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bagirrra123 [75]
2 years ago
10

Fred and Gina are playing tennis. The first player to win 2 sets wins their match. Fred has a 3/5 chance to win each set while G

ina has a 2/5 chance. What is the probability that the match is decided in only two sets?
Mathematics
2 answers:
Anna71 [15]2 years ago
5 0

From the information below the games is over in 3 sets. For Gina to win the match there are 3 possibilities.

1) Gina wins the first 2 sets with probability

2)Gina wins the first set. Looses the next set. Wins the third set.

3)Gina looses the first set. Wins the next 2 sets.

The required probability that Gina wins is the sum of the above 3 probabilities . That is the probability that Gina wins the match is

Nookie1986 [14]2 years ago
3 0

Answer:P=\frac{9+4}{25}=\frac{13}{25}

Step-by-step explanation:

Given

Fred has a chance of 3/5 to win

and Gina has a chance of 2/5 to win

Probability(P) that the match is decided in only two sets is when

Either Fred or Gina win both matches continuously

P=P(Fred win both match)+P(Gina win both match)

P=\frac{3\times 3}{5\times 5}+\frac{2\times 2}{5\times 5}

P=\frac{9+4}{25}=\frac{13}{25}

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give an example on an addition problem in which you would and would not group the addends differently to add
Murrr4er [49]
Addends are any of the numbers added together in an equation. 

The only time their grouping would matter would be if there were parentheses used to alter the normal Order of Operations. 

For ex:
2 - (8 + 3)  here, the 8 and 3 have to be grouped together before doing the subtraction.

Any addition problem without parentheses can be used for one where the grouping doesn't matter
3 0
2 years ago
The area of the rectangle is 64.8 square centimeters. What is the perimeter of the rectangle? One group of ten tenths and one gr
9966 [12]

Answer:

(a) Perimeter = 32.2\ cm

(b) 100

Step-by-step explanation:

Solving (a):

Given

Shape: Rectangle

Area = 64.8

Required

Calculate the perimeter.

Area is calculated as:

Area = L * W

Where

L = Length and W = Width

Substitute 64.8 for Area

64.8 = L * W

Make L the subject:

L = \frac{64.8}{W}

Perimeter is calculated as:

P = 2 * (L + W)

Substitute 64.8/W for L

P = 2 * (\frac{64.8}{W} + W)

P = \frac{129.6}{W} + 2W

To solve further, we take the derivative of P and set it to 0, afterwards.

dP = -\frac{129.6}{W^2} + 2

Set to 0

0 = -\frac{129.6}{W^2} + 2

Collect Like Terms

\frac{129.6}{W^2} = 2

Cross Multiply:

2W^2= 129.6

Divide through by 2

W^2 = 64.8

Take square roots

W = \sqrt{64.8

W = 8.05

Recall that:

L = \frac{64.8}{W}

So:

L= \frac{64.8}{8.05}\\

L= 8.05

The perimeter is:

Perimeter = 2 * (8.05 + 8.05)

Perimeter = 2 * (16.10)

Perimeter = 32.2\ cm

Solving (b):

Given

((1 Group of 10 tenths) and (1 group of 8 tenths))/(6 groups of 3 tenths)

Required

Solve

Tenths = \frac{1}{10}

So, the expression becomes:

((1 Group of 10 * \frac{1}{10}) and (1 group of 8 * \frac{1}{10}))/(6 groups of 3 * \frac{1}{10})

This gives:

((1 Group of \frac{10}{10}) and (1 group of \frac{8}{10}))/(6 groups of \frac{3}{10})

Group means product, so the expression becomes:

\frac{(1 * \frac{10}{10} \ and\ 1 * \frac{8}{10})}{6 * \frac{3}{10}}

And, as used here means addition

\frac{(1 * \frac{10}{10} +  1 * \frac{8}{10})}{6 * \frac{3}{10}}

Simplify:

\frac{(1 * 1 +  1 * 0.80)}{6 *0.30}

\frac{(1 +  0.80)}{1.80}

\frac{1.80}{1.80}

= 1.00

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It would most possibly be 38.05 or less

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