A simple random sample of 100 8th graders at a large suburban middle school indicated that 84% of them are involved with some type of after school activity. Find the 98% confidence interval that estimates the proportion of them that are involved in an after school activity.

3x + 4y -25 =0 represents the amount of change sianna should recieve

where is the graph? need more info

ending amount = beginning amount / (2^n)

where "n" = number of half-lives (or elapsed time / half-life)

ending amount = 50 mg / (2 ^10 / 5.3)

ending amount = 50 mg / (2 ^ 1.8867924528)

ending amount = 50 m / 3.6981210597

ending amount = 13.5203794557 mg

or about 13.5 mg remain

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