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3 years ago

The probability of Asher's soccer team winning its next game is 7/10. What is probability of not winning their next game?

Mathematics

2 answers:

Daniel [21]3 years ago

7 0

Answer:

Step-by-step explanation:

There are 36 total possible outcomes.

Rolling a sum of 11 or higher, there are 3 possible rolls, to make a 3/36 = 1/12 probability.

Rolling a sum of 4 there are also 3 possibilities, so the chance would be the same.

Rolling a sum of 9, there are 4 possibilities, which is a better chance.

Rolling a sum less than 5, there is 6 possibilities, which is a better chance.

Rolling greater than 5 but less than 7 means rolling a sum of 6, there are 5 chances, which is a better chance.

Rolling greater than 9 but less than 11, means rolling a 10, there are 3 possibilities, which is the same.

Rolling greater than 2 and less than 4 means rolling a 3, there are 2 possibilities, which is less.

The answers would be:

Rolling a sum of 9,

Rolling a sum less than 5

Rolling greater than 5 but less than 7

Kipish [7]3 years ago

4 0

3/10 or 30% I think since if they have 70% to win they would have 30% to lose

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Hoochie [10]

15.49 and 15.48 any number below 15.50 and above 15.45 will round up to 15.5

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4 years ago

PLZ HELP I DONT UNDERSTAND A password for a security system must consist of a eater followed by two digits. How many possible pa

Talja [164]

Answer:

7020

Step-by-step explanation:

The letter could be any 1 of the 26 and it could be placed in any 1 of 3 places so this makes 26*3 = 78 possibilities.

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= 10P2 = 10! / 8! = 10*9 = 90.

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3 years ago

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GenaCL600 [577]

Answer:

Step-by-step explanation:

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3 years ago

the letters of the word independent are written on individual card and the cards are put into a box a card is selected and then

puteri [66]

Answer:

<em>The probability of obtaining the letter p twice is 1/121</em>

Step-by-step explanation:

<u>Probability of Recurring Events</u>

There are 11 letters in the word 'independent', one of which is the letter 'p'.

When those letters are written on individual cards and they are put into a box, there are 11 different choices to pick at random.

This means the individual probability of getting a 'p' is:

$\displaystyle P_1=\frac{1}{11}$

The card is reinserted into the box, leaving the sample space unaltered, thus the second card has the same probability:

$\displaystyle P_2=\frac{1}{11}$

We'll use the multiplication rule. Just multiply the probability of the first event by the second.

$\displaystyle P=P_1*P_2=\frac{1}{11}*\frac{1}{11}$

$\displaystyle P=\frac{1}{121}$

The probability of obtaining the letter p twice is 1/121

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WITCHER [35]

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or as

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