He finds that the mean is 79.0 and the median is 78.0


for these six quiz scores. If John earns a 100% on the


next exam, what will the new mean and median be?


The new mean will be 82.0 and the new median will


be 78.0.


O The new mean will be 82.0 and the new median will


be 80.0


The new mean will be 89.5 and the new median will


be 78.0.


The new mean will be 89.5 and the new median will


be 80.0

The most apt description of expected value is: The long-term average outcome of an experiment.

Expected value refers to the long-term average outcome of an experiment. It is a mathematical concept that represents the average value of a random variable over an infinite number of trials.

In other words, it is the sum of all possible outcomes of an experiment, weighted by their probability of occurring.

Expected value is a crucial concept in probability theory and statistics, as it helps in predicting the likely outcome of an experiment or event. It is often used in decision-making processes, such as in finance and insurance, to calculate the potential risks and rewards associated with different options.

The other options listed in the question are not correct definitions of expected value. The standard deviation is a measure of the variability or spread of a random variable, while the best result and median are specific values that may or may not be related to the expected value.

What you should expect to happen in an experiment every time may be related to the expected value, but it is not a precise definition.

Overall, expected value is a fundamental concept in probability theory and statistics that represents the long-term average outcome of an experiment or event.

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Answer:

3

Step-by-step explanation:

Since y_1(0) = 1, it satisfies the initial condition. However, y_2(0) = 0 does not satisfy the initial condition, as it should be y(0) = 1. Therefore, only y_1(t) is a solution of the initial value problem.

To verify that both y_1(t) = 1 - t and y_2(t) = -t^2/4 are solutions of the initial value problem, we first need to understand what the problem is. An initial value problem is a differential equation that includes an initial condition. In this case, we can assume that the initial condition is y(0) = 1.
Now, let's substitute both y_1(t) and y_2(t) into the differential equation and see if they satisfy the initial condition. The differential equation is not provided, but assuming it is y'(t) = -t/2, we have:
y_1'(t) = -1
y_2'(t) = -t/2
Substituting y_1(t) and y_2(t) into the differential equation gives:
y_1'(t) = -1

= -t/2 (when t = 2)
y_2'(t) = -t/2

= -t/2 (for all t)
Thus, both y_1(t) and y_2(t) satisfy the differential equation. Now, let's check if they satisfy the initial condition.
y_1(0) = 1 - 0

= 1
y_2(0) = -0^2/4

= 0
In conclusion, y_1(t) = 1 - t is the only solution that satisfies the differential equation and initial condition, while y_2(t) = -t^2/4 is not a solution since it does not satisfy the initial condition.

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Answer: D.

Step-by-step explanation:

You can start out with the form AX = B and solve for matrix X that would yield the answer

Then X = (A^-1)(B)

A = [1    -1]

      [1     1]

A^-1 = (1/(1 - (1)(-1)))*[1    1]

                             [-1   1]

which can be written as (1/2) * [1    1]

                                                  [-1   1]

B = [26]

     [6]

(A^-1)(B) = (1/2)*[1    1] [6]

                        [-1   1] [26]

= (1/2)*[32]

          [20]

= [16]

  [10]

Answer:

The initial value of the boat was $20,000

The percent decrease per year of the value of the boat is 10%.

The interval on which the value of the boat is decreasing while Harry has it is (0,10)

Step-by-step explanation:

It is given that the boat costs $20,000 in 2010 when x = 0. So, the initial value of the boat was $20,000.

Next, find the percent decrease per year of the value of the boat. Consider the general form of an exponential equation, y = a(b)x, where a is the initial value of the boat, b is the base of the exponent, x is the number of years after 2010, and y is the value of the boat, in dollars.

Consider the point (1 , 18,000) which lies on the graph of this situation. Substitute x = 1, y = 18,000, and a = 20,000 into the exponential equation and isolate b.

Recall that for exponential decay, b = 1 - r where r represents the decay rate. Substitute b = 0.9 into this equation and solve for r.

So, the decay rate is 0.1, and the percent decrease per year of the value of the boat is 10%.

Harry bought the boat in 2010, and plans on selling it after 10 years in 2020. Therefore, the interval on which the value of the boat is decreasing while Harry has it is [0, 10].

Answer:

Step-by-step explanation:

Answer:

13

Step-by-step explanation:

5 · x + 3 = 5 · 2 + 3 = 10 + 3 = 13

is good answer the x value and y value

Answer:

No, the inverse function does not pass the horizontal line test.

Step-by-step explanation:

Answer:

The answer is 11x - 1

Step-by-step explanation:

I did the distributive property and it should look like this 6x - 10 + 5x + 9

After that I just combined like terms and got 11x - 1

pls mark brainliest

Answer:

2(3x - 5) + 3 (2x + 3)

6x - 10 + 6x + 9

6x - 1 + 6x

12x -1

Step-by-step explanation:

Hope this helps!!

Answer:

True

Step-by-step explanation:

6^3, or 6x6x6, is equal to 216.

---

hope it helps

Answer:

true

Step-by-step explanation:

Answer:

D.

Step-by-step explanation:

We know that to find unit price, we divide the unit in the numerator by the quantity in the denominator.

For the first option, we must put $1.49 as the numerator and 8 as the denominator. It will then look like this:

For our second option, we must put $2.29 as the numerator, and 12 as the denominator. It will look like this:

For our third option, we must put $2.75 as the numerator, and 15 as the denominator. It will look like this:

For our final option, we must put $3.89 as the numerator, and 20 as the denominator. It will look like this:

Therefore, the last option is the best buy because $0.18625 > $0.18333.

9514 1404 393

Answer:

  a) profit has a typical curve: negative at low volume, and again at very high volume.

  b) $239,407

Step-by-step explanation:

a) The graph is negative below x=1.58, so the break-even point is 158 toys sold. The profit declines steeply above 533 toys sold, to again go negative for 774 toys sold. This sort of curve seems fairly typical.

__

b) The profit is a maximum of $239,407 when 533 toys are sold.

exact form

x=71/8

Decimal Form:

x=8.875

Mixed Number Form:

x=  8    7/8

Answer:

Step-by-step explanation:

8 7/8

The portfolio weights for stock A and stock B are approximately 55.69% and 44.31%, respectively.

To calculate the portfolio weights for stock A and stock B, we need to determine the total value of the portfolio and the proportion of each stock's value within that total.

The total value of the portfolio can be calculated as follows:

Total value = (Number of shares of stock A × Price per share of stock A) + (Number of shares of stock B × Price per share of stock B)

Total value = (135 × $84) + (110 × $82)

Total value = $11,340 + $9,020

Total value = $20,360

Now, we can calculate the portfolio weights for each stock:

Weight of stock A = (Value of stock A ÷ Total value) × 100%

Weight of stock A = (($84 × 135) ÷ $20,360) × 100%

Weight of stock A = $11,340 ÷ $20,360

Weight of stock A ≈ 0.5569 or 55.69%

Weight of stock B = (Value of stock B ÷ Total value) × 100%

Weight of stock B = (($82 × 110) ÷ $20,360) × 100%

Weight of stock B = $9,020 ÷ $20,360

Weight of stock B ≈ 0.4431 or 44.31%

Therefore, the portfolio weights for stock A and stock B are approximately 55.69% and 44.31%, respectively.

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Step-by-step explanation:

-Apply pesticides in the evening. Many pesticides are extremely toxic to honey Bees and other beneficial insects.

-Choose the appropriate formulation.

-Use less toxic, rapidly degradable pesticides.

-Alter application method.

-Establish apiaries in safe locations.

I hope I helped.

Answer: 28282

Step-by-step explanation:

I think

The conditions required for a random variable X to follow a Poisson process. The probability of success is the sme for aany two intervals of equal length. The probability of two or more successes in any sufficiently small subinterval is 0.

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The nine fundamental solutions to the differential equation are:

\(e^{(-3+i)t}, e^{(-3-i)t}, e^t, te^t,\) 1, t, t²/2!, t³/3!, \(t^4\)/4!, \(e^{(-5+i)t}, ~and ~e^{(-5-i)t}\)

We have,

The characteristic equation of the given differential equation is:

\((r^2 + 6r + 10)^2 \times r^2 (r - 1)^3 = 0\)

We can find the fundamental solutions by looking at the roots of the characteristic equation.

The roots can be categorized as follows:

Roots of multiplicity 2 = -3 + i and -3 - i

Roots of multiplicity 2 =  1

Root of multiplicity 1 =  0

Root of multiplicity 2 = -5 + i and -5 - i

For each of these roots, we need to find the corresponding fundamental solution.

For the roots (-3 + i) and (-3 - i), the corresponding fundamental solutions are:

\(e^{(-3+i)t}~ and~ e^{(-3-i)t}\)

For root 1, the corresponding fundamental solutions are:

\(e^t~and~te^t\)

For the root 0, the corresponding fundamental solutions are:

1, t, t²/2!, t³/3!, ..., \(t^8\)/8!

For the roots (-5 + i) and (-5 - i), the corresponding fundamental solutions are:

\(e^{(-5+i)t} ~and~e^{(-5-i)t}\)

Therefore,

The nine fundamental solutions to the differential equation are:

\(e^{(-3+i)t}, e^{(-3-i)t}, e^t, te^t,\) 1, t, t²/2!, t³/3!, \(t^4\)/4!, \(e^{(-5+i)t}, ~and ~e^{(-5-i)t}\)

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Answer: Row 3

Row 2

6 and 10

Step-by-step explanation: C B A

Suppose that a certain population obeys the logistic equation dy/dx= ry[1-(y/k)]

a) So if y0= K/3, then

dy/dx=  ry[1-(y/k)]

The above statement is equation 1.

Equation 2 is in the form of

dy/dx + P(t)y = yⁿQ(t)

which is a Bernoulli's differential equation with n=2. So 1- 2= -1

To solve it we put

Then dz/dx= -y⁻²(dy/dx)

This above statement is equation 3.

P(t)= -r, Q(t)= -(r/K)

and integrating factor is given by

IF= e ∫(1-n)P(t)dt= e ∫(-1)(-r)dt= ert

Then the solution is given by

z(IF)= ∫(1-n)Q(t)(IF)dt+C

If given y0= K/3, then

3/K= 1/K+C

Thus the solution to this part is

y⁻1ert= ert/K+2/K= 1/K(2+ert)

To find the time at which the initial population doubled that is 2K/3

Thus

2K/3= Kert/(2+ert)

For r=0.025, then r= 55.452

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In the language of modular arithmetic, we're given

\(x-3\equiv0\pmod7\implies x\equiv3\pmod7\)

\(y+3\equiv0\pmod7\implies y\equiv-3\equiv4\pmod7\)

Then x = 7a + 3 and y = 7b + 4 for integers a and b.

Substitute these into the quadratic expression and simplify:

\(x^2+xy+y^2+n\equiv0\pmod7\)

\((7a+3)^2+(7a+3)(7b+4)+(7b+4)^2+n\equiv0\pmod7\)

\(49a^2+42a+9+49ab+28a+21b+12+49b^2+56b+16+n \equiv 0\pmod7\)

\(37+n\equiv 0\pmod7\)

\(n\equiv-2\equiv5\pmod7\)

which means the smallest positive integer n we are looking for is 5.

The lines which are parallel are none.

The slope is the ratio of the vertical changes to the horizontal changes between two points of the line.

m = ( y₂ - y₁ ) / ( x₂ - x₁ )

where (x₁, y₁) and (x₂, y₂) are the two points that you are trying to find the slope between.

Given;

Coordinates of three lines

a;(1,5) and (-2,-4)

b;(3,2) and (1,-4)

c;(6,1) and (-4,2)

Now, slopes of the lines

a= -4-5/-2-1

=10/3

b=-4-2/1-3

=-3

c=2-1/-4-6

=1/-10

Therefore, by slopes of the line none of them are parallel.

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For n ≥ 459, Algorithm A is better than Algorithm B when B uses n√n operations.

To determine the value of n₀ for which Algorithm A is better than Algorithm B when B uses n√n operations, we need to find the point at which the number of operations for Algorithm A is less than the number of operations for Algorithm B.

Algorithm A: 10n log n operations

Algorithm B: n√n operations

Let's set up the inequality and solve for n₀:

10n log n < n√n

Dividing both sides by n gives:

10 log n < √n

Squaring both sides to eliminate the square root gives:

100 (log n)² < n

To solve this inequality, we can use trial and error or graph the functions to find the intersection point. After calculating, we find that n₀ is approximately 459. Therefore, For n ≥ 459, Algorithm A is better than Algorithm B when B uses n√n operations.

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R-1.3: For \($n \geq 14$\), Algorithm A is better than Algorithm B when B uses \($n^2$\) operations.

R-1.4: Algorithm A is always better than Algorithm B when B uses \($n\sqrt{n}$\) operations.

R-1.3:

Algorithm A: \($10n \log n$\) operations

Algorithm B: \($n^2$\) operations

We want to determine the value of \($n_0$\) such that Algorithm A is better than Algorithm B for \($n \geq n_0$\).

We need to compare the growth rates:

\($10n \log n < n^2$\)

\($10 \log n < n$\)

\($\log n < \frac{n}{10}$\)

To solve this inequality, we can plot the graphs of \($y = \log n$\) and \($y = \frac{n}{10}$\) and find the point of intersection.

By observing the graphs, we can see that the two functions intersect at \($n \approx 14$\). Therefore, for \($n \geq 14$\), Algorithm A is better than Algorithm B.

R-1.4:

Algorithm A: \($10n \log n$\) operations

Algorithm B: \($n\sqrt{n}$\) operations

We want to determine the value of \($n_0$\) such that Algorithm A is better than Algorithm B for \($n \geq n_0$\).

We need to compare the growth rates:

\($10n \log n < n\sqrt{n}$\)

\($10 \log n < \sqrt{n}$\)

\($(10 \log n)^2 < n$\)

\($100 \log^2 n < n$\)

To solve this inequality, we can use numerical methods or make an approximation. By observing the inequality, we can see that the left-hand side \($(100 \log^2 n)$\) grows much slower than the right-hand side \($(n)$\) for large values of \($n$\).

Therefore, we can approximate that:

\($100 \log^2 n < n$\)

For large values of \($n$\), the left-hand side is negligible compared to the right-hand side. Hence, for \($n \geq 1$\), Algorithm A is better than Algorithm B when B uses \($n\sqrt{n}$\) operations.

So, for R-1.4, the value of \($n_0$\) is 1, meaning Algorithm A is always better than Algorithm B when B uses \($n\sqrt{n}$\) operations.

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B: the relationship is proportional

The  worst-case time complexity to determine whether two particular vertices are adjacent in this graph is O(n).

In the worst-case scenario, we need to examine all the edges incident to both vertices to determine whether they are adjacent or not. Since each vertex has degree n/2, there are n/2 edges incident to each vertex. Therefore, in the worst case, we need to examine n/2 + n/2 = n edges.

In the adjacency list representation, we can find the list of adjacent vertices for a given vertex in O(n) time by scanning the adjacency list for that vertex. Therefore, to examine all the edges incident to both vertices, we need to scan the adjacency lists of both vertices, which takes O(n) time for each vertex.

Thus, the worst-case time complexity to determine whether two particular vertices are adjacent in this graph is O(n).

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Answer:

3.27

Step-by-step explanation:

Answer:

yea

Step-by-step explanation: