Step-by-step explanation:
<em>option</em><em> </em><em>D </em><em>is </em><em>correct</em><em>,</em>
<em>because</em><em> </em><em>the </em><em>orientation</em><em>/</em><em>postion</em><em> </em><em>of </em><em>the </em><em>vertices </em><em>does </em><em>got </em><em>changed.</em><em>.</em><em>.</em>
<em>that's</em><em> </em><em>why </em><em>it </em><em>is </em><em>not </em><em>true</em>
<em>hope </em><em>this </em><em>answer </em><em>helps </em><em>you </em><em>dear </em><em>take</em><em> </em><em>care</em>
27.034%
Let's define the function P(x) for the probability of getting a parking space exactly x times over a 9 month period. it would be:
P(x) = (0.3^x)(0.7^(9-x))*9!/(x!(9-x)!)
Let me explain the above. The raising of (0.3^x)(0.7^(9-x)) is the probability of getting exactly x successes and 9-x failures. Then we shuffle them in the 9! possible arrangements. But since we can't tell the differences between successes, we divide by the x! different ways of arranging the successes. And since we can't distinguish between the different failures, we divide by the (9-x)! different ways of arranging those failures as well. So P(4) = 0.171532242 meaning that there's a 17.153% chance of getting a parking space exactly 4 times.
Now all we need to do is calculate the sum of P(x) for x ranging from 4 to 9.
So
P(4) = 0.171532242
P(5) = 0.073513818
P(6) = 0.021003948
P(7) = 0.003857868
P(8) = 0.000413343
P(9) = 0.000019683
And
0.171532242 + 0.073513818 + 0.021003948 + 0.003857868 + 0.000413343
+ 0.000019683 = 0.270340902
So the probability of getting a parking space at least four out of the nine months is 27.034%
Answer:
when we take the square root of 500 we get number with decimal value so its not a perfect square
Step-by-step explanation:
x^2=500
x=square root of 500
The similarity
ratio of a cube with volume 729 m3 to a cube with a volume of 3,375 m3.iR² is 3:5. The correct answer between
all the choices given is the second choice or letter B. I am hoping that this
answer has satisfied your query and it will be able to help you, and if you
would like, feel free to ask another question.
