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Cloud [144]
3 years ago
7

PLEASE HELP ME WITH EMPIRICAL RULE!! ASAP!

Mathematics
1 answer:
Nat2105 [25]3 years ago
8 0

Answer:

hahahahahahahahahahhahahahahhahahahahhahagagagaggagagagahhahahahahha

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What is the gcf of the following expression
vovikov84 [41]
3. 6/3 equals 2 and 9 divided by 3 equals 3, so a simplified expresssion would say (2n^2)-3.
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Does 7(y−8)=7y+42 have one solution, no solution, or infinite solutions
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No solution is your answer
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I need help I don’t know what is the scoring rate
Semmy [17]
It’s 21 because 252 divided by 12 equals 21
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3 years ago
Andrea runs 4 miles every day, but she wants to increase her distance in order to run a 26 mile marathon. She decides to add 2 m
Doss [256]

Answer:

By the time she achieves her 26 mile goal Andrea will have run 176 miles.

Step-by-step explanation:

Since Andrea runs 4 miles every day, but she wants to increase her distance in order to run a 26 mile marathon and she decides to add 2 miles each day to her distance until she achieves her goal, to determine, if she starts with 6 miles today, how many miles will she have run in total, by the time she achieves her 26 mile goal the following calculation should be performed:

6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 + 26 = X

14 + 22 + 30 + 38 + 46 + 26 = X

36 + 68 + 72 = X

36 + 140 = X

176 = X

So, by the time she achieves her 26 mile goal Andrea will have run 176 miles.

6 0
3 years ago
How many subsets of {1, 2, 3, 4, 6, 8, 10, 15} are there for which the sum of the elements is 15?
stepladder [879]

Answer:

512

Step-by-step explanation:

Suppose we ask how many subsets of {1,2,3,4,5} add up to a number ≥8. The crucial idea is that we partition the set into two parts; these two parts are called complements of each other. Obviously, the sum of the two parts must add up to 15. Exactly one of those parts is therefore ≥8. There must be at least one such part, because of the pigeonhole principle (specifically, two 7's are sufficient only to add up to 14). And if one part has sum ≥8, the other part—its complement—must have sum ≤15−8=7

.

For instance, if I divide the set into parts {1,2,4}

and {3,5}, the first part adds up to 7, and its complement adds up to 8

.

Once one makes that observation, the rest of the proof is straightforward. There are 25=32

different subsets of this set (including itself and the empty set). For each one, either its sum, or its complement's sum (but not both), must be ≥8. Since exactly half of the subsets have sum ≥8, the number of such subsets is 32/2, or 16.

8 0
4 years ago
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