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Sindrei [870]
3 years ago
15

A paper cup is dropped and it’s landing position is recorded. The cup can land in the side, on the open end, or on the closed en

d. The results of 20 trials are shown in the table below: based on the table, which of the following best compares and experimental probability of the cup landing on its open end with the experimental probability of the cup landing on its closed end?
A. The probabilities are equal
B. The probability of landing on the open end is greater
C. The probability of landing in the closed end is greater
D. No conclusion can be made

Mathematics
1 answer:
Likurg_2 [28]3 years ago
5 0

Answer:

The probability of landing in the closed end is greater, choice C.

Step-by-step explanation:

Experimental probability is evaluated as; the number of favorable outcomes/number of trials. Therefore, the experimental probability of the cup landing on its open end is given by; 5/20 = 0.25 while the experimental probability of the cup landing on its closed end is given by; 8/20 = 0.4 which is greater than 0.25.

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katen-ka-za [31]

Answer:

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Step-by-step explanation:

36/60 = 0.6

6/10=0.6

4 0
3 years ago
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zvonat [6]

Answer:

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Step-by-step explanation:

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5 0
3 years ago
Can anyone help me with the 7th and 8th question which I have taken a bad photo of?
Vsevolod [243]
<u>Question 7:

Define x :
</u>
Let the digit in the ones place be x.
The digit in the tens place is  x+2.

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<u>Find x:</u>

10(x + 2) + x + 10x + (x + 2) = 66
10x  + 20 + x + 10x + x + 2 = 66
22x + 22 = 66
22x = 44
x = 2

<u>Find the number:</u>

The number in the ones place = x= 2
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Answer : The number is 42.

<u>Question 8:</u>

<u>Define x:</u>

Let the digit in the tens place be x.
The digit in the ones place is x+7.

The number  = 10x + x + 7 = 11x + 7

If the digit interchange,
the number = 10(x + 7) + x = 10x + 70 + x = 11x + 70

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</u>
9(11x + 7) = 2(11x + 70)
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x = 1

<u>Find the number:
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x = 1
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Answer: The number is 18.

7 0
3 years ago
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nadya68 [22]

Answer:

A and B I believe

Step-by-step explanation:

6 0
3 years ago
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6 0
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