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Misha Larkins [42]
3 years ago
10

Help me with #4 pleasedeeee

Mathematics
1 answer:
Romashka [77]3 years ago
4 0

Answer:

\dfrac{dy}{dx}=-\dfrac{2x-y-1}{2y-x-1}

Step-by-step explanation:

Differentiating, you get ...

... x·dy +y·dx +dx +dy = 2x·dx +2y·dy

Collecting terms gives ...

... dy(x +1 -2y) = dx(2x -y -1)

... dy/dx = -(2x -y -1)/(2y -x -1)

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Help me with this question pleaseeee
Lerok [7]

Answer:

a=69

Step-by-step explanation:

121-52 is equal to 69

3 0
2 years ago
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Please help! Will Mark if you are right!!
slavikrds [6]

Answer:

sorry I wish I could help):

5 0
2 years ago
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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
Can someone answer this question please WBA’s wee it correctly if it’s correct I will mark you brainliest
oksano4ka [1.4K]

Answer:

\frac{1}{6}*\frac{4}{13}

Step-by-step explanation:

A 6 sided die has 6 outcomes.

Your sister picks from a 13 - slice plate meaning the fact that there are 13 outcomes.

There are 4 yellow slices so picking a yellow slice has a \frac{4}{13} chance

You need to mutiply both displayed values to solve this problem.

\frac{1}{6}*\frac{4}{13}

6 0
3 years ago
What is the slope of this line?
olga2289 [7]

Answer:

3

Step-by-step explanation:

In the Slope-Intercept Formula, <em>y</em><em> </em><em>=</em><em> </em><em>mx</em><em> </em><em>+</em><em> </em><em>b</em><em>,</em><em> </em><em>m</em><em> </em>is the <em>Rate</em><em> </em><em>of</em><em> </em><em>Change</em><em> </em>[<em>Slope</em>]. Anyway, starting from the <em>y-intercept</em><em> </em>of [0, 2], move 3 units <em>north</em><em> </em>over 1 unit <em>east</em><em>.</em><em> </em>That is called <em>rise</em><em>\</em><em>run</em><em> </em><em>→</em><em> </em>3\1 = 3.

I am joyous to assist you anytime.

7 0
2 years ago
Read 2 more answers
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