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prohojiy [21]
3 years ago
7

90,175,087,309 in expanded form

Mathematics
2 answers:
NemiM [27]3 years ago
8 0
90,000,000,000
+ 0
+ 100,000,000
+ 70,000,000
+ 5,000,000
+ 0
+ 80,000
+ 7,000
+ 300
+ 0
+ 9

Expanded Factors Form:
9 × 10,000,000,000
+ 0 × 1,000,000,000
+ 1 × 100,000,000
+ 7 × 10,000,000
+ 5 × 1,000,000
+ 0 × 100,000
+ 8 × 10,000
+ 7 × 1,000
+ 3 × 100
+ 0 × 10
+ 9 × 1

Expanded Exponential Form:
9 × 1010
+ 0 × 109
+ 1 × 108
+ 7 × 107
+ 5 × 106
+ 0 × 105
+ 8 × 104
+ 7 × 103
+ 3 × 102
+ 0 × 101
+ 9 × 100

Word Form:

ninety billion, one hundred seventy-five million, eighty-seven thousand, three hundred nine
lorasvet [3.4K]3 years ago
3 0
90,000,000,000 + 0 + 100,000,000 + 70,000,000 + 5,000,000 + 0 + 80,000 + 7,000 + 300 + 0 + 9
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Given:

30-hour review course average a score of 620 on that exam.

70-hour review course average a score of 749.

To find:

The linear equation which fits this data, and use this equation to predict an average score for persons taking a 57-hour review course.

Solution:

Let x be the number of hours of review course and y be the average score on that exam.

30-hour review course average a score of 620 on that exam. So, the linear function passes through the point (30,620).

70-hour review course average a score of 749. So, the linear function passes through the point (70,749).

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y-620=\dfrac{749-620}{70-30}(x-30)

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y-620=\dfrac{129}{40}(x)-\dfrac{129}{40}(30)

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Adding 620 on both sides, we get

y=\dfrac{129}{40}x-\dfrac{387}{4}+620

y=\dfrac{129}{40}x+\dfrac{2480-387}{4}

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We need to find the y-value for x=57.

y=\dfrac{129}{40}(57)+\dfrac{2093}{4}

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y\approx 707.1

Therefore, the required linear equation for the given situation is y=\dfrac{129}{40}x+\dfrac{2093}{4} and the average score for persons taking a 57-hour review course is 707.1.

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