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Naya [18.7K]
3 years ago
14

Plssss helppp me plsss

Mathematics
2 answers:
gavmur [86]3 years ago
5 0
I think it’s 1 too, for this answer
adelina 88 [10]3 years ago
4 0
6 I think but maybe 1
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Quesnon
xenn [34]

Answer:

C. 44 in.

Step-by-step explanation:

= 2 \times \pi \times r \\  = 2 \times  \frac{22}{7}  \times 7 \\  = 44 \: in

5 0
3 years ago
How do you solve this absolute value Q.2{5g+3π}=234
Naddika [18.5K]
Absolute values are generally written as || instead of {}

|x| = 1 is the equivalent of 2 equations
x=1 and x=-1
3 0
3 years ago
Read 2 more answers
In which equation is x the independent variable?<br> A. x=7y<br> B. y=7x<br> C. y+7=x<br> D. y/7=x
Arturiano [62]

Answer:

B is the answer.

A) No because the value of x depends on what 7•y is

C) No because x depends on the value of y+7

D) No because the x depends on the value of y/7

In other words, on B, x doesn't depend on anything, your just given the value of it. Don't know how else to explain it, hope you get it.

7 0
3 years ago
Read 2 more answers
Abjejwjjdjejwjdhhrigvhhyrwwtiij itessfjhvxx
Pachacha [2.7K]
Sjdjdkzjdkdncmdkdjfnkrmd, sjdjdjdjd dkdkcekdkdk djdjjdkdjdj ..
3 0
2 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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