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TiliK225 [7]
3 years ago
10

Can anyone help with this problem

Mathematics
1 answer:
vovikov84 [41]3 years ago
3 0
I hope this helps you

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Write the equation of the line that passes through the point (2,-9)<br> and has a slope of -5.
balu736 [363]

Step-by-step explanation:

Find equation

y - y1 = m(x - x1)

y + 9 = -5(x - 2)

y = -5x + 10 - 9

y = -5x + 1 → Equaton

7 0
2 years ago
Help click on the pick for the question​
Maksim231197 [3]
The answer is £2.02
8 0
4 years ago
Read 2 more answers
There are 30 people in a room. You want to predict the number of people in the room who share birthdays with others in the room.
Sergeeva-Olga [200]

There are 30 people in a room. You want to predict the number of people in the room who share birthdays with others in the room. How could you begin to set up a simulation for this scenario is given below

Step-by-step explanation:

First, I’m going to walk through a step-by-step of solving it, and I’ll provide a short explanation at the bottom for why this is the case.

To figure this stat, let’s first realize that, excluding twins, each of the 30 people has an equivalent 365 days of the year that could be their birthday. Therefore, the total combination of all the possible probabilities of birthdays for all of the 30 people is 365 * 365 * 365 * … 30 times or, better expressed, 365^30.

An easier way to solve this problem than solving for the probability that any 2 or more of the 30 people share a birthday is to solve for the probability that all of the people have unique birthdays (non-shared) and subtracting that from 100%.

The first person has 365 possible days that could be their birthday without sharing with someone else. Then, the second person has 364 days that could be their birthday without sharing with someone else, because person 1’s birthday is one of those days. This process goes on for all of the 30 people, until the 30th person has 336 possible days that could be their birthday.

A better way to express the total possible combinations of days in which none of the 30 people share a birthday is 365 * 364 * 363 * … until 336 or, better expressed, 365!/335!

To solve for the total probability that, out of the original 365^30 days, there are 365!/335! of them where no one shares a birthday, we simply divide the latter by the former.

(365!/335!)/(365^30)

If you do this immense calculation, you can solve that the probability that none of the 30 people will share a birthday is 29.36837573%. If you subtract this from 100%, you get the probability that a minimum of two people do share a birthday, which is 70.63162427% or roughly 7/10.

While at first glance, the answer may seem obvious as 30/365 or 335/365 or any other quick calculation, you have to realize that this calculation is an example of stacking probability. While the probability that the first two people don’t share a birthday is quite minuscule, this probability stacks, so to speak, and grows exponentially for every additional person whose birthday you must consider. An easier way of comprehending this is recognizing that, for every additional person, you have to calculate the probability that their birthday does not match with any of the others’ birthdays. For numbers going past just a handful, this number does grow quickly, as each person has a certain number of people with whom they can’t share a birthday, and this is true for every one of that certain number of people.

While this may be hard to wrap your head around, it is simple when done by calculation. It’s just an example of how the intuitive part of your brain tries to solve this problem by going for the quick, easy solution that may not always be accurate. You have to force the deep thinking part of your brain to actually analyze the problem for what it truly is and see that it is a complex probability.

8 0
4 years ago
Read 2 more answers
The sum of two numbers is 5.036. one number has a 4 in the tenths place and 7 in the thousandths place. The other number has a 1
UNO [17]
   1.x2x
 +x.4x7
   5.036  (sorry the underline function is no longer available)

In the thousandth column we have x +7 and we wind up with a 6 in that column.
 Since 7 is already more than 6 that means the only number that will work is 9 and there will be a carry of one into the hundredths place.

  1.x29
+x.4x7
  5.036

Now let's look at the tenths column.  2 becomes 3 in the answer and there was a carry of 1 from the thousandths so the other number must be a zero
   1.x29
+ x.407
   5.036   Now I need a number that goes with 4 and puts a zero in the answer.
                That number has to be 6 and does cause a carry of one into the ones
                 place

  1.629
+x.407
  5.036  

Finally (remembering there is a carry of one in the ones place)  I have two (1 + 1 carried) toward my five in the answer.  I need three more.

Here is the final example:
    1.629
  +3.407
    5.036
 
7 0
3 years ago
How many three-digit numbers are there such that the first two digits are even and the last digit is odd? Answer is a whole numb
Genrish500 [490]
221
223
225
227
229
241
243
245
247
249
261
263
265
267
269
281
283
285
287
289
421
423
425
427
429
441
443
445
447
449
461
463
465
467
469
481
483
485
487
489
621
623
625
627
629
641
643
645
647
649
661
663
665
667
669
681
683
685
687
689
821
823
825
827
829
841
843
845
847
849
861
863
865
867
869
881
883
885
887
889
80 three-digit numbers 
(just to make sure, check)

5 0
3 years ago
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