Kindly refer to attachment for solution.
Hope it helps ^_^
Answer:
C. unlikely
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
A probability is said to be extremely likely if it is 95% or higher, and extremely unlikely if it is 5% or lower. A probabilty higher than 50% and lower than 95% is said to be likely, and higher than 5% and lower than 50% is said to be unlikely.
In this problem, we have that:

How likely is it that a single survey would return a mean of 30%?
We have to find the pvalue of Z when X = 0.30.



has a pvalue of 0.1587.
So the correct answer is:
C. unlikely
You may recall that you can't take the square root of a negative number (if you want a real number). So, the area under the square root, also known as the radicand, must be zero or positive. This is how we restrict the domain.
We write the inequality

.
B
R=S*0.5^(t/8)
<span>R is the remaining amount </span>
<span>S is the starting amount (500) </span>
<span>0.5^ is for the HALF in half-life </span>
<span>t/8 show that every 8 ts (every 8 hours), it will be halved once </span>
<span>...so plug in 500mg for the general solution... </span>
<span>R=(500)*(0.5)^(t/8) </span>
<span>... plug in 24h to solve for after 24h </span>
<span>R=(500)*(0.5)^(24/8) </span>
<span>R=(500)*(0.5)^(3) </span>
<span>R=(500)*(0.125) </span>
<span>R=(0.0625) </span>
<span>...therefore there with be 0.0625 mg of the dose remaining</span>