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umka2103 [35]
3 years ago
15

In a dance competition, a participant has to score a total of at least 30 points in the first four rounds combined to move on to

the fifth and final round. Steward scored 5 points in the first round. He then went on to score additional points in the second, third, and fourth rounds. In each of those rounds, his score was identical. Which inequality best shows the number of points, p, that Steward scored in each of the second, third, and fourth rounds if he earned a place in the finals?
5 + 3p ≥ 30
5 + 3p ≤ 30
5p + 3 ≥ 30
5p + 3 ≤ 30
Mathematics
1 answer:
77julia77 [94]3 years ago
3 0
<span>So Steward scored 5 for the first round, that will be our constant. There is 3 more rounds for him to score at least 30. at least tells us this is a greater than or equal to 5+3p<30</span>

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Consider the linear expression and a first step of simplification shown below.
wlad13 [49]

Answer:

Commutative Property

Step-by-step explanation:

From step 1 to step 2, we are changing the order, that means we are using the Commutative Property.  The Commutative Property means to move around, or a+b = b+a


6 0
3 years ago
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

​​

8 0
3 years ago
5x - y ÷ 8 when x = 5 and y = 8
stiks02 [169]

Answer:

24

Step-by-step explanation:

5x - y ÷ 8

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PEMDAS

multiply and divide from left to right

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6 0
3 years ago
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3 hope that can help you
4 0
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WITCHER [35]
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4 0
4 years ago
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