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Pani-rosa [81]
3 years ago
10

I need help with this :/ pls help

Mathematics
1 answer:
emmainna [20.7K]3 years ago
8 0

Answer:

When you divide by 10^3, the decimal point moves three places to the right. When you multiply by 0.001, the decimal point moves three places to the right. So, both equal .0777

Step-by-step explanation:

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The stock price for International Business Machines (IBM) historically has followed an approximately normal distribution (when a
olganol [36]

Answer:

0.494 is the probability that on a selected day the stock price is between $186.26 and $192.47.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $188.876

Standard Deviation, σ = $4.6412

We are given that the distribution of stock price is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P(stock price is between $186.26 and $192.47)

P(186.26 \leq x \leq 192.47) = P(\displaystyle\frac{186.26 - 188.876}{4.6412} \leq z \leq \displaystyle\frac{192.47-188.876}{4.6412}) = P(-0.5636 \leq z \leq 0.7743)\\\\= P(z \leq 0.7743) - P(z < -0.5636)\\= 0.781 - 0.287 = 0.494 = 49.4\%

P(186.26 \leq x \leq 192.47) = 49.4\%

0.494 is the probability that on a selected day the stock price is between $186.26 and $192.47.

5 0
3 years ago
Perform the indicated operation and write the result in the form a + bi. i(5 - i) + 2i (i + 6) the
Nikitich [7]

i(5 - i) + 2i (i + 6) =

distribute

5i - i^2 +2i^2 +12i

5i - (-1) +2 (-1) +12i

5i +1 -2 +12i

-1+17i

8 0
3 years ago
Read 2 more answers
Why might Razis method not be a good way to estimate the probability? My team won 18 of their last 20 matches, so the probabilit
FrozenT [24]

Answer:

It is a really idealistic way to estimate probability.

It usually works well for random events where there is no restriction like:

"There are N elements in a set, the probability of randomly selecting any element is the same (so there is no restriction) meaning that for every single element, the probability is 1/N.

Now suppose that there are K elements (K < N) with a given characteristic, then the probability of randomly selecting one of the K elements out of the sett with N elements is equal to K times the probability of each element, this is:

P = K*(1/N) = K/N"

Then if a dice has 6 possible outputs, and we want to find the probability of getting a 3 or a 5, we need to find the quotient between the outcomes where we get a 3 or a 5 (2) and the total number of outcomes (6)

P = 2/6

Now let's go to the case in the problem:

We know that out of the last 20 games, your team won 18.

Then yes, your team won (18/20)*100% = 90% of their played games.

Does this mean tath the probability of winning the next game is 90%?

Well no, because this is not a random selection, there are a lot of other things needed to analyze, as the win rate of the other team (suppose that the other team won 14 of their last 20 games, then their win rate is (16/20)*100% = 70%

Does this mean that the probability where our team wins is 90%, and the probability of the enemy team winning is 70%?

Well no, this makes no sense at all, and this is why this method is not a good way to estimate probability in complex cases like this one.

5 0
3 years ago
Use the figure at the right to answer the following questions.
Lisa [10]

<h3><u>Answers:</u></h3>

1. Name all the line segments.

<u>Ans.</u><u> </u> PL, LA, AY, YS,

2. How many line segments were formed?

<u>Ans.</u><u> </u> 4 (four)

3. Name all rays

<u>Ans.</u><u> </u>AM, AN

4. How many rays were formed?

<u>Ans.</u> 2 (two)

5. Name the opposite rays :

<u>Ans.</u><u> </u>AM, AN

6. Is point A between MN ?

<u>Ans.</u><u> </u>Yes

7. Find the line segments and rays formed by 5 points.

<u>Ans.</u><u> </u>

  • Line segment = PL, LA, AY, YS,
  • Rays = AM, AN

<h3>Hope it helps you.</h3>
7 0
2 years ago
A right triangle has a leg of length 4.6 inches and a hypotenuse of length 5.4 inches. Find the approximate length of the other
SpyIntel [72]

The answer is 2.8in

7 0
3 years ago
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