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vaieri [72.5K]
3 years ago
6

Find the solution of given expression

rmula1" title=" \sqrt{100 \times 16} " alt=" \sqrt{100 \times 16} " align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
enyata [817]3 years ago
5 0

Answer:

\sqrt{100 \times 16}=\sqrt{10² \times 4²}=±(10×4)=±40 is your answer

Bond [772]3 years ago
4 0

Answer:

40 will be your answer!!!

Step-by-step explanation:

Thanks!!

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Please help this was due yesterday!
CaHeK987 [17]

Answer:

a) <

b) <

c) >

d) <

e) >

Step-by-step explanation:

5 0
3 years ago
Tickets for a concert cost $29. 00. The greatest number of tickets one person can buy is 9.If 28 peopleeach buy 9 tickets for $2
Blababa [14]

Answer: 7,308


Step-by-step explanation: 28 X 9 = 252

252 X 29 = 7,308


6 0
3 years ago
Read 2 more answers
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.

<u><em>1) The probability of having a face card in the first turn</em></u>

In this case, the chances are 12 in 52:

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

<u><em>2) The probability of having a face card in the second turn, having a non-face card in the first turn.</em></u>

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

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

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

<u><em>4) The probability of having a face card in the fourth turn, having a non-face card in the previous turns.</em></u>

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.

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3 years ago
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Vesna [10]
So the distance formula: d = rad (x2 - x1)^2 + (y2 - y1)^2
it doesn't matter what order you do, you just have to make sure it's a y coordinate for y and an x coordinate for x.
rad (9-5)^2 + ((-6)-1)^2
rad 4^2 + (-7)^2
rad 16 + 49
rad 65
4 0
2 years ago
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Which way will the decimal go if you are dividing with decimals by a power of 10
astraxan [27]

Answer: Multiplying and Dividing Decimals by Powers of 10

When you divide a decimal by a power of 10, simply move the decimal place to the left as many places as there are 0s in the power of 10.

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2 years ago
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