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Nikolay [14]
3 years ago
14

Compare each whole number in the following pairs. Insert the symbol < or > to make the statement true. a. 15 _____ 8 b. 4

_____ 134 c. 3 _____ 5 d. 351 _____ 215 Replace each ? with the symbol = or ≠ to make the statement true. a. 54 ? 53 b. 8 ? 4 + 4 c. 23 + 6 ? 6 + 23 d. 64 ? 99 Compare the fraction by using the symbol < , <, or =. a. 4⁄9 _____ 5⁄9 b. 4⁄5 _____ 8⁄10 c. 7⁄12 _____ 7⁄15 d. 3⁄4 _____ 6⁄7 Compare the decimals by using the symbol < or >. a. 0.531 _____ 0.456 b. 0.63 _____ 0.47 c. 0.267 _____ 0.539 d. 0.07 _____ 0.7 In each of the following pairs of numbers, how much larger is the first number than the second? a. 154 and 27 b. 25 and 12 c. 135 and 127 d. 46 and 24 Phyllis has finally found the TV she wants to buy. At Store A, the TV costs $498. At Store B, the same style TV costs $425. How much will Phyllis save by buying the TV at Store B? Both Dale and Chan went running for an hour. Dale ran 5 miles and Chan ran 2.5 miles. How did Dale’s distance compare to Chan’s? (Use division to compare.) Compare the numbers in the following sequence and try to determine what the next number should be. 6, 9, 18, 21, 42, 45, ?
Mathematics
1 answer:
katrin2010 [14]3 years ago
3 0
Insert the symbol < or > to make the statement true.
a) >
b) <
c) <
d) >

Replace each ? with the symbol = or ≠ to make the statement true.
a) ≠
b) =
c) =
d) ≠
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p =  -  \frac{1}{4}

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xeze [42]

Answer:

The probability is   \frac{6}{91} ≅ 0.0659

Step-by-step explanation:

Let's analyze the question.

There are 15 students in the 8th grade.

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We are looking for the probability that Trevor, Terry and Evan will be in the same algebra class.

One possible way to solve this question is to think about the product probability rule.

We can use it because we are in an equiprobable space. (And  also the events are independent).

Let's set for example a class for Evan.

The probability that Evan will be in a class is 1

Then for Terry there are 4 places out of 14 that puts Terry in the Evan's class.

We write 1.\frac{4}{14}

Finally for Trevor there are 3 places out of the remaining 13 that puts Trevor in the same class with Evan and Terry.

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1.\frac{4}{14}.\frac{3}{13}=\frac{6}{91}

The probability of the event is \frac{6}{91} ≅ 0.0659

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4 years ago
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