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Dovator [93]
3 years ago
10

A soccer team estimates that they will score on 77​% of the corner kicks. In next​ week's game, the team hopes to kick 1414 corn

er kicks. What are the chances that they will score on 22 of those​ opportunities?
Mathematics
1 answer:
andrezito [222]3 years ago
6 0

Answer:

18.66521%

Step-by-step explanation:

It seems all of the numbers are duplicated, the correct number should be 7%, 14 corner kicks, and 2 opportunities.  

The team has 7%(x=0.07) chance to score so that means the chance to not scoring will be: y= 1-x = 100%-7%= 93%. There are 14 opportunities and we want to know the probability to get exactly 2 scores. The calculation will be:

P(x=2)= 2C14 * x^2 * y^12

P(x=2)=91 * 0.07^2 * 0.93^12= 18.66521%

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2 years ago
There are 20 sweets in a box.
natima [27]

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probability \: of \: getting \: red \: sweets  =  \frac{x}{20}  \\ probability \: of \: getting \: yellow \: sweets \:  =  \frac{(20 - x)}{20}

5 0
2 years ago
A copy machine makes 24 copies per minute. how many copies does it make in 5 minutes and 30 seconds
enot [183]
In 1 minute the copy machine copies = 24
That means
In 60 seconds the copy machine copies = 24
First we need to convert 5 minutes and 30 seconds to seconds
Then
5 minutes and 30 seconds = (5 * 60) + 30
                                           = 300 + 30
                                           = 330 seconds
So
In 330 seconds the copy machine will copy = (24/60) * 330
                                                                     = 4 * 33
                                                                     = 132
So in 5 minutes and 30 seconds the copy machine will copy 132 copies.
6 0
3 years ago
Read 2 more answers
How do we use unit rate on every day life? What does unit rate show
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5 0
3 years ago
the numbers 1,2,3,4, and 5 are written on slips of paper, and 2 slips are drawn at random one at a time without replacet. find t
Tresset [83]

Consider such events:

A - slip with number 3 is chosen;

B - the sum of numbers is 4.

You have to count Pr(A|B).

Use formula for conditional probability:

Pr(A|B)=\dfrac{Pr(A\cap B)}{Pr(B)}.

1. The event A\cap B consists in selecting two slips, first is 3 and second should be 1, because the sum is 4. The number of favorable outcomes is exactly 1 and the number of all possible outcomes is 5·4=20 (you have 5 ways to select 1st slip and 4 ways to select 2nd slip). Then the probability of event A\cap B is

Pr(A\cap B)=\dfrac{1}{20}.

2. The event B consists in selecting two slips with the sum 4. The number of favorable outcomes is exactly 2 (1st slip 3 and 2nd slip 1 or 1st slip 1 and 2nd slip 3) and the number of all possible outcomes is 5·4=20 (you have 5 ways to select 1st slip and 4 ways to select 2nd slip). Then the probability of event B is

Pr(B)=\dfrac{2}{20}=\dfrac{1}{10}.

3. Then

Pr(A|B)=\dfrac{\frac{1}{20} }{\frac{1}{10} }=\dfrac{1}{2}.

Answer: \dfrac{1}{2}.

5 0
3 years ago
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