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3 years ago

EQUILATERAL ISOSELES SCALENE

Mathematics

2 answers:

GaryK [48]3 years ago

7 0

Answer:

isoseles triangle

explanation: none of the sides are the same

hope this helps

Yanka [14]3 years ago

5 0

Answer: SCALENE

Step-by-step explanation:

All sides are unequal, ISOSELES has 2 equal sides and EQUILATERAL all sides are equal.

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Suppose I ask you to pick any four cards at random from a deck of 52, without replacement, and bet you one dollar that at least

Tatiana [17]

Answer:

a) No, because you have only 33.8% of chances of winning the bet.

b) No, because you have only 44.7% of chances of winning the bet.

Step-by-step explanation:

a) Of the total amount of cards (n=52 cards) there are 12 face cards (3 face cards: Jack, Queen, or King for everyone of the 4 suits: clubs, diamonds, hearts and spades).

The probabiility of losing this bet is the sum of:

- The probability of having a face card in the first turn

- The probability of having a face card in the second turn, having a non-face card in the first turn.

- The probability of having a face card in the third turn, having a non-face card in the previous turns.

- The probability of having a face card in the fourth turn, having a non-face card in the previous turns.

1) The probability of having a face card in the first turn

In this case, the chances are 12 in 52:

$P_1=P(face\, card)=12/52=0.231$

2) The probability of having a face card in the second turn, having a non-face card in the first turn.

In this case, first we have to get a non-face card (there are 40 in the dech of 52), and then, with the rest of the cards (there are 51 left now), getting a face card:

$P_2=P(non\,face\,card)*P(face\,card)=(40/52)*(12/51)=0.769*0.235=0.181$

3) The probability of having a face card in the third turn, having a non-face card in the first and second turn.

In this case, first we have to get two consecutive non-face card, and then, with the rest of the cards, getting a face card:

$P_3=(40/52)*(39/51)*(12/50)\\\\P_3=0.769*0.765*0.240=0.141$

4) The probability of having a face card in the fourth turn, having a non-face card in the previous turns.

In this case, first we have to get three consecutive non-face card, and then, with the rest of the cards, getting a face card:

$P_4=(40/52)*(39/51)*(38/50)*(12/49)\\\\P_4=0.769*0.765*0.76*0.245=0.109$

With these four probabilities we can calculate the probability of losing this bet:

$P=P_1+P_2+P_3+P_4=0.231+0.181+0.141+0.109=0.662$

The probability of losing is 66.2%, which is the same as saying you have (1-0.662)=0.338 or 33.8% of winning chances. Losing is more probable than winning, so you should not take the bet.

b) If the bet involves 3 cards, the only difference with a) is that there is no probability of getting the face card in the fourth turn.

We can calculate the probability of losing as the sum of the first probabilities already calculated:

$P=P_1+P_2+P_3=0.231+0.181+0.141=0.553$

There is 55.3% of losing (or 44.7% of winning), so it is still not convenient to bet.

5 0

4 years ago

Given that , the _____ justifies the conclusion that ∆ABD ≅ ∆AEC.

trasher [3.6K]

AAS Postulate

It is given that CE = BD so we know "S" (representing side) has to be in the three letter postulate.
It is also given that angle DBA and angle CEA are right angles, so therefore they are congruent. Now we know that an "A" must also be in the postulate.

Lastly, we know that the triangles have a second angle, EAB, in common because they share it overlappingly. So there must be another "A" in the postulate.

Now we need to look at the order in which it is presented. The order follows Angle, Angle, Side so the postulate must be the AAS postulate. Hope this helps!

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3 years ago

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Need help on my math only have today till I turn int tomorrow

Aneli [31]

Hope this helps explain what was needed to be solved! :)

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3 years ago

Find the inverse of f(x)=-1/3x-5

andreyandreev [35.5K]

-1.6666 hope this helps you

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3 years ago

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In the 2009-2010 school year in country A, there were 92,000 foreign students from country B. This number I 23% more than the nu

max2010maxim [7]

Answer:

$\approx74796$

Step-by-step explanation:

GIVEN: In the $2009-10$ school year in country A, there were $92,000$ foreign students from country B. This number is $23\%$ more than the number of students from country C.