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iVinArrow [24]
3 years ago
13

I need to know what the answer is plz help

Mathematics
2 answers:
balu736 [363]3 years ago
3 0
What’s the question ?
kykrilka [37]3 years ago
3 0
Could you say the question i cant answer it?
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When the numbers from 1 to 100 are written, in how many different numbers does the digit 3 appear?​
inessss [21]
19 Times.
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8 0
4 years ago
What is 56.000,000,000 in scienfic notaion?
dusya [7]

Answer:

56,000 = 5.6 x 10^**4

Step-by-step explanation:

56,000 = 56 x 10,000

Inserting decimal:

56 x 1,000 = 5.6 x 10,000

Answer:

56,000 = 5.6 x 10^**4

( btw ^** means exponent )

6 0
3 years ago
A map has a scale in which 1.25 inches represents 250 miles
olga2289 [7]

What question are trying to ask?

6 0
3 years ago
Which list shows the in greatest to least?<br><br>(NEED ASAP)
NeX [460]

The answer is B.

\frac{14}{5} = \frac{28}{10}

\frac{5}{2} = \frac{25}{10}

1\frac{3}{10} = \frac{13}{10}

\frac{3}{5} = \frac{6}{10}

If you find a common denominator, the process becomes much easier. I hope this was helpful.

8 0
3 years ago
A 5-card hand is dealt from a perfectly shuffled deck. Define the events: A: the hand is a four of a kind (all four cards of one
TiliK225 [7]

In a hand of 5 cards, you want 4 of them to be of the same rank, and the fifth can be any of the remaining 48 cards. So if the rank of the 4-of-a-kind is fixed, there are \binom44\binom{48}1=48 possible hands. To account for any choice of rank, we choose 1 of the 13 possible ranks and multiply this count by \binom{13}1=13. So there are 624 possible hands containing a 4-of-a-kind. Hence A occurs with probability

\dfrac{\binom{13}1\binom44\binom{48}1}{\binom{52}5}=\dfrac{624}{2,598,960}\approx0.00024

There are 4 aces in the deck. If exactly 1 occurs in the hand, the remaining 4 cards can be any of the remaining 48 non-ace cards, contributing \binom41\binom{48}4=778,320 possible hands. Exactly 2 aces are drawn in \binom42\binom{48}3=103,776 hands. And so on. This gives a total of

\displaystyle\sum_{a=1}^4\binom4a\binom{48}{5-a}=886,656

possible hands containing at least 1 ace, and hence B occurs with probability

\dfrac{\sum\limits_{a=1}^4\binom4a\binom{48}{5-a}}{\binom{52}5}=\dfrac{18,472}{54,145}\approx0.3412

The product of these probability is approximately 0.000082.

A and B are independent if the probability of both events occurring simultaneously is the same as the above probability, i.e. P(A\cap B)=P(A)P(B). This happens if

  • the hand has 4 aces and 1 non-ace, or
  • the hand has a non-ace 4-of-a-kind and 1 ace

The above "sub-events" are mutually exclusive and share no overlap. There are 48 possible non-aces to choose from, so the first sub-event consists of 48 possible hands. There are 12 non-ace 4-of-a-kinds and 4 choices of ace for the fifth card, so the second sub-event has a total of 12*4 = 48 possible hands. So A\cap B consists of 96 possible hands, which occurs with probability

\dfrac{96}{\binom{52}5}\approx0.0000369

and so the events A and B are NOT independent.

4 0
3 years ago
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