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coldgirl [10]
3 years ago
5

Math quick i need help hurry 15 points :)

Mathematics
1 answer:
schepotkina [342]3 years ago
4 0

Answer:

B

Step-by-step explanation:

I don't know how to explain it but

10/2=5 (Driver A)

20/2= 10 (Driver B)

Sorry if it's not right

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an online furniture store sells chairs for $50 each and tables for $300 each. Every day, the store can ship no more than 53 piec
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If 9 chairs were sold,  the store has to sell 11 tables in order to meet the requirements.

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3 years ago
Goals scored by a hockey team in successive matches are 5,7,4,2,4,0,5,5 and 3. What is the number of goals, the team must score
Stels [109]

Answer:

5

Step-by-step explanation:

Total goals /total no. of matches = 4

5+7+4+2+4+0+5+5+3+x / 10 = 4

35 + x = 4 × 10

x = 40 - 35

x = 5

So, the team must score 5 inorder to get the average of 4 goals per match.

6 0
2 years ago
What is 12x + 6y = 24 for ?
EastWind [94]
It is fore multiple things. if thats your quiestion. y can =4 and x=0, or x can = 2 and y=0.
6 0
3 years ago
Read 2 more answers
1. 7x = -x + 24<br> Solve for the variable
dedylja [7]
7x = -x + 24
+x. +x
8x = 24
X = 3
6 0
3 years ago
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Four cards are dealt from a standard fifty-two-card poker deck. What is the probability that all four are aces given that at lea
elena-s [515]

Answer:

The probability is 0.0052

Step-by-step explanation:

Let's call A the event that the four cards are aces, B the event that at least three are aces. So, the probability P(A/B) that all four are aces given that at least three are aces is calculated as:

P(A/B) =  P(A∩B)/P(B)

The probability P(B) that at least three are aces is the sum of the following probabilities:

  • The four card are aces: This is one hand from the 270,725 differents sets of four cards, so the probability is 1/270,725
  • There are exactly 3 aces: we need to calculated how many hands have exactly 3 aces, so we are going to calculate de number of combinations or ways in which we can select k elements from a group of n elements. This can be calculated as:

nCk=\frac{n!}{k!(n-k)!}

So, the number of ways to select exactly 3 aces is:

4C3*48C1=\frac{4!}{3!(4-3)!}*\frac{48!}{1!(48-1)!}=192

Because we are going to select 3 aces from the 4 in the poker deck and we are going to select 1 card from the 48 that aren't aces. So the probability in this case is 192/270,725

Then, the probability P(B) that at least three are aces is:

P(B)=\frac{1}{270,725} +\frac{192}{270,725} =\frac{193}{270,725}

On the other hand the probability P(A∩B) that the four cards are aces and at least three are aces is equal to the probability that the four card are aces, so:

P(A∩B) = 1/270,725

Finally, the probability P(A/B) that all four are aces given that at least three are aces is:

P=\frac{1/270,725}{193/270,725} =\frac{1}{193}=0.0052

5 0
3 years ago
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