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ivann1987 [24]
3 years ago
8

I need help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Maksim231197 [3]3 years ago
7 0

Answer:

c) I think'

Step-by-step explanation:

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You write each of the 26 letters of the alphabet on separate index cards. If you choose 3 cards at random without replacing them
natali 33 [55]

Answer:

1/15600

Step-by-step explanation:

This is a great example of what people call the Multiplication Principle.

First, find out how many ways the cards can be chosen.  You're picking 3 cards from a set of 26 without replacement.  The number of choices for the first card is 26, for the second card 25, for the third card 24.  The cards can be chosen in 26 x 25 x 24 = 15600 ways.

Now for the specific choice "A B C" the number of ways that can happen is

1 x 1 x 1  (one choice for the first card, 1 for the 2nd, 1 for the 3rd)

The probability is 1/15600

8 0
3 years ago
Write using only positive exponents. 36x^-4y^2z^0/5x^2y^-3z^-2
mamaluj [8]
36y²(1)y³z²
_________
x⁴5x²

36y^5z² / 5x^6
4 0
3 years ago
Help what's is 16÷ 264
Jet001 [13]
The answer is 16 remadier 5

4 0
3 years ago
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How is the number five and two hundred three thousandths written in standard form?
Andru [333]

Answer:

0.05203*10 to the power 2

Step-by-step explanation:

To get the standard form you divide a number in order to multiply it by a power of 10 to get the same number. So 5.203 in standard form is o.o5203 * 10 to the power 2. The 10 to the power 2 comes from the number of palces you move the comma/ decimal point to the left

HOPE IT HELPED

5 0
3 years ago
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A padlock has a four-digit code that includes digits from 0 to 9, inclusive. What is the probability that the code does not cons
Eduardwww [97]
Since there is no repetition allowed, there are 10 possibilities for the 1st digit, 9 for the 2nd, 8 for the 3rd, and 7 for the 4th. This gives a total of (10)(9)(8)(7) = 5040 four-digit codes.
For all odd digits to be used, there are 5 possibilities for the 1st digit (1,3,5,7,9), 4 for the 2nd, 3 for the 3rd, 2 for the 4th. This gives a total of (5)(4)(3)(2) = 120 codes that only use odd digits.
Therefore there are 5040 - 120 = 4920 codes that do not consist of all odd digits. The probability is 4920/5040 = 41/42.

4 0
3 years ago
Read 2 more answers
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