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GrogVix [38]
3 years ago
6

D. 10 + 11k can someone plzz help meeee plzzzz

Mathematics
1 answer:
torisob [31]3 years ago
8 0

Answer:

(A).  \frac{15}{2} + \frac{33}{4} <em>k</em>

Step-by-step explanation:

\frac{3}{4} × 2(2 + 4k + 3 + \frac{3}{2} k ) = \frac{3}{2} ( 5 + \frac{11}{2} k ) = \frac{15}{2} + \frac{33}{4} k

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Rezolvati problema prin metoda figurativa:
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Ne este dat numărul total de elevi = 30.

Să presupunem că numărul de băieți este egal cu numărul de fete.

Dacă împărțim 30 cu 2, obținem 15.

Deci, numărul de băieți = 15 și numărul de fete = 15 dacă ambele sunt egale.

"Dar numărul de fete este mai mult de 2 băieți."

Astfel putem scădea 1 din 15 și adăugăm 1 din 15.

După adăugarea a 1 din 15, ajungem

15 + 1 = 16.

Și scăzând de la 1 la 15, ajungem acolo

15-1 = 14.

Deci, numărul de băieți este de 14 și numărul de fete este de 16

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49. From 3 coin tosses, there are 8 possible outcomes:

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Likewise, all but the last listed outcome have at least one tail. The problem is symmetrical that way when the coin is fair. 87.5% of outcomes have at least one tail.

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Perhaps you can tell I read your question as having two parts. If your question is the probability of getting at least one head AND at least one tail, you can see that condition includes 6 of the 8 outcomes, or 75%, matching selection d.

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