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Natasha_Volkova [10]
3 years ago
15

Suzy's soccer team has played 6 games this year. The mode of the number of goals her team scored is 2 goals. How will the mode b

e affected if Suzy's team scores 2 goals in the seventh game?
A
The mode will decrease.

B
The mode will increase.

C
The mode will not change.

D
There is not enough information.

Mathematics
2 answers:
goldfiish [28.3K]3 years ago
8 0

Answer:

I am not 100% sure but I think it will increase.

Step-by-step explanation:

DaniilM [7]3 years ago
5 0
C THE MODE WILL NOT CHANGE BECAUSE IT WILL STILL BE 2 . BRAINLIEST ???
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Suppose we have three urns, namely, A B and C. A has 3 black balls and 7 white balls. B has 7 black balls and 13 white balls. C
professor190 [17]

Answer:

a. 11/25

b. 11/25

Step-by-step explanation:

We proceed as follows;

From the question, we have the following information;

Three urns A, B and C contains ( 3 black balls 7 white balls), (7 black balls and 13 white balls) and (12 black balls and 8 white balls) respectively.

Now,

Since events of choosing urn A, B and C are denoted by Ai , i=1, 2, 3

Then , P(A1 + P(A2) +P(A3) =1 ....(1)

And P(A1):P(A2):P(A3) = 1: 2: 2 (given) ....(2)

Let P(A1) = x, then using equation (2)

P(A2) = 2x and P(A3) = 2x

(from the ratio given in the question)

Substituting these values in equation (1), we get

x+ 2x + 2x =1

Or 5x =1

Or x =1/5

So, P(A1) =x =1/5 , ....(3)

P(A2) = 2x= 2/5 and ....(4)

P(A3) = 2x= 2/5 ...(5)

Also urns A, B and C has total balls = 10, 20 , 20 respectively.

Now, if we choose one urn and then pick up 2 balls randomly then;

(a) Probability that the first ball is black

=P(A1)×P(Back ball from urn A) +P(A2)×P(Black ball from urn B) + P(A3)×P(Black ball from urn C)

= (1/5)×(3/10) + (2/5)×(7/20) + (2/5)×(12/20)

= (3/50) + (7/50) + (12/50)

=22/50

=11/25

(b) The Probability that the first ball is black given that the second ball is white is same as the probability that first ball is black (11/25). This is because the event of picking of first ball is independent of the event of picking of second ball.

Although the event picking of the second ball is dependent on the event of picking the first ball.

Hence, probability that the first ball is black given that the second ball is white is 11/25

​​

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A leaky pool loses 2 and one-fourth inches of water each day. Jermaine and Mildred are asked to determine the change in the amou
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Answer:

  Mildred is correct because she distributed 5 correctly

Step-by-step explanation:

We cannot tell exactly which solution goes with which student, but we believe you are saying ...

Mildred’s Solution:

  5 (Negative 2 and one-fourth) = 5 (negative 2 minus one-fourth)

  = negative 10 minus StartFraction 5 over 4 EndFraction

  = negative 11 and one-fourth

This solution describes the correct distribution of 5. Whoever it belongs to gave the correct solution.

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4 years ago
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