QR Factorization is a useful technique when the normal equations
for a least squares problem are
ill-conditioned. What does ill-conditioned
mean?
(Limit your answer to 25 words of less.)

Answers

Answer 1

Ill-conditioned means that the problem or system being considered is sensitive to small changes in the input or data, leading to unstable or inaccurate results.

When solving a least squares problem using the normal equations, ill-conditioning refers to situations where the matrix involved is nearly singular or has a high condition number.

This means that small perturbations or errors in the data can result in large changes in the computed solution.

In the context of QR factorization, if the normal equations for a least squares problem are ill-conditioned, it implies that the matrix being decomposed using QR factorization is close to being singular or has a high condition number. QR factorization can help in such cases by providing a more stable and accurate solution compared to directly solving the normal equations.

QR factorization decomposes a matrix into the product of an orthogonal matrix Q and an upper triangular matrix R. This factorization can help mitigate the effects of ill-conditioning by providing a numerically stable way to solve the least squares problem.

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Related Questions

A chocolate muffin recipe serves 12 people. An oatmeal raisin cookie recipe serves 36 people. A lemon cake recipe serves 16 people. Each recipe needs 2 eggs. The muffins need 3 cups of flour, while the cookies need 2 cups. The cake uses 2 lemons and the cookies use 1 cup of raisins and oatmeal. Each recipe needs 1 cup of sugar and milk. If we need to make cookies for 48 people, how much flour is needed?

Answers

To make cookies for 48 people, the recipe requires 6 cups of flour.

The oatmeal raisin cookie recipe serves 36 people and requires 2 eggs and 2 cups of flour. Since we need to make cookies for 48 people, we can calculate the amount of flour required as follows:

36 people → 2 cups of flour

1 person → (2 cups of flour) / (36 people) = (1/18) cups of flour

To make cookies for 48 people:

48 people × (1/18) cups of flour = 2.67 cups of flour

Therefore, to make cookies for 48 people, we need approximately 2.67 cups of flour.

Note: Since the result is a fraction of a cup, it is advisable to round up to the nearest whole number, so in this case, 3 cups of flour would be needed.

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HELP HELP PLEASE WRING ANSWERS OR LINKS WILL BE REPORTED ​

Answers

Answer:

$0.24 per orange

Step-by-step explanation:

$3.84 / 16 = 0.24

A cube with an edge of length s has a volume of 27 units.
What is the length of s?

Answers

Answer:

s = 3

Step-by-step explanation:

The volume formula for a cube is V = s^3, where “s” is the edge length. Since we know the volume and need to find “s,” we just do the inverse operation for an exponent, which is a radical. The cubed root of 27 is 3, so there’s your answer! Hope this is helpful & accurate. Best wishes.

I WILL GIVE BRAINLIEST!!!

consider the polynomial function q(x)=-2x^8+5x^6-3x^5+50

end behavior

Answers

Answer:

Use the degree and the leading coefficient to determine the behavior.

Falls to the left and falls to the right

Step-by-step explanation:








Use the Laplace transform to solve the initial-value problem x" + 4 = f(t), x(0)=0, x'(0) = 0, if t < 5 f(t) = t25. 3 sin(t-5) if t > 5.

Answers

By applying the initial conditions and inverse Laplace transforming, we can obtain the solution x(t) = (1 - cos(2t))u(t-5) + (3 sin(t-5))u(t-5), where u(t) is the unit step function. Therefore, the solution to the initial-value problem is x(t) = (1 - cos(2t))u(t-5) + (3 sin(t-5))u(t-5)

Taking the Laplace transform of the given differential equation x" + 4 = f(t), we obtain the algebraic equation in the Laplace domain: s^2X(s) + 4sX(s) + 4 = F(s), where X(s) is the Laplace transform of x(t) and F(s) is the Laplace transform of f(t).

Next, applying the initial conditions x(0) = 0 and x'(0) = 0, we get X(0) = 0 and sX(0) = 0. Substituting these initial conditions into the Laplace domain equation, we have s^2X(s) + 4sX(s) + 4 = F(s), with X(0) = 0 and sX(0) = 0.

Now, let's consider the Laplace transform of f(t) using the given piecewise function. For [tex]t < 5, f(t) = t^2/5, and for t > 5, f(t) = 3sin(t-5).[/tex]Taking the Laplace transform of f(t) in each interval, we have [tex]F(s) = (1/s^3) + (3/s^2) for t < 5 and F(s) = (3/s^2) * (1/(s^2+1)) for t > 5.[/tex]

Substituting these Laplace transforms into the equation[tex]s^2X(s) + 4sX(s) +[/tex]4 = F(s), we can solve for X(s). Simplifying, we obtain [tex]X(s) = (1/s^3) + (3/s^2) / (s^2 + 4s + 4) + (3/s^2) * (1/(s^2+1)).[/tex]

To find the inverse Laplace transform of X(s), we can split it into partial fractions and apply the inverse Laplace transform formula. The solution is x(t) = (1 - cos(2t))u(t-5) + (3 sin(t-5))u(t-5), where u(t) is the unit step function.

Therefore, the solution to the initial-value problem is x(t) = (1 - cos(2t))u(t-5) + (3 sin(t-5))u(t-5), where u(t) is the unit step function that ensures the piecewise function is activated at t = 5.

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A stone is dropped from the upper observation deck of a tower, 900 m above the ground. (Assume g=9.8 m/s2.) (a) Find the distance (in meters) of the stone above ground level at time t, h(t)= (b) How long does it take the stone to reach the ground? (Round your answer to two decimal places.) (c) With what velocity does it strike the ground? (Round your answer to one decimal place.) m/s (d) If the stone is thrown downward with a speed of 3 m/s, how long does it take to reach the ground? (Round your answer to two decimal places.)

Answers

After considering the given data we conclude that the distance between stone and ground level [tex]h(t) = -4.9t^2 + 900[/tex], time taken for the stone to reach the ground 18.22 seconds,the velocity with which it strikes the ground 178.76 m/s, if thrown with a down ward speed of 3m/s then the duration needed is 18.47 seconds.

A stone is dropped from the upper observation deck of a tower, 900 m above the ground. We can use the kinematic equations of motion to answer the following questions:
a) The distance of the stone above ground level at time t can be found using the equation:
[tex]h(t) = -1/2gt^2 + v_0t + h_0[/tex]
where g is the acceleration due to gravity (9.8 m/s²), v0 is the initial velocity (0 m/s), h0 is the initial height (900 m), and t is the time elapsed. Plugging in the values, we get:
[tex]h(t) = -4.9t^2 + 900[/tex]
b) To find how long it takes for the stone to reach the ground, we need to find the time when h(t) = 0:
[tex]-4.9t^2 + 900 = 0[/tex]
Solving for t, we get:
[tex]t = \sqrt(900/4.9) = 18.22 seconds[/tex]
Therefore, it takes the stone 18.22 seconds to reach the ground.
c) To find the velocity with which the stone strikes the ground, we can use the equation:
[tex]v(t) = -gt + v_0[/tex]
where v(t) is the velocity at time t. Plugging in the values, we get:
[tex]v(t) = -9.8(18.22) + 0 = -178.76 m/s[/tex]
Therefore, the stone strikes the ground with a velocity of 178.76 m/s.
d) If the stone is thrown downward with a speed of 3 m/s, we can use the same equation [tex]v(t) = -9.8(18.22) + 0 = -178.76 m/s[/tex] to find how long it takes to reach the ground. This time, [tex]v_0[/tex] is -3 m/s (since it is thrown downward) and [tex]h_0[/tex] is still 900 m. Plugging in the values, we get:
[tex]-4.9t^2 - 3t + 900 = 0[/tex]
Solving for t, we get:
t = 18.47 seconds
Therefore, it takes the stone 18.47 seconds to reach the ground when thrown downward with a speed of 3 m/s.
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Are there outliers in the set of data below? Hint: Use your formulas from the lesson

52, 58, 62, 66, 67, 68, 68, 70, 70, 72, 73, 74, 76, 84, 90



Question 1 options:

52 and 90


68 and 70


There are no outliers

pls help

Answers

I think 68 and 70 is an outliers

Gerald and Wheatly, Applied Numerical Analysis ▶6. If e¹.3 is approximated by Lagrangian interpolation from the values for eº = 1, el = 2.7183, and e² = 7.3891, what are the minimum and maximum estimates for the error? Compare to the actual error.

Answers

Lagrangian interpolation is used to approximate the value of e¹.3 using three known values: eº = 1, el = 2.7183, and e² = 7.3891. We can find the minimum and maximum estimates for the error.

To approximate e¹.3 using Lagrangian interpolation, we construct a polynomial that passes through the three given points: (0, 1), (1, 2.7183), and (2, 7.3891). We can then evaluate this polynomial at x = 1.3 to estimate the value of e¹.3.

Using Lagrangian interpolation, the polynomial P(x) is given by:

P(x) = 1 * L₀(x) + 2.7183 * L₁(x) + 7.3891 * L₂(x),

where L₀(x), L₁(x), and L₂(x) are the Lagrange basis polynomials associated with the three data points.

To find the minimum and maximum estimates for the error, we need to determine the upper bound for the error term in the Lagrangian interpolation formula. The error term is given by:

E(x) = f(x) - P(x),

where f(x) is the actual function we are approximating (in this case, e^x).

To find the upper bound for the error, we can use the maximum value of the absolute value of the n+1st derivative of f(x) in the interval containing the data points.

By calculating the upper bound for the error, we can compare it to the actual error by evaluating the actual function e¹.3 and subtracting the approximation P(1.3) obtained from Lagrangian interpolation.

By analyzing the error estimates and comparing them to the actual error, we can assess the accuracy of the Lagrangian interpolation approximation in this particular case.

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1. Find the equation of the parabola satisfying the given conditions.

Focus: (3,6); Directrix: x=−1

A. (x−1)2=8(y−6)

B. (y−6)2=8(x−1)

C. (x−1)2=−8(y−6)

D. (y−6)2=−8(x−1)


2. Find the equation of the parabola satisfying the given conditions.

Focus: (−6,3); Directrix: y=1

A. (y−2)2=4(x+6)

B. (x+6)2=4(y−2)

C. (x+6)2=−4(y−2)

D. (y−2)2=−4(x+6)


3. Find the equation of an ellipse that has foci at (−1,0) and (4,0), where the sum of the distances between each point on the ellipse and the two foci is 9.

A. (x+1)2+y2−−−−−−−−−−−√+(x−4)2+y2−−−−−−−−−−−√=9

B. (x−1)2+y2−−−−−−−−−−−√+(x+4)2+y2−−−−−−−−−−−√=9

C. (x+1)2+y2−−−−−−−−−−−√+(x−4)2+y2−−−−−−−−−−−√=81

D. (x−1)2+y2−−−−−−−−−−−√+(x+4)2+y2−−−−−−−−−−−√=81


4. Find the equation of a hyperbola that has foci at (−1,0) and (4,0), where the difference of the distances between each point on the ellipse and the two foci is 5.

A. (x+1)2+y2−−−−−−−−−−−√−(x−4)2+y2−−−−−−−−−−−√=25

B. (x−1)2+y2−−−−−−−−−−−√−(x+4)2+y2−−−−−−−−−−−√=5

C. (x−1)2+y2−−−−−−−−−−−√−(x+4)2+y2−−−−−−−−−−−√=25

D. (x+1)2+y2−−−−−−−−−−−√−(x−4)2+y2−−−−−−−−−−−√=5

Answers

Answer:

CABD

Step-by-step explanation:

The Taylor polynomial P, = (-10 * 9 about x = 0 is used to approximate the value of the function f at x=1 Find the value that verifies 5p (1)-(1)-500 n=1 n! Pa 1 384 1 384 0 ਤਕ OP PA 1 6144 PA 6144

Answers

The value that verifies 5p (1)-(1)-500 is -124.04

To approximate the value of the function f at x = 1 using the Taylor polynomial Pₙ = (-10)^n/ n! about x = 0, we need to find the value of P₅(1).

First, let's compute the derivatives of f(x) = e^x up to the fifth derivative:

f'(x) = e^x

f''(x) = e^x

f'''(x) = e^x

f''''(x) = e^x

f⁽⁵⁾(x) = e^x

Now, let's evaluate these derivatives at x = 0:

f(0) = e^0 = 1

f'(0) = e^0 = 1

f''(0) = e^0 = 1

f'''(0) = e^0 = 1

f''''(0) = e^0 = 1

f⁽⁵⁾(0) = e^0 = 1

Using these values, we can compute the Taylor polynomial P₅(x):

P₅(x) = f(0) + f'(0)(x - 0) + f''(0)(x - 0)²/2! + f'''(0)(x - 0)³/3! + f''''(0)(x - 0)⁴/4! + f⁽⁵⁾(0)(x - 0)⁵/5!

P₅(x) = 1 + 1x + 1x²/2! + 1x³/3! + 1x⁴/4! + 1x⁵/5!

Now, let's evaluate P₅(1):

P₅(1) = 1 + 1(1) + 1(1)²/2! + 1(1)³/3! + 1(1)⁴/4! + 1(1)⁵/5!

P₅(1) = 1 + 1 + 1/2 + 1/6 + 1/24 + 1/120

P₅(1) = 227/120

Therefore, the value that verifies 5P₅(1) - (1) - 500 is:

5P₅(1) - (1) - 500 = 5 * (227/120) - 1 - 500

= 1135/120 - 1 - 500

= 1135/120 - 120/120 - 60000/120

= (1135 - 120 - 60000)/120

= -59485/120

= -124.04

So, the value that verifies 5P₅(1) - (1) - 500 is approximately -124.04.

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NO LINKS!!!

A mechanic charges $45 to inspect your heater, plus $80 per hour to work on it. You owe the mechanic a total of $385. Write and solve an equation to find the amount of time (h) (in hours) the mechanic works on your heater.

What is the equation and and the answer?

Answers

Answer:

385=80x+45

x=4.25 hours

all I know is that it took him 4 hours and 32 minutes to work on it. 385-45=340/80=4.32

[LOOK AT THE PICTURE URGENT]

Answers

Answer:B

Step-by-step explanation:if it's a negative 3/4x then in be there is a -4 and there is a -7 so we can do -7 -4 but 4 is a negative so it turns into a positive so it's _

-7+4 and it become a smaller negative so -7+4 = -3. So it has to be B. Hope this gets brainliest

HELPPPPPPPPPPPPPPP!!!!!!!

Answers

i think it’s -2 and 10 but sorry if i am wrong btw the 10 is on the left and -2 on the right

Answer:

I feel like the answers are 10 and -2

Step-by-step explanation:

Hope this helps! Have a great day and good luck! :)

One angle measures 19° and another angle measures (4d − 9)°. If the angles are complementary, what is the value of d?

d = 7
d = 20
d = 25
d = 42.5

Answers

Answer:

d = 20

Step-by-step explanation:

90-19 = 71

(4d-9) = 71

4d = 80

d = 20

Answer:

d = 20

Step-by-step explanation:

Complementary angles are two angles that add up to 90°.

We know that one angle is 19° and the other is (4d − 9)°. So, we can set up the equation:

19 + (4d − 9) = 90.

Solving for d, we get:

19 + (4d − 9) = 90

19 + 4d − 9 = 90

4d + 10 = 90

4d = 80

d = 20

Therefore, the value of d is 20.

With which information can you construct more than one triangle?
A the measurements of two angles
B the measurements of two angles and the length of the included side
C the measurements of all the angles
D the lengths of two sides and the measurement of the included angle

Answers

B because side length and angle measure is given

Answer:

B and D

Explanation:

Solve the following differential equation by using Laplace transform method. y" +2y' +y = cos2t where y(0)=1 y'(O)=1.

Answers

The solution to the given differential equation with the initial conditions y(0) = 1 and y'(0) = 1 is:

[tex]y(t) = -e^{-t}/10 + (11/10)t*e^{-t} + (1/10)cos(2t) + (3/10)sin(2t)[/tex]

To solve the given differential equation using Laplace transform, we will apply the Laplace transform to both sides of the equation and then solve for the transformed variable.

Let's denote the Laplace transform of y(t) as Y(s).

Taking the Laplace transform of both sides of the differential equation, we get:

[tex]s^2Y(s) + 2sY(s) + Y(s) = (s^2 + 2s + 1)/(s^2 + 4)[/tex]

Now, let's solve for Y(s):

[tex]Y(s)(s^2 + 2s + 1) = (s^2 + 2s + 1)/(s^2 + 4)\\Y(s) = (s^2 + 2s + 1)/(s^2 + 4)(s^2 + 2s + 1)[/tex]

Factoring the denominator:

[tex]Y(s) = (s^2 + 2s + 1)/((s + 1)^2(s^2 + 4))[/tex]

Now, we need to decompose the fraction into partial fractions. Let's express the numerator in terms of A, B, C, and D:

[tex]s^2 + 2s + 1 = A/(s + 1) + B/(s + 1)^2 + (Cs + D)/(s^2 + 4)[/tex]

To find the values of A, B, C, and D, we can equate the numerators:

[tex]s^2 + 2s + 1 = A(s + 1)(s^2 + 4) + B(s^2 + 4) + (Cs + D)(s + 1)^2[/tex]

Expanding and equating coefficients:

[tex]s^2 + 2s + 1 = A(s^3 + 5s^2 + 4s) + B(s^2 + 4) + (C(s^2 + 2s + 1) + D(s + 1)^2)[/tex]

Simplifying:

[tex]s^2 + 2s + 1 = (A + C)s^3 + (5A + C + D)s^2 + (4A + 2C + D)s + (4A + D)[/tex]

Equating coefficients:

A + C = 0 (coefficient of [tex]s^3[/tex])

5A + C + D = 1 (coefficient of [tex]s^2)[/tex]

4A + 2C + D = 2 (coefficient of s)

4A + D = 1 (constant term)

Solving these equations simultaneously, we find A = -1/10, B = 11/10, C = 1/10, and D = 3/10.

Now, substituting these values back into Y(s):

[tex]Y(s) = (-1/10)/(s + 1) + (11/10)/(s + 1)^2 + (1/10)(s + 3)/(s^2 + 4) + (3/10)/(s^2 + 4)[/tex]

To find y(t), we need to take the inverse Laplace transform of Y(s). Fortunately, we can use a Laplace transform table to find the inverse Laplace transform of each term.

The inverse Laplace transform of (-1/10)/(s + 1) is [tex]-e^{-t}/10.[/tex]

The inverse Laplace transform of (11/10)/(s + 1)² is (11/10)t*[tex]e^{-t}.[/tex]

The inverse Laplace transform of (1/10)(s + 3)/(s² + 4) is (1/10)cos(2t).

The inverse Laplace transform of (3/10)/(s² + 4) is (3/10)sin(2t).

Combining these results, the solution y(t) is:

[tex]y(t) = -e^{-t}/10 + (11/10)t*e^{-t} + (1/10)cos(2t) + (3/10)sin(2t)[/tex]

Therefore, the solution to the given differential equation with the initial conditions y(0) = 1 and y'(0) = 1 is:

[tex]y(t) = -e^{-t}/10 + (11/10)t*e^{-t} + (1/10)cos(2t) + (3/10)sin(2t)[/tex]

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Assume that the playbook contains 16 passing plays and 12 running plays. The coach randomly selects 8 plays from the playbook. What is the probability that the coach selects at least 3 passing plays and at least 2 running plays?

Answers

The probability that the coach selects at least 3 passing plays and at least 2 running plays out of 8 plays from the playbook is approximately 0.4914 or 49.14%. This means there is a 49.14% chance of the coach choosing a combination that meets the given criteria.

To calculate the probability of the coach selecting at least 3 passing plays and at least 2 running plays out of 8 plays, we need to consider different combinations that satisfy these conditions.

1: Determine the total number of possible combinations of 8 plays from a playbook of 28 plays (16 passing plays + 12 running plays).

Total Combinations = C(28, 8) = 28! / (8! * (28-8)!) = 3,395,685

2: Calculate the number of combinations that have at least 3 passing plays and at least 2 running plays.

First, we calculate the number of combinations with exactly 3 passing plays and 2 running plays:

Number of Combinations with 3 passing and 2 running = C(16, 3) * C(12, 2) = (16! / (3! * (16-3)!) * (12! / (2! * (12-2)!) = 560 * 66 = 36,960

Next, we calculate the number of combinations with exactly 4 passing plays and 2 running plays:

Number of Combinations with 4 passing and 2 running = C(16, 4) * C(12, 2) = (16! / (4! * (16-4)!) * (12! / (2! * (12-2)!) = 1,820 * 66 = 120,120

Finally, we calculate the number of combinations with 5 passing plays and at least 2 running plays:

Number of Combinations with 5 passing and 2 or more running = C(16, 5) * (C(12, 2) + C(12, 3) + C(12, 4) + C(12, 5) + C(12, 6) + C(12, 7) + C(12, 8)) = (16! / (5! * (16-5)!) * (C(12, 2) + C(12, 3) + C(12, 4) + C(12, 5) + C(12, 6) + C(12, 7) + C(12, 8)) = 4368 * (66 + 220 + 495 + 792 + 924 + 792 + 495) = 4368 * 3786 = 16,530,048

Total Number of Combinations with at least 3 passing and 2 running plays = Number of Combinations with 3 passing and 2 running + Number of Combinations with 4 passing and 2 running + Number of Combinations with 5 passing and 2 or more running = 36,960 + 120,120 + 16,530,048 = 16,687,128

3: Calculate the probability.

Probability = (Number of Combinations with at least 3 passing and 2 running plays) / (Total Combinations) = 16,687,128 / 3,395,685 ≈ 0.4914

Therefore, the probability that the coach selects at least 3 passing plays and at least 2 running plays out of 8 plays is approximately 0.4914 or 49.14%.

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PLS HELP WITH WORKINGSSSSSSSS​

Answers

Answer:

60°

Step-by-step explanation:

90°-30°…

according to your question

AB
Round your answer to the nearest hundredth.
А
50°
6
12
B

Answers

Answer:

h = 7.832

Step-by-step explanation:

This is a right angled triangle so, taking 50 as reference angle,

hypotenuse = ?

perpendicular = 6

The ratio for p and h is given by

Sin 50 = p/h

Sin 50 = 6 /h

h = 6 / Sin 50

h = 7.832

The students in a club are selling flowerpots to raise money.Each flowerpot sells for $15.

Part A

Write an expression that represents The total amount of money, in dollars, The students raise from selling flowerpots.
Answer your expression in the box provided. Enter only your expression.Please hurry!

Answers

Answer:

y = 15x

Step-by-step explanation:

For every flower pot purchased (y), the quantitity of the price (x) will go up by $15.

Answer:

y = 15x

Step-by-step explanation:

The change in water level of a lake is modeled by a polynomial function, W(x). Describe how to find the x-intercepts of W(x) and how to construct a rough graph of W(x) so that the Parks Department can predict when there will be no change in the water level. You may create a sample polynomial of degree 3 or higher to use in your explanations.
(Dividing and Solving Polynomials)

Answers

First. Finding the x-intercepts of  

Let  be the change in water level. So to find the x-intercepts of this function we can use The Rational Zero Test that states:

To find the zeros of the polynomial:

We use the Trial-and-Error Method which states that a factor of the constant term:

can be a zero of a polynomial (the x-intercepts).

So let's use an example: Suppose you have the following polynomial:

where the constant term is . The possible zeros are the factors of this term, that is:

.

Thus:

From the foregoing, we can affirm that  are zeros of the polynomial.

Second. Construction a rough graph of  

Given that this is a polynomial, then the function is continuous. To graph it we set the roots on the coordinate system. We take the interval:

and compute  where  is a real number between -2 and -1. If , the curve start rising, if not, the curve start falling. For instance:

Therefore the curve start falling and it goes up and down until  and from this point it rises without a bound as shown in the figure below

What is the Quartile 1 for the Box & Whisker Plot below?


PLSS HELP

Answers

Answer:

17

Step-by-step explanation:

The lower quartile Q₁ is positioned at the left side of the box.

The value at the left is Q₁ is 17

suppose 40% of adults in the u.s. say they get their financial advice from family members. a random sample of 8 adults is selected. what is the probability at least 5 of the 8 say they get their financial advice from family members?

Answers

The probability that at least 5 of the 8 adults say they get their financial advice from family members can be calculated using binomial probability distribution.

Formula for binomial probability distribution is:P(X=k) = nCk * pk * (1-p)n-kwhere, P(X=k) is the probability of k successes in n independent trials, p is the probability of success in one trial, q=1-p is the probability of failure in one trial, nCk is the combination of k successes in n independent trials.

In this case, the probability of success is p=0.4 as 40% of adults in the US say they get their financial advice from family members.

Therefore, the probability of failure is q=1-0.4=0.6.So, P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)P(X=k) = nCk * pk * (1-p)n-kwhere, n=8, p=0.4, q=0.6For k = 5, nCk = 8C5P(X = 5) = 8C5 * (0.4)5 * (0.6)3= 0.2787

For k = 6, nCk = 8C6P(X = 6) = 8C6 * (0.4)6 * (0.6)2= 0.1960For k = 7, nCk = 8C7P(X = 7) = 8C7 * (0.4)7 * (0.6)1= 0.0575For k = 8, nCk = 8C8P(X = 8) = 8C8 * (0.4)8 * (0.6)0= 0.0030Therefore,P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.2787 + 0.1960 + 0.0575 + 0.0030 = 0.5352

Hence, the probability that at least 5 of the 8 adults say they get their financial advice from family members is 0.5352.

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Given information that 40% of adults in the US say they get their financial advice from family members. We have to find the probability of at least 5 of the 8 say they get their financial advice from family members.

Hence, the required probability is 0.7530.

The probability of getting financial advice from family members is 40%. Let X be the number of people out of 8, who get their financial advice from family members. Here, X follows a binomial distribution with parameters n = 8 and p = 0.4. The probability of getting atleast 5 people getting their financial advice from family members is

P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

Using binomial distribution formula, we get

P(X = x) = ${n\choose x}p^xq^{n-x}$

Where, n = 8, p = 0.4, q = 0.6

The probability of getting exactly 5 people out of 8 getting their financial advice from family members

P(X = 5) = ${8\choose 5} 0.4^5 (0.6)^{8-5}$

= 0.27869

The probability of getting exactly 6 people out of 8 getting their financial advice from family members

P(X = 6) = ${8\choose 6} 0.4^6 (0.6)^{8-6}$

= 0.29360

The probability of getting exactly 7 people out of 8 getting their financial advice from family members

P(X = 7) = ${8\choose 7} 0.4^7 (0.6)^{8-7}$

= 0.16493

The probability of getting exactly 8 people out of 8 getting their financial advice from family members

P(X = 8) = ${8\choose 8} 0.4^8 (0.6)^{8-8}$

= 0.01678

Therefore, the probability of getting at least 5 people getting their financial advice from family members is

P(X ≥ 5) = 0.27869 + 0.29360 + 0.16493 + 0.01678

= 0.7530

Hence, the required probability is 0.7530.

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Let M = {a E Ra > 1). Then M is a vector space under standard addition and scalar multiplication of real numbers. False True * Let W = {a + 2x + bx² € Pz: a, b E R} with the standard operations in P2. Which of the following statements is true? 1+xEW W is a subspace of P2. The above is true W is not a subspace of P.

Answers

The statement "W is a subspace of P2" is true because the set W, defined as W = {a + 2x + bx² ∈ P2: a, b ∈ R}, is a subspace of P2.

To determine if the set W = {a + 2x + bx² ∈ P2: a, b ∈ R} is a subspace of P2, we need to check if it satisfies three conditions: closure under addition, closure under scalar multiplication, and contains the zero vector.

Closure under addition: For any two polynomials p(x) = a + 2x + bx² and q(x) = c + 2x + dx² in W, their sum p(x) + q(x) = (a + c) + 4x + (b + d)x² is also a polynomial in W. This shows that W is closed under addition.

Closure under scalar multiplication: For any polynomial p(x) = a + 2x + bx² in W and any scalar c, the scalar multiple c * p(x) = ca + 2cx + cbx² is also a polynomial in W. Therefore, W is closed under scalar multiplication.

Contains the zero vector: The zero vector in P2 is the polynomial 0x² + 0x + 0, which can be expressed as a + 2x + bx² with a = 0 and b = 0. Since this polynomial satisfies the conditions of W, W contains the zero vector.

Since W satisfies all three conditions, it is a subspace of P2.

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pls help im pretty sure its easy i just forgot

Answers

the average can be calculated by adding the scores together and then dividing by the number of scores.

we can set up an equation:

let x = test score needed on next test

(72+72+80+x)/4 = 71

multiply both sides by 4

72+72+80+x = 284

add like terms

224+x=284

subtract 224 from both sides

x=60

she will need a 60 for her average to be 71

She needs to score a 60 to have a average score of 71,

72+72+80+60=284

284/4=71

what would be the distance between (-35,20) and (15,20).

Answers

Answer:

Well you just have to follow the formula to find the distance between 2 linear plots.

d= √(x2−x1)^2+(y2−y1)^2

Step-by-step explanation:

If you follow it the answer would be

D=50

Please answer correctly! I will mark you Brainliest!

Answers

Answer:

V=385 cubic units

Step-by-step explanation:

The volume of a rectangular prism is given by the formula [tex]V=lwh[/tex], where l is the length, w is the width, and h is the height. Our dimensions are 5, 7, and 11. So, we have to multiply them to find the volume.

5 × 7 × 11 = 385

Thus, the volume of the rectangular prism is 385 cubic units.

Solve the given system of differential equations by systematic elimination.
(D − 1)x+ (D² + 1)y = 1
(D² − 1)x+ (D + 1)y = 2

(x(t), y(t)) = (e^-t/2 [-5/3cos(√47/2)t - 125/3 sin(√47/2)t]+ 20/3 cos (3t) + 20/3 sin (3t)

Answers

Given system of differential equation is(D − 1)x+ (D² + 1)y = 1 ...

(i)(D² − 1)x+ (D + 1)y = 2 ...(ii)By using systematic elimination method, we have(D²+1)(D²−1)x+(D+1)(D−1)y=D²+1×1-(D+1)×1=0Simplifying the above equation, we get(D⁴-1)x=-(D-1)y...(iii)Applying D on both sides of (iii), we get D(D⁴-1)x=-(D-1 )DyD⁵x- Dx=-(Dy-y)or D⁵x+Dy=y ... (iv)Now applying D on (i), we get(D−1)Dx+(D²+1)Dy=0or D(D²+1)y=(1-D)x ...(v)Now applying D on (ii), we get(D²−1)Dx+(D+1)Dy=0or D(D+1)x=(1+D)y ...(vi)Now, substituting the value of x and y from equations (v) and (vi) in equation (iv), we getD⁵x+(1+D)Dx=(1-D)Dy D⁵x+(1+D)Dx=-(1-D)x ...(vii) Simplifying the above equation, we getD⁶x+2D⁴x+D²x+x=0or D²(D⁴+1)x+D²x=-x ...(viii)or D²(D⁴+2)x=-xor D⁴x+2x=-xor D⁴x=-3xNow using D on both sides, we get D⁵x=-3Dxor D⁶x=-3D²x

Now, substituting the value of D²x from equation (iii) in equation (i), we get(D-1)x+(D²+1)y=1 ...(i)⇒ (D-1)x+y=1 ...(ix)Now, substituting the value of D²x from equation (iii) in equation (ii), we get(D²-1)x+(D+1)y=2 ...(ii)⇒ -(D+1)x+y=0or (D+1)x-y=0 ...(x)From equation (ix) and (x), we have2x=1or x=1/2Now, substituting the value of x in equation (ix), we have D(1/2)+y=1or y=1-1/2=1/2Thus, the solution of the given system of differential equation is(x(t), y(t))=(e^(-t/2))[(-5/3)cos((sqrt(47)/2)t)-(125/3)sin((sqrt(47)/2)t)]+(20/3)cos(3t)+(20/3)sin(3t), (1/2)

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Ariana orders 4 large pizzas and 1 order of breadsticks The total for her order is $34.46. Emily orders 2 large pizzas and 1 order of breadsticks. $18.48 is the total for her order. Determine the order for 1 large pizza and 1 order of breadsticks?

Answers

4p + 1b = 34.46
2p + 1b = 18.48

Solve either equation for b

2p + b = 18.48
b = 18.48 - 2p

Substitute into first equation
4p + 18.48 - 2p = 34.46
2p + 18.48 = 34.46
2p = 15.98
p = $7.99

Plug back into either equation to find cost of breadsticks

2(7.99) + b = 18.48
15.98 + b = 18.48
b = $2.50

One pizza + breadstick
$7.99 + $2.50
$10.49 :)

hurry !!!! I need help​

Answers

Answer:

x = 66

Step-by-step explanation:

m<5 + m<6 = 180

2x + 48 = 180

2x = 132

x = 66

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