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3 years ago

Please help me out? tysm! (Answer the bottom 3 or all except for the ones already answered)

Mathematics

2 answers:

Alexxandr [17]3 years ago

8 0

1 - 9.5L
2 - 500L
3 - 2L

valina [46]3 years ago

6 0

Step-by-step explanation:

The first one on the left : 9.5 L

the one in the middle : 500ml

the one on the right : 2L

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4 1/3×2 3/4. this is 6 grade math

elena-14-01-66 [18.8K]

11 11/12 is the answer

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3 years ago

WILL MARK BRAINLEST TO THE CORRECT ANSWER (55 POINTS!!!)

nikitadnepr [17]

Answer:

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Step-by-step explanation:

6 0

2 years ago

26.68 is 58% of what number?

LekaFEV [45]

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4 0

2 years ago

You are dealt two cards successfuly without replacement from a shuffled deck of 52 playing card. Find the probability that first

katen-ka-za [31]

The simplified expression is $(\frac{4}{663})$

Step-by-step explanation:

Here, the total number of cards in a given deck = 52

let E : Event of drawing a first card which is King

Total number of kings in the given deck = 4

So, $P(E) = \frac{\textrm{The total number of king}}{\textrm{The total number of cards}}$ = $\frac{4}{52} = (\frac{1}{13} )$

Now, as the picked card is NOT REPLACED,

So, now the total number of cards = 52 - 1 = 51

Total number of queen in the deck is same as before = 13

let K : Event of drawing a second card which is queen

So, $P(K) = \frac{\textrm{The total number of queen}}{\textrm{The total number of cards}}$ = $(\frac{4}{51} )$

Now, the combined probability of picking first card as king and second as queen = P(E) x P(K) = $(\frac{1}{13}) \times(\frac{4}{51}) = (\frac{4}{663} )$

Hence, the simplified expression is $(\frac{4}{663})$

6 0

3 years ago

The drama club was selecting which carnival booths to sponsor at the fall carnival from a list of nine. How many different ways

denis-greek [22]

Answer:

Option B

Step-by-step explanation:

Here we have to apply " combination and permutation. " It is given that the drama club had to choose three booths from a selection of 9, considering the possible ways to choose so. This is a perfect example of combination. In nCr, n corresponds to 9, respectively r corresponds to 3.

$\mathrm{n\:choose\:r},\\nCr=\frac{n!}{r!\left(n-r\right)!},\\\\\frac{9!}{3!\left(9-3\right)!} =\\\frac{9!}{3!\cdot \:6!} =\\\\\frac{9\cdot \:8\cdot \:7}{3!} =\\\frac{504}{6} =\\\\84\\\\Solution = Option B$

Hope that helps!

8 0

3 years ago