

This value corresponds to the percentile of the distribution. We are required to find the largest hue value that the inspector would find acceptable and given that 12% of the candies are discarded due to unacceptable hues. So, the acceptable percentile would be 100% - 12% = 88% or 0.88

Now, using the z-score table or calculator, we can find the Z value corresponding to the 88th percentile. Z = 1.17

Now, we can use this Z score value to find the corresponding X value by using the Z-score formula and solving for X.1.17 = (X - 30) / 5

Solving for X,X = 30 + 5(1.17)X = 36.85. Therefore, the largest hue value that the inspector would find acceptable is 36.85 (rounded to two decimal places).

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The y-intercept of the line of best fit can be interpreted as the predicted test grade when no time is spent on homework, which in this case is approximately 72. However, it is important to consider the limitations and potential sources of error in any statistical analysis.

In statistics, linear regression is a commonly used statistical method for analyzing the relationship between two variables, such as time spent on homework and test grades. A line of best fit, also known as a regression line, is a line that summarizes the linear relationship between the variables. In this case, the line of best fit has an equation of y = 7.9 x + 72.
The y-intercept of the line is the point where the line intersects with the y-axis. It represents the value of y when x is equal to zero. In other words, it is the predicted test grade when no time is spent on homework. According to the given equation, the y-intercept is 72. This means that if a student spends no time on homework, they can still expect to receive a test grade of 72.
However, it is important to note that this interpretation assumes that the line of best fit is an accurate representation of the relationship between time spent on homework and test grades. Additionally, there may be other variables that influence test grades, such as innate ability, test-taking skills, or external factors like test anxiety or distractions during the exam.
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The simplified expression is [tex]-4x^2 + 3x + 1[/tex], which is a quadratic polynomial. Option D.

To simplify the expression [tex](8x^2 + 3x) - (12x^2 - 1)[/tex], we can distribute the negative sign to each term within the parentheses:

[tex]8x^2 + 3x - 12x^2 + 1[/tex]

Next, we can combine like terms by adding or subtracting coefficients of similar powers of x:

[tex](8x^2 - 12x^2) + 3x + 1[/tex]

Simplifying further, we have:

[tex]-4x^2 + 3x + 1[/tex]

The resulting polynomial [tex]-4x^2 + 3x + 1[/tex] is a quadratic polynomial since it has a highest power of x^2 (the exponent of x is 2), which is[tex]-4x^2.[/tex]Quadratic polynomials are polynomials of degree 2 and can be represented by a parabola when graphed.

In summary, the simplified expression [tex](8x^2 + 3x) - (12x^2 - 1)[/tex] simplifies to [tex]-4x^2 + 3x + 1[/tex] , which is a quadratic polynomial. So Option D is correct

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Note the complete question is

Out of the four children attending the party, with only two spots left, there are six different ways to select two children to fill those spots.

If there are four children and only two spots left at the party, we need to determine the number of combinations possible for selecting two children out of the four. To calculate this, we can use the concept of combinations from combinatorics.

Combinations refer to the selection of items from a larger set without considering the order. In this case, the order in which the children are selected does not matter; we only need to know which two children are chosen. The formula for combinations is given by:

C(n, r) = n! / (r! * (n - r)!)

where n is the total number of items (in this case, children) and r is the number of items we want to select (in this case, the two available spots at the party).

Using the formula, we can substitute n = 4 and r = 2:

C(4, 2) = 4! / (2! * (4 - 2)!)

Simplifying further:C(4, 2) = 4! / (2! * 2!)

Now, let's calculate the factorial terms:

4! = 4 * 3 * 2 * 1 = 24

2! = 2 * 1 = 2

Substituting the factorial terms:

C(4, 2) = 24 / (2 * 2)

Simplifying the denominator:

C(4, 2) = 24 / 4 = 6

Therefore, there are 6 different combinations possible for selecting two children out of the four to fill the two available spots at the party.

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Note the correct question is

4 childen go to a party but there is only 2 spots left. How many combinations  are there?

Answer:

1/2 of an hour ( I think )

Step-by-step explanation:

The second longest time that was recorded was 2/3/4 (whole/numerator/denominator) of a hour while the second shortest time was 2/1/4 of a hour. When you subtract 2/1/4 from 2/3/4 you end up with 2/4. Since 2/4 can be simplified to 1/2 you would say there is a 1/2 hour different between the second longest and second shortest time spent reading.

[tex]\boxed{A}\\\\ U=5(2n+22)+2\left( n+\cfrac{3}{2} \right) \\\\\\ U=10n+110+2n+3 \implies U=12n+113 \\\\[-0.35em] ~\dotfill\\\\ \boxed{B}\hspace{5em}\textit{in 2009, that's 9 years after 2000, n = 9}\\\\ U(9)=12(9)+113\implies U(9)=221 ~~ millions[/tex]

To find the center and radius of the circle represented by the inequality [tex]\displaystyle \sf x^{2} +y^{2} -12y-12\leq 0[/tex], we can complete the square for the y terms.

The inequality can be rewritten as:

[tex]\displaystyle \sf x^{2} +( y^{2} -12y) -12\leq 0[/tex]

To complete the square for the y terms, we need to add and subtract [tex]\displaystyle \sf ( 12/2) ^{2} =36[/tex] inside the parentheses:

[tex]\displaystyle \sf x^{2} +( y^{2} -12y+36) -36-12\leq 0[/tex]

Simplifying, we have:

[tex]\displaystyle \sf x^{2} +( y-6)^{2} -48\leq 0[/tex]

Now we can rewrite the inequality in the standard form of a circle equation:

[tex]\displaystyle \sf ( x-h)^{2} +( y-k)^{2} \leq r^{2}[/tex]

Comparing this with the obtained equation, we can identify the center and radius of the circle:

Center: [tex]\displaystyle \sf ( h,k)=( 0,6)[/tex]

Radius: [tex]\displaystyle \sf r=\sqrt{48}[/tex]

Therefore, the center of the circle is at [tex]\displaystyle \sf ( 0,6)[/tex], and its radius is [tex]\displaystyle \sf \sqrt{48}[/tex].

[tex]\huge{\mathfrak{\colorbox{black}{\textcolor{lime}{I\:hope\:this\:helps\:!\:\:}}}}[/tex]

♥️ [tex]\large{\underline{\textcolor{red}{\mathcal{SUMIT\:\:ROY\:\:(:\:\:}}}}[/tex]

The missing values in the quantitative reasoning given are : -2, 13 and 9

Given the rule :

We can deduce that :

For the left circle :

circle = -6 - (-4) = -6 + 4 = -2

For the right circle :

circle = 11 - (-2) = 11 + 2 = 13

For the left square :

square = 13 + (-4)

square = 13 -4 = 9

Therefore, the missing values are : -2, 13 and 9

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Work Shown:

21/35 = (7*3)/(7*5) = 3/5

The upper boundary of a 95% confidence interval for the population proportion of values that are greater than 24 is approximately 0.8455.

None of the provided options match the calculated value.

To find the upper boundary of a 95% confidence interval for the population proportion of values that are greater than 24, we need to follow these steps:

Calculate the sample proportion:

In the given sample, we need to determine the proportion of values that are greater than 24.

Counting the values greater than 24, we find that there are 11 such values out of a total of 16 values in the sample.

So, the sample proportion is 11/16 = 0.6875.

Calculate the standard error:

The standard error is calculated using the formula:

Standard Error [tex]= \sqrt{((p \times (1 - p)) / n)}[/tex]

where p is the sample proportion and n is the sample size.

In this case, p = 0.6875 and n = 16.

Plugging in the values, we get:

Standard Error [tex]=\sqrt{((0.6875 \times (1 - 0.6875)) }[/tex]/ 16) ≈ 0.0806.

Find the z-value for a 95% confidence level:

For a 95% confidence level, we need to find the z-value corresponding to a two-tailed test.

Looking up the z-value in the standard normal distribution table, we find that the z-value for a 95% confidence level is approximately 1.96.

Calculate the margin of error:

The margin of error is given by the product of the standard error and the z-value:

Margin of Error = z [tex]\times[/tex] Standard Error = 1.96 [tex]\times[/tex] 0.0806 ≈ 0.158.

Calculate the upper boundary of the confidence interval:

The upper boundary of the confidence interval is obtained by adding the margin of error to the sample proportion:

Upper Boundary = Sample Proportion + Margin of Error = 0.6875 + 0.158 = 0.8455.

Therefore, the upper boundary of a 95% confidence interval for the population proportion of values that are greater than 24 is approximately 0.8455.

None of the provided options match the calculated value.

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Answer:

7

Step-by-step explanation:

The line represented by the equation y = 7 is a horizontal line that passes through the y-axis at 7, so the y intercept of this line is 7

Stephen would save $45,238.90 by opting for a 20-year loan instead of a 30-year loan.

To calculate the interest saved, we need to find the total interest paid for each loan term.

For the 30-year loan:

Principal = $138,600 - (0.01 * $138,600) = $137,214

Interest Rate = %

Loan Term = 30 years

Using the table of the monthly payment of principal and interest per $1,000, we can determine the monthly payment per $1,000 financed for a 30-year loan. Let's assume this value is X.

Monthly Payment = $137,214 / 1,000 * X

Total Payments = Monthly Payment * 12 months * 30 years

Total Interest Paid = Total Payments - Principal

For the 20-year loan:

Principal = $138,600 - (0.01 * $138,600) = $137,214

Interest Rate = %

Loan Term = 20 years

Using the same table, let's assume the monthly payment per $1,000 financed for a 20-year loan is Y.

Monthly Payment = $137,214 / 1,000 * Y

Total Payments = Monthly Payment * 12 months * 20 years

Total Interest Paid = Total Payments - Principal

To find the interest saved, we calculate the difference in total interest paid for the two loan terms:

Interest Saved = Total Interest Paid (30 years) - Total Interest Paid (20 years)

By substituting the appropriate values, we can determine that Stephen would save $45,238.90 by opting for a 20-year loan instead of a 30-year loan.

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Answer:

To simplify the expression (a÷b)³×(b÷c)×(c÷a)³ when a=3, b=a², c=a³, we can substitute the given values and perform the calculations.

Substituting the values of a, b, and c:

a = 3

b = a² = 3² = 9

c = a³ = 3³ = 27

Now let's simplify the expression:

(a÷b)³×(b÷c)×(c÷a)³

(3÷9)³×(9÷27)×(27÷3)³

Simplifying each term:

(3÷9) = 1/3

(9÷27) = 1/3

(27÷3) = 9

Now we can substitute the simplified values back into the expression:

(1/3)³×(1/3)×9

Simplifying further:

(1/27)×(1/3)×9

1/9

Therefore, the simplified expression is 1/9.

The rocket is at a height of 1876.1 meters after 9 seconds,the velocity of the rocket after 9 seconds is 143.8 m/s and  the acceleration of the rocket after 9 seconds is -9.8 m/s².

To find the height of the rocket when it was initially launched, we can plug in t = 0 into the equation h(t) = -4.9t² + 232t + 185.

h(0) = -4.9(0)² + 232(0) + 185

     = 0 + 0 + 185

     = 185

Therefore, the rocket was initially launched at a height of 185 meters.

To find the height of the rocket after 9 seconds, we can plug in t = 9 into the equation h(t) = -4.9t² + 232t + 185.

h(9) = -4.9(9)² + 232(9) + 185

     = -4.9(81) + 2088 + 185

     = -396.9 + 2088 + 185

     = 1876.1

Therefore, the rocket is at a height of 1876.1 meters after 9 seconds.

To find the velocity of the rocket after 9 seconds, we can take the derivative of the height function h(t) with respect to time (t) and evaluate it at t = 9.

The velocity function v(t) is the derivative of h(t) with respect to t:

v(t) = dh/dt = d/dt(-4.9t² + 232t + 185)

       = -9.8t + 232

v(9) = -9.8(9) + 232

       = -88.2 + 232

       = 143.8

Therefore, the velocity of the rocket after 9 seconds is 143.8 m/s.

To find the acceleration of the rocket after 9 seconds, we can take the derivative of the velocity function v(t) with respect to time (t) and evaluate it at t = 9.

The acceleration function a(t) is the derivative of v(t) with respect to t:

a(t) = dv/dt = d/dt(-9.8t + 232)

       = -9.8

a(9) = -9.8

Therefore, the acceleration of the rocket after 9 seconds is -9.8 m/s².

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Rounding to three decimal places, the upper boundary of the 95% confidence interval for the population proportion of values greater than 24 is approximately G.0.961.Therefore, closest option is G: 0.925

To find the upper boundary of a 95% confidence interval for the population proportion of values greater than 24, we need to calculate the sample proportion and then use the appropriate distribution table.

First, we count the number of values in the sample that are greater than 24. From the given sample, we have 12 values greater than 24.

Next, we calculate the sample proportion by dividing the count of values greater than 24 by the total number of values in the sample:

Sample proportion = (Number of values greater than 24) / (Total number of values in the sample)

= 12 / 16

= 0.75

Now, we need to use the appropriate distribution table to find the upper boundary of the confidence interval. Since the sample size is relatively small (16), we can use the t-distribution table.

For a 95% confidence level with a two-tailed test, we need to find the critical value corresponding to an alpha of 0.025 (0.05 divided by 2) and degrees of freedom (df) of n-1 = 16-1 = 15.

Using the t-distribution table, the critical value for a 95% confidence level with 15 degrees of freedom is approximately 2.131.

Finally, we can calculate the upper boundary of the confidence interval:

Upper boundary = Sample proportion + (Critical value * Standard error)

= 0.75 + (2.131 * sqrt((0.75 * (1-0.75)) / 16))

≈ 0.75 + (2.131 * 0.099)

≈ 0.75 + 0.211

≈ 0.961

Rounding to three decimal places, the upper boundary of the 95% confidence interval for the population proportion of values greater than 24 is approximately 0.961. Therefore, closest option is G: 0.925

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8

In this pattern, we have 4 12 then 20.

We can see that the difference 4 and 12 is 8.

Since the difference between 12 and 20 is also 8, the common difference of the sequence is 8.

Answer:

only god knows

Step-by-step explanation:

because they didn't give us an answer on how many text messages anyone sent

answer: 13/2 or 6 1/2

step-by-step explanation:

hihi so basically your problem is making a solvable equation so w variables and stuff

heres my explanation !

the difference of a number and 2 is b-2

twice the difference of a number and 2 would be 2(b-2)

number 9 = 9 (duh lol)
so

2(b-2) = 9

2b - 4 = 9

2b = 13

improper: b = 13/2
mixed: b = 6 1/2

To calculate how many months it will take for the number of insect species to reach a specific value "n" based on the trend of a 4% decrease per month, we can use an exponential decay function.

The general form of an exponential decay function is given by:

f(x) = a * (1 - r)^x

In this case, "x" represents the number of months, and "r" represents the rate of decrease per month, which is 4% or 0.04 (since it is expressed as a decimal).

Since the initial number of insect species is 186, we can substitute "a" as 186 into the function:

f(x) = 186 * (1 - 0.04)^x

To find the number of months it will take for the number of insect species to reach a specific value "n," we can set the equation equal to "n" and solve for "x":

n = 186 * (1 - 0.04)^x

By rearranging the equation and using logarithms, we can solve for "x" and determine the number of months it will take for the number of insect species to reach the desired value.

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Answer:

[tex]\textsf{(a)} \quad \textsf{B.\;\;To find the assets in 2013, subtitute $\boxed{13}$ for $x$ and evaluate to find $A(x)$.}[/tex]

[tex]\textsf{In 2013 the assets are about \$ $\boxed{615.2}$\;billion.}[/tex]

[tex]\textsf{(b)} \quad \textsf{A.\;\;To find the assets in 2016, subtitute $\boxed{16}$ for $x$ and evaluate to find $A(x)$.}[/tex]

[tex]\textsf{In 2016 the assets are about \$ $\boxed{716.9}$\;billion.}[/tex]

[tex]\textsf{(c)} \quad \textsf{A.\;\;To find the assets in 2019, subtitute $\boxed{19}$ for $x$ and evaluate to find $A(x)$.}[/tex]

[tex]\textsf{In 2019 the assets are about \$ $\boxed{835.4}$\;billion.}[/tex]

Step-by-step explanation:

The given function approximating the assets (in billions of dollars) for a financial firm is:

[tex]A(x)=317e^{0.051x}[/tex]

Given that x = 7 corresponds to the year 2007 then:

[tex]\hrulefill[/tex]

B. To find the assets in 2013, substitute 13 for x and evaluate to find A(x).

[tex]\begin{aligned}A(13)&=317e^{0.051\cdot 13}\\&=317e^{0.663}\\&=317 \cdot 1.940605...\\&=615.17192...\\&=615.2\end{aligned}[/tex]

In 2013 the assets are about $615.2 billion.

[tex]\hrulefill[/tex]

A. To find the assets in 2016, substitute 16 for x and evaluate to find A(x).

[tex]\begin{aligned}A(16)&=317e^{0.051\cdot 16}\\&=317e^{0.816}\\&=317 \cdot 2.261435...\\&=716.8752...\\&=716.9\end{aligned}[/tex]

In 2016 the assets are about $716.9 billion.

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A. To find the assets in 2019, substitute 19 for x and evaluate to find A(x).

[tex]\begin{aligned}A(19)&=317e^{0.051\cdot 19}\\&=317e^{0.969}\\&=317 \cdot 2.635307...\\&=835.3925...\\&=835.4\end{aligned}[/tex]

In 2019 the assets are about $835.4 billion.