All categories

3 years ago

Write which goes in each box plz help ASAP!!!

Mathematics

2 answers:

kupik [55]3 years ago

5 0

Answer:

Command Economy: China during the Cold War, North Korea during the Cold War.

Market Economy: Japan during the Cold War, South Korea during the Cold War, Taiwan during the Cold War

jeka943 years ago

3 0

Answer:

Command Economy:

China during the Cold War.

North Korea during the Cold War.

Market Economy:

Japan during the Cold War.

South Korea during the Cold War.

Taiwan during the Cold War.

<u>Mongolia during the Cold War</u><u>:</u>

As when I was taking History, Mongolia was not really a topic during the World War, but I believe that Mongolia is a market economy due to it's fairly bad relations with the USSR.

You might be interested in

Hazel has band practice every other day and volunteers every seven days. if she had a band practice and volunteered on oct 23, w

LuckyWell [14K]

Cjdfb hjkbftbbui yuftg

3 0

3 years ago

Pls help me calculate the inverse of y equals 2x over x-7

NeTakaya

Y = 2x/x-7

x = 2y/y-7

xy - 7x = 2y

-7x = 2y - xy

-7x = y(2-x)

y = -7x / 2-x

y = 7x / x-2

f-¹(x) = 7x / x-2

5 0

3 years ago

Is (-3,11) a solution to the system of equation shown below.

natita [175]

No, when you substitute it into the first equation for x and y it does not equal 36 this it's not a solution

(-3)-(3)(11)=36
-3-33=36
-36 =/ 36

8 0

3 years ago

Let P = 0.50.30.50.7 be the transition matrix for a Markov chain with two states. Let x0 = 0.50.5 be the initial state vector fo

pav-90 [236]

Answer:

Probability distribution vector = $\left(\begin{array}{c}0.375\\ 0.625 \end{array} \right)$

Step-By-Step Explanation

If $P=\left(\begin{array}{cc}0.5&0.3\\ 0.5&0.7 \end{array} \right)$ is the transition matrix for a Markov chain with two states.

$x_{0}=\left(\begin{array}{c}0.5\\ 0.5 \end{array} \right)$ be the initial state vector for the population.

$X_{1}=P x_{0}=\left(\begin{array}{cc}0.5&0.3\\ 0.5&0.7 \end{array} \right) \left(\begin{array}{c}0.5\\ 0.5 \end{array} \right) =\left(\begin{array}{c}0.4\\ 0.6 \end{array} \right)$

$X_{2}=P^{2} x_{0}=\left(\begin{array}{c}0.38\\ 0.62 \end{array} \right)$

$X_{3}=P^{3} x_{0}=\left(\begin{array}{c}0.38\\ 0.62 \end{array} \right)$

$X_{30}=P^{30} x_{0}=\left(\begin{array}{c}0.37499\\ 0.625 \end{array} \right)$

In the long run, the probability distribution vector Xm approaches the probability distribution vector $\left(\begin{array}{c}0.375\\ 0.625 \end{array} \right)$ .

This is called the steady-state (or limiting,) distribution vector.

4 0

2 years ago

Write the equation of the ellipse graphed below.

Lana71 [14]

Answer:

Step-by-step explanation:

This is a horizontally stretched ellipse. That means that the horizontal line which is represented by the x-axis is the longer one. The equation for this type of ellipse is

$\frac{(x-h)^2}{a^2} +\frac{(y-k)^2}{b^2} =1$

In an ellipse, the a value is ALWAYS bigger than the b value, and since the x-axis represents the longer axis, the a goes under the x-squared term.

To solve for the h, the k, the a, and the b, we simply have to do some counting. The center of the ellipse, the (h, k) our of equation, is sitting at (4, 6). Put them where they belong in the equation. The a axis (the horizontal one) is 8 units long, and the b axis (the vertical one) is 4 units long. Square both of them and put them where they belong in the equation:

$\frac{(x-4)^2}{64}+\frac{(y-6)^2}{16} =1$

There you go!

8 0

3 years ago