Aleading dental journal claims that 63% of young adults do not brush their teeth regularly. a dental hygienist wants to conduct a survey to verify this with her young adult patients, and wants to do so with a margin of error (me) of ± 9%. what minimum sample size, rounded to the nearest whole person, does the hygienist need to use?

Margin of error = Z*Sqrt [p(1-p)/n]

Where
Z = Z-value at a given confidence interval (assume 1.96 for 95%)
p = point estimate = 63% = 0.63
n = sample size

Substituting;
0.09 = 1.96* Sqrt [0.63(1-0.63)/n]
(0.09/1.96)^1 = 0.63(1-0.63)/n = 0.2331/n
n = 0.2331/[(0.09/1.96)^2] = 110.55 ≈ 111

Therefore, minimum sample at 95% confidence interval is 111 young adults.

Answer for 2: his profit would be $2.88 explanation: you have to first multiply 72 and 0.12 together to get $8.64 once you do that you know how much money the shopkeeper made from selling the candy bars individually. so you have to do 8.64-5.76=2.88 hope that ; )answer for 3: yeah i’m sorry i have no idea what number 3 is. i’m sorry i couldn’t with that.

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