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Diano4ka-milaya [45]
3 years ago
11

Plz help im struggling....

Mathematics
1 answer:
worty [1.4K]3 years ago
7 0
Plug in 0 into x of the equation y=2x+3, solve it, then take that answer and use that as your y then plug in the answer for y and solve for x. (i realized i am very bad at explaining this over this. Sorry I tried)
You might be interested in
The lifespan of a guinea pig is 8 years shorter than the lifespan of a puma. Write a subtraction equation that you could use to
KiRa [710]

Answer: p-8

Step-by-step explanation:first thins problem is impossible without knowing the lifespan of a puma. But if you knew it oh let’s say it was 20 you would subtract the puma by 8.

7 0
2 years ago
Pls help me
gizmo_the_mogwai [7]

Answer:

no

no

no

yes yes

Step-by-step explanation:

not 100% but it should be right

3 0
2 years ago
Help please I need it know
elena-14-01-66 [18.8K]
On 15 angle 1 and 2 would be 150 degrees i think, i could be wrong.
On 16 angle 1 would be 140 degrees and angle 2 would be 40 degrees. hope this helps
8 0
3 years ago
Assume that male and female births are equally likely and that the birth of any child does not affect the probability of the gen
Aleksandr-060686 [28]

Let X be the number of boys in n selected births. Let p be the probability of getting baby boy on selected birth.

Here n=10. Also the male and female births are equally likely it means chance of baby boy or girl is 1/2

P(Boy) = P(girl) =0.5

p =0.5

From given information we have n =10 fixed number of trials, p is probability of success which is constant for each trial . And each trial is independent of each other.

So X follows Binomial distribution with n=10 and p=0.5

The probability function of Binomial distribution for k number of success, x=k is given as

P(X=k) = (10Ck) 0.5^{k} (1-0.5)^{10-k}

We have to find probability of getting 8 boys in n=10 births

P(X=8) = (10C8) 0.5^{8} (1-0.5)^{10-8}

= 45 * 0.0039 * 0.25

P(X = 8) = 0.0438

The probability of getting exactly 8 boys in selected 10 births is 0.044

8 0
3 years ago
How can 12 people share an 8 segment gummy worm? find two ways
Artemon [7]
Okay, so this question has a lot to do with DISTINGUISHABILITY which basically means: Can you tell the difference between each of the 12 people and can you tell the difference between each of the 8 gummy worm segments?

There are 4 possibilities: The people and gummies are both distinguishable, the people and gummies are both indistinguishable, the people are distinguishable and the gummies are not, or the people are not distinguishable and the gummies are.

<u>The people and gummies are both distinguishable. (This is the easiest case)</u>
There are 8 gummies. Each gummy can go to 12 different people. So the total number of ways we can give 8 distinct gummies to 12 distinct people is 8^12.
<u>
</u><u>The people and gummies are both indistinguishable (Much harder)
</u><u />This case is not necessarily harder, but it is very tedious and requires lots of careful, boring work. You would have to list out every single possible case, 1 by 1, there are no easier shortcuts... for example
8 0 0 0 0 0 0 0 0 0 0 0
7 1 0 0 0 0 0 0 0 0 0 0
6 2 0 0 0 0 0 0 0 0 0 0
6 1 1 0 0 0 0 0 0 0 0 0
5 3 0 0 0 0 0 0 0 0 0 0
5 2 1 0 0 0 0 0 0 0 0 0
5 1 1 1 0 0 0 0 0 0 0 0
4 4 0 0 0 0 0 0 0 0 0 0
...
...
And this would keep on continuing until you're sure you have counted every single one. But remember:
4 4 0 0 0 0 0 0 0 0 0 0
is the same as
4 0 0 4 0 0 0 0 0 0 0 0
because the people are indistinguishable.
<u>Lifehack: Use Wolfram Alpha for questions like this. I tried this just now and Wolfram Alpha stubbornly only counts partitions with all members greater than 1... so for this particular problem, WA won't work
</u><u>
</u><u>The people are distinguishable but the gummies are not</u>
This requires a bit more experience with combinatorics...
My comments above replying to the question talk about this. It is essentially 19C11 = 19C8.
<u>
</u><u>The people are indistinguishable but the gummies are distinguishable</u>
This is by far the hardest case. Remember our second case (<u>The people and gummies are both indistinguishable (Much harder)).</u>
Now we have to add on an extra step on top of all that partitioning. For each and every partition we have to calculate how many ways there are to make that partition. For the
8 0 0 0 0 0 0 0 0 0 0 there is only 1 possible way, all the gummies go to one person
For the
7 1 0 0 0 0 0 0 0 0 0 there is 8 possible ways, we choose one from the 8 gummies that goes to a new person (instead of the person that already has 7 gummies).

This brings everything to an end. I covered the general techniques for each of the 4 distinguishability cases. Hopefully this helps. If it does, please consider making this the brainliest answer. It took me quite a while to write this! :)


7 0
3 years ago
Read 2 more answers
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