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Sidana [21]
4 years ago
10

Item 17 The cities Portland, Oregon; Tacoma, Washington; and Seattle, Washington, in respective order, lie approximately in a st

raight line. The distance from Portland to Tacoma is 120 miles. The distance from Portland to Seattle is 145 miles. Find the distance from Tacoma to Seattle.

Mathematics
1 answer:
muminat4 years ago
3 0

Answer:

Distance: 25 miles

Step-by-step explanation:

we have distances between cities as follows:

From Portland to Tacoma: 120 miles

from Portland to Seattle: 145 miles

120+x=145 ⇒ x = 145 - 120 = 25 miles

from Tacoma to Seattle : 25 miles (according graph)

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Conditional Distribution, Marginal Distribution, Joint Distribution. <br> What’s the difference?
lutik1710 [3]

Explanation:

Marginal distribution: This distribution gives the probability for each possible value of the Random variable ignoring other random variables. Basically, the values of other variables is not considered in the marginal distribution, they can be any value possible. For example, if you have two variables X and Y, the probability of X being equal to a value, lets say, 4, contemplates every possible scenario where X is equal to 4, independently of the value Y has taken. If you want the probability of a dice being a multiple of 3, you are interested that the dice is either 3 or 6, but you dont care if the dice is even or odd.

Conditional distribution: This distribution contrasts from the previous one in the sense that we are restricting the universe of events to specific condition for other variable, making a modification of our marginal results. If we know that throwing a dice will give us a result higher than 2, then to in order to calculate the probability of the dice being a multiple of 3 using that condition, we have two favourable cases (3 and 6) from 4 total possible results (3,4,5 and 6) discarding the impossible values (1 and 2) from this universe since they dont match the condition given (note that the restrictions given can also reduce the total of favourable cases).

The joint distribution calculates the probabilities for two different events (related to two different random variables) occuring simultaneously. If we want to calculate the joint probability of a dice being multiple of 3 and greater than 2 at the same time, our possible cases in this case are 3 and 6 from 6 possible results. We are not discarding 1 or 2 as possible results because we are not assuming, that the dice is greater than 2, that is another condition that we should met in the combination of events.

3 0
3 years ago
b)suppose only 1 in every 1,000 stars in the observable universe has a planetary system. how many planetary system are there? Fi
rosijanka [135]

Answer:

Please read below :)

Step-by-step explanation:

so you have 1 in a 1000, or 1/1000. But you're missing a part of the answer. If you post the whole answer I can give you the correct answer :)

5 0
3 years ago
Ashley wants to find out how much money she has in her piggy bank. She has x number of quarters and y number of dimes and says t
patriot [66]
Amount from dimes would be: 0.1y
Amount from quarters = 0.25x

So, total amount = 0.25x + 0.1y

So, she is not correct

Hope this helps!
5 0
3 years ago
The graph of this function is shifted 7 units right and shifted 5 units up.
mojhsa [17]
<h2><u>Graphing</u></h2>

<h3>The graph of this function is shifted 7 units right and shifted 5 units up.</h3>

<h3>f(x) = -2/3(x - 7)² - 5</h3>

  • True
  • False

<u>Answer:</u>

  • <u>False</u>

<u>Explanation:</u>

  • Why false? As I noticed, in the equation f(x) = -2/3(x - 7)² - 5, the vertex is (7, -5). The value of a, which is -2/3, is negative. So the graph opens downwards. The graph shifts 7 units to the right, which is correct, but graphing 5 units up is wrong. Why? 5 is negative, so it should be shifted 5 units down.

Wxndy~~

6 0
2 years ago
I need help I can’t figure it out
dusya [7]

Answer:

\large\boxed{\dfrac{41p}{70q^4(r-9)^2}}

Step-by-step explanation:

\dfrac{2p^4}{5q^5(r-9)^3}\cdot\dfrac{41q(r-9)}{28p^3}\\\\=\dfrac{2\!\!\!\!\diagup^1\cdot41}{5\cdot28\!\!\!\!\!\diagup_{14}}\cdot\dfrac{q\!\!\!\!\diagup}{q^{5\!\!\!\!\diagup^4}}\cdot\dfrac{(r\!\!\!\!\!\!{--}-9)\!\!\!\!\!\!{--}}{(r-9)^{3\!\!\!\!\diagup^2}}\cdot\dfrac{p^{4\!\!\!\!\diagup}}{p^3\!\!\!\!\!\!\diagup}\\\\=\dfrac{41}{70}\cdot\dfrac{1}{q^4}\cdot\dfrac{1}{(r-9)^2}\cdot\dfrac{p}{1}\\\\=\dfrac{41p}{70q^4(r-9)^2}

7 0
3 years ago
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