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larisa [96]
3 years ago
14

A dealer selects a card at random from a standard deck of 52 cards.

Mathematics
1 answer:
Oksanka [162]3 years ago
3 0

Answer:

a) you are correct about a.

b)2/13 chance

c) 1/2 chance

please mark me as brainliest and give me a thanks and five stars please

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Multiply (2+3i)(4+6i)
DochEvi [55]

Answer:

-10+24i

hope that helps

7 0
3 years ago
Read 2 more answers
The data below represents the number of essays that students in Mr. Ji's class wrote. 2,3,5,5,6,7,8,8,11 Which box plot correctl
sweet-ann [11.9K]

Answer:

C

Step-by-step explanation:

2,3,5,5,<u>6</u>,7,8,8,11

6 is the middle line, 8 is the end of the box, and 4 is the beginning of the bow.

Hope this helps.

5 0
3 years ago
Rewrite the equation in Ax+By=C form.<br> Use integers for A, B, and C.<br> y-3=6(x-5)
Mama L [17]

Answer:

y-6x=-27

Step-by-step explanation:

simplify with distributive property

y-3=6x-30

add 3 to both sides

y=6x-27

take away 6x from both sides

y-6x=-27

P.S. please give brainliest

7 0
3 years ago
Question 31 pts Prove the statement is true using mathematical induction: 2n-1 ≤ n! Use the space below to write your answer. To
Sladkaya [172]

Answer:

P(3) is true since 2(3) - 1 = 5  < 3! = 6.  

Step-by-step explanation:

Let P(n) be the proposition that 2n-1 ≤ n!. for n ≥ 3

Basis: P(3) is true since 2(3) - 1 = 5  < 3! = 6.  

Inductive Step: Assume P(k) holds, i.e., 2k - 1 ≤ k! for an arbitrary integer k ≥ 3. To show that P(k + 1) holds:

2(k+1) - 1 = 2k + 2 - 1

≤ 2 + k! (by the inductive hypothesis)

= (k + 1)! Therefore,2n-1 ≤ n! holds, for every integer n ≥ 3.

4 0
3 years ago
Hank has eight action figures this is 12 less than the quotient of 60 and the number figures palo has
Maru [420]

Answer:

<em>6</em>

Step-by-step explanation:

<em>hold up hold up so you are saying 60 less than what quotient I'm guessing 68 divided by 12 so the answer would be 6</em>

5 0
3 years ago
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