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yuradex [85]
3 years ago
12

What is eight dozen in standard form?

Mathematics
1 answer:
Amiraneli [1.4K]3 years ago
8 0
Assuming 8 dozen is 96 not 8000000000000 it would be 9.6x10¹. If it is 8000000000000 then it's 8x10¹²
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Please help will give brainliest it’s for a test<br><br> Round 29.20 to 30 <br> Show work please
djverab [1.8K]

Answer:

you cant round that to 30

Step-by-step explanation

29.20

29.2 2 is less than 5

so you cant round you can only round if the .tens place is 5 or more

4 0
2 years ago
How do you find the pattern of 0,2,6,12,20???
professor190 [17]
0x1=0
1x2=2
2x3=6
3x4=12
4x5=20
3 0
3 years ago
Which expressions are equivalent to 2-(-6+3)+4c
Artyom0805 [142]

Answer:

c. none of the above

Step-by-step explanation:

(-6+3)= -3

2--3=2+3=5

5+4c cant add them since they arent like terms

final answer 5+4c and that option isnt here

3 0
2 years ago
Please give me an explanation and your work
pishuonlain [190]

Answer:

-0.3*0.2*12, multiply -0.3 by 0.2 to get -0.06, now multiply 12 by it to get, -0.72.

Step-by-step explanation:

8 0
3 years ago
Suppose we have 3 cards identical in form except that both sides of the first card are colored red, both sides of the second car
Nikolay [14]

Answer:

probability that the other side is colored black if the upper side of the chosen card is colored red = 1/3

Step-by-step explanation:

First of all;

Let B1 be the event that the card with two red sides is selected

Let B2 be the event that the

card with two black sides is selected

Let B3 be the event that the card with one red side and one black side is

selected

Let A be the event that the upper side of the selected card (when put down on the ground)

is red.

Now, from the question;

P(B3) = ⅓

P(A|B3) = ½

P(B1) = ⅓

P(A|B1) = 1

P(B2) = ⅓

P(A|B2)) = 0

(P(B3) = ⅓

P(A|B3) = ½

Now, we want to find the probability that the other side is colored black if the upper side of the chosen card is colored red. This probability is; P(B3|A). Thus, from the Bayes’ formula, it follows that;

P(B3|A) = [P(B3)•P(A|B3)]/[(P(B1)•P(A|B1)) + (P(B2)•P(A|B2)) + (P(B3)•P(A|B3))]

Thus;

P(B3|A) = [⅓×½]/[(⅓×1) + (⅓•0) + (⅓×½)]

P(B3|A) = (1/6)/(⅓ + 0 + 1/6)

P(B3|A) = (1/6)/(1/2)

P(B3|A) = 1/3

5 0
3 years ago
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