8 1/3 times 4 1/5 = ?

Mathematics

2 answers:

Levart [38]3 years ago

5 0

<h2>Answer:</h2>

<u><em>I believe the answer is 35. </em></u>

<h2>Step-by-step explanation:</h2>

<em>If I am incorrect please correct me but this is what I got. Also, you never told me if it's in simplified form or in any form it should be in. Hope this somewhat helped and have a good one!</em>

arlik [135]3 years ago

3 0

Answer:

Step-by-step explanation:

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kirill115 [55]

Answer:

there aint a question to answer lol

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

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) = $(\frac{1}{13}) \times(\frac{4}{51}) = (\frac{4}{663} )$