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3 years ago

Find the median of the following data set. 1 1/4,5/8 ,3/5 ,1/2 ,1 1/2, 1 3/4

Mathematics

1 answer:

pashok25 [27]3 years ago

8 0

Answer:

1. 7/8

Step-by-step explanation:

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Seven less than one-third of a number is 5

miskamm [114]

22 would be the answer. You reverse the equation so you multiply 3 and add 7 5x3 = 15 15+7= 22

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3 years ago

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Which answer choice

joja [24]

Answer: b

Step-by-step explanation: I did this on a test before

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2 years ago

An online furniture store sells chairs for $200 each and tables for 166o ach. Every

ladessa [460]

Answer:

15 tables

Step-by-step explanation:

Chairs = $200

Tables = $600

Maximum items = 39

Minimum sales = $13,000

If 20 chairs are sold, revenue is $4,000. At least $9,000 in additional revenue is required. To earn $9,000, a minimum of 15 tables would need to be sold, which is fine since a maximum of 19 tables is permitted to be sold.

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2 years ago

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A particle moving in a planar force field has a position vector x that satisfies x'=Ax. The 2×2 matrix A has eigenvalues 4 and 2

andrey2020 [161]

Answer:

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

Step-by-step explanation:

Consider the provided matrix.

$v_1=\begin{bmatrix}-3\\1 \end{bmatrix}$

$v_2=\begin{bmatrix}-1\\1 \end{bmatrix}$

$\lambda_1=4, \lambda_2=2$

The general solution of the equation $x'=Ax$

$x(t)=c_1v_1e^{\lambda_1t}+c_2v_2e^{\lambda_2t}$

Substitute the respective values we get:

$x(t)=c_1\begin{bmatrix}-3\\1 \end{bmatrix}e^{4t}+c_2\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-3c_1e^{4t}-c_2e^{2t}\\c_1e^{4t}+c_2e^{2t} \end{bmatrix}$

Substitute initial condition $x(0)=\begin{bmatrix}-6\\1 \end{bmatrix}$

$\begin{bmatrix}-3c_1-c_2\\c_1+c_2 \end{bmatrix}=\begin{bmatrix}-6\\1 \end{bmatrix}$

Reduce matrix to reduced row echelon form.

$\begin{bmatrix} 1& 0 & \frac{5}{2}\\ 0& 1 & \frac{-3}{2}\end{bmatrix}$

Therefore, $c_1=2.5,c_2=1.5$

Thus, the general solution of the equation $x'=Ax$

$x(t)=2.5\begin{bmatrix}-3\\1\end{bmatrix}e^{4t}-1.5\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

6 0

2 years ago

How do I find the endpoint of a segment when given the first endpoint (-6, 5) and the midpoint(-1.5, 1)?

Anika [276]

Answer:

(3, -3)

Step-by-step explanation:

Use the midpoint formula.

x = (x₁ + x₂) / 2

-1.5 = (-6 + x) / 2

-3 = -6 + x

x = 3

y = (y₁ + y₂) / 2

1 = (5 + y) / 2

2 = 5 + y

y = -3

The other endpoint is (3, -3).

Graph: desmos.com/calculator/aze920ns2h

3 0

3 years ago