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1 year ago

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Bipasha can eat 24 rasgullas in 12 minutes. Her friend Sujata takes DOUBLE the time to eat the same number.One day they buy 24 r

asgullas and start eating them. If they eat at their normal speeds till the rasgullas are over, how many rasgullas would Sujata have eaten?

1 answer:

Crazy boy [7]1 year ago

5 0

Answer:

12

Step-by-step explanation:

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Write (3p^3)^2 without exponents

rodikova [14]

Answer:

3p*3p*3p*3p*3p*3p

Step-by-step explanation:

Multiply exponents

3p^6

expand

3p*3p*3p*3p*3p*3p = 3p^6

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2 years ago

Which scenario is most likely the one shown on the graph?

nadezda [96]

Hello,

I don't see a graph you can edit your answer by putting a graph then I can help you.

Your friend --- Young Sinatra

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1 year ago

Read 2 more answers

Player V and Player M have competed against each other many times. Historical data show that each player is equally likely to wi

Helen [10]

Answer:

0.46 (46%)

Step-by-step explanation:

We have the following data:

- Probability that player V wins the first set:

$p(1)=0.50$

(because the text says the two players are equally likely to win the first set)

- Probability that player V wins the 2nd set if he has won the 1st set:

$p(2|1)=0.60$

So, the probability that player V wins the first 2 sets is:

$p(12)=p(1)\cdot p(2|1)=(0.50)(0.60)=0.30$ (1)

Instead, the probability that player V loses the 2nd set if he has won the 1st set is 0.40 (=1-0.60), so

$p(2^c|1)=0.40$

So, the probabiity that player V winse the 1st set but loses the 2nd set is

$p(12^c)=p(1)\cdot p(2^c|1)=(0.50)(0.40)=0.20$ (2)

Also, we have:

- Probability that player V loses the 1st set:

$p(1^c)=0.50$

- Probability that she will lose the 2nd set in this case is 0.70, it means that the probability that she will win the 2nd set if she lost the 1st set is 0.30, so:

$p(2|1^c)=0.30$

So, the probability that she will lose the 1st set and win the 2nd set is:

$p(1^c2)=p(1^c)\cdot p(2|1^c)=(0.50)(0.30)=0.15$ (3)

Combined together (2) and (3), this means that the probability that player V wins exactly 1 set out of the first two sets is:

$p(1/2)=p(12^c)+p(1^c2)=0.20+0.15=0.35$ (4)

At this point, the probability that she will win the 3rd set is

$p(3)=0.45$

This means that the overall probability that she will win the 3rd set if she won exacty 1 of the first 2 sets is:

$p(1/2,3) = p(1/2)\cdot p(3)=(0.35)(0.45)=0.16$ (5)

So, the overall probability that player V will win a match against player M is the sum of (1) and (5):

$p(W)=p(12)+p(1/2,3)=0.30+0.16=0.46$

5 0

2 years ago

ACD is a triangle and B is a point on AC. AB = 8cm and BC is 6cm. Angle BCD = 48° and angle BDC = 50°. (a) Find the length of BD

FromTheMoon [43]

Answer:

5.8206 cm
10.528 cm
23.056 cm^2

Step-by-step explanation:

(a) The Law of Sines can be used to find BD.

BD/sin(48°) = BD/sin(50°)

BD = (6 cm)(sin(48°)/sin(60°)) ≈ 5.82064 cm

__

(b) We can use the Law of Cosines to find AD.

AD^2 = AB^2 +BD^2 -2·AB·BD·cos(98°) . . . . . angle ABD = 48°+50°

AD^2 ≈ 110.841

AD ≈ √110.841 ≈ 10.5281 . . . cm

__

(c) The area of ∆ABD can be found using the formula ...

A = ab·sin(θ)/2 . . . . . where a=AB, b=BD, θ = 98°

A = (8 cm)(5.82064 cm)sin(98°)/2 ≈ 23.0560 cm^2

_____

Angle ABD is the external angle of ∆BCD that is the sum of the remote interior angles BCD and BDC. Hence ∠ABD = 48° +50° = 98°.

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2 years ago

Antonia is keeping track of the number of pages she reads. She started the school year having already read 95 pages her book, an

cupoosta [38]

F(x) means "function of" and x represents the day number, example: day 6 would be f(6)=(20×6)+95

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2 years ago