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sergey [27]
3 years ago
14

2.Solve x squared - 4x by Factoring

Mathematics
1 answer:
777dan777 [17]3 years ago
8 0

Answer:

Step-by-step explanation:

hello:

x²-4x=0  means : x(x-4)=0

x=0 or x-4=0

conclusion : x=0 and x= 4   ...answer : B

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The price of an item has been reduced by <br> 45% <br> . The original price was <br> $17
Artist 52 [7]
Price is reduced by 45%. Which means that the new price is 55% of the original.
Therefore, (55/100)*17 = $9.35 =new price
8 0
4 years ago
A city hosted a music festival that included three concerts. According to the sales database, 28% of the audience attended the f
Vitek1552 [10]

Answer:

P(X\cap Y\cap Z)=0.05

Step-by-step explanation:

From the question we are told that:

Percentage of audience in first concert P(X)=0.28

Percentage of audience in second concert P(Y)=0.42

Percentage of audience in third concert P(Z)=0.30

Audience Percentage at at-least one concert P(X \cup Y \cup Z)=0.80

Percentage of audience at first & second concert P(X \cap Y)=0.10

Percentage of audience in first & third concert P(X \cap Z)=0.08

Percentage of audience in second & third concert P(Y\cap Z)=0.07

 

Generally the equation for probability of attending all concerts P(X\cap Y\cap Z)is mathematically given by

P(X \cup Y \cup Z)=P(X)+P(Y)+P(Z)-P(X \cap Y)-P(X \cap Z)-P(Y\cap Z)+P(X\cap Y\cap Z)

P(X\cap Y\cap Z)=P(X \cup Y \cup Z)-P(X)-P(Y)-P(Z)+P(X \cap Y)+P(X \cap Z)+P(Y\cap Z)

P(X\cap Y\cap Z)=0.80-0.28-0.42-0.30+0.10+0.80+0.70

P(X\cap Y\cap Z)=0.05

Therefore the probability that a randomly selected audient attended all the concerts

P(X\cap Y\cap Z)=0.05

8 0
3 years ago
Use the Distributive Property on the<br> equation below.<br> 3(4y + 3) = [ ? ]y +[ ]
olga2289 [7]

Answer:

12y+9

Step-by-step explanation:

4 0
3 years ago
You are dealt 5 cards from a standard deck of 52 playing cards. In how many ways can you get.....
tatiyna

Answer:

Probability of Four of a Kind

The probability of being dealt four of a kind is 0.0002400960384. On average, four of a kind is dealt one time in every 4,165 deals.

Probability of Two Pair

On average, players get two pair about one time in every 21 deals.

Step-by-step explanation:

Probability of Four of a Kind

Let's execute the analytical plan described above to find the probability of four of a kind.

First, we count the number of five-card hands that can be dealt from a standard deck of 52 cards. This is a combination problem. The number of combinations is n! / r!(n - r)!. We have 52 cards in the deck so n = 52. And we want to arrange them in unordered groups of 5, so r = 5. Thus, the number of combinations is:

52C5 = 52! / 5!(52 - 5)! = 52! / 5!47! = 2,598,960

Hence, there are 2,598,960 distinct poker hands.

We count the number of ways that five cards can be dealt to produce four of a kind. It requires three independent choices to produce four of a kind:

Choose the rank of the card that appears four times in the hand. A playing card can have a rank of 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king, or ace. For four of a kind, we choose 1 rank from a set of 13 ranks. The number of ways to do this is 13C1.

Choose one rank for the fifth card. There are 12 remaining ranks, from which we choose one. The number of ways to do this is 12C1.

Choose a suit for the fifth card. There are four suits, from which we choose one. The number of ways to do this is 4C1.

The number of ways to produce four of a kind (Num4) is equal to the product of the number of ways to make each independent choice. Therefore,

Num4 = 13C1 * 12C1 * 4C1 = 13 * 12 * 4 = 624

Conclusion: There are 624 different ways to deal a poker hand that can be classified as four of a kind.

Finally, we compute the probability. There are 2,598,960 unique poker hands. Of those, 624 are four of a kind. Therefore, the probability of being dealt four of a kind (P4) is:

P4 = 624 / 2,598,960 = 0.0002400960384

The probability of being dealt four of a kind is 0.0002400960384. On average, four of a kind is dealt one time in every 4,165 deals.

Probability of Two Pair

To find the probability for two pair, we execute the same analytical plan that we've used to compute the other probabilities.

There are 2,598,960 distinct poker hands.

Next, count the number of ways that five cards can be dealt to produce two pair. It requires five independent choices to produce two pair:

Choose the rank for cards of matching rank. A playing card can have a rank of 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king, or ace. For two pair, we choose 2 ranks from a set of 13 ranks. The number of ways to do this is 13C2.

Choose the rank of the remaining non-matching card. There are 11 remaining ranks, from which we choose one. The number of ways to do this is 11C1.

Choose suits for the first two-card combination. There are four suits, from which we choose two. The number of ways to do this is 4C2.

Choose suits for the second two-card combination. There are four suits, from which we choose two. The number of ways to do this is 4C2.

Choose a suit for the non-matching card. There are four suits, from which we choose one. The number of ways to do this is 4C1.

The number of ways to produce two pair (Numtp) is equal to the product of the number of ways to make each independent choice. Therefore,

Numtp = 13C2 * 11C1 * 4C2 * 4C2 * 4C1

Numtp = 78 * 11 * 6 * 6 * 4 = 123,552

Finally, compute the probability. There are 2,598,960 unique poker hands. Of those, 123,552 are two pair. Therefore, the probability of being dealt two pair (Ptp) is:

Ptp = 123,552 / 2,598,960 = 0.04753901561

On average, players get two pair about one time in every 21 deals.

5 0
3 years ago
Solve the two step equation to find the value of d: 30=-5d-5
mylen [45]
D is equal to 7 because 7x5=35-5=30
5 0
3 years ago
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