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andrew-mc [135]
3 years ago
15

PLZ HELP................................................................................................................

Mathematics
2 answers:
Phoenix [80]3 years ago
8 0

Answer:

7) 4a,  5b, 11b, 3a, 9, 21

8) 4, 5, 11, 3

9) a, b, b, a

amm18123 years ago
5 0

Answer:

o how do you do that work my brain would blow up

Step-by-step explanation:

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A teacher polled only the 8th grade students in English classrooms on what their favorite color is. She found that the favorite
amid [387]

Answer:

This is a biased sample.

Step-by-step explanation:

This is because she gave a random sample, but the she had the convenience of the students being with her when she surveyed. So it'e not actually random. She could also worded the survey or gave a biased response.

4 0
3 years ago
PLEASE HELP MEE (40 points) will Mark BRAINLIEST!
Lady_Fox [76]
Need more information on the choice, please provide
4 0
3 years ago
The equation which correctly represents this situation is
uranmaximum [27]

Answer:

C) x2 + x = 306

Step-by-step explanation:

x2 + x = 306

x2 + x - 306  = 0

x2  + 18x - 17x + 18 * (-17) = 0

x ( x + 18) - 17 (x +18) =0

(x + 18) (x - 17 ) = 0

x = -18 or 17

5 0
4 years ago
HELP PLEASE!!! I need help with 94 if you could show the steps that would be very helpful!
aksik [14]
A combination is an unordered arrangement of r distinct objects in a set of n objects. To find the number of permutations, we use the following equation:

n!/((n-r)!r!)

In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.

5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5

5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10

5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10

5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5

Notice that discarding 1 or discarding 4 have the same number of combinations, as do discarding 2 or 3. This is being they are inverses of each other. That is, if we discard 2 cards there will be 3 left, or if we discard 3 there will be 2 left.

Now we add together the combinations

1 + 5 + 10 + 10 + 5 + 1 = 32 choices combinations to discard.

The answer is 32.

-------------------------------

Note: There is also an equation for permutations which is:

n!/(n-r)!

Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.

We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.

I hope this helps! If you have any questions, let me know :)








7 0
3 years ago
I need help with 5 & 6 please
amid [387]
For number 5, the answer is B.
For number 6, the answer is B.
Y are welcome for the help and have a nice day!
4 0
3 years ago
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