If the sample probability of an individual traveling to the statue is 25%, the number of trials is and the number of successful trials is . The probability that exactly 7 out of 10 individuals will visit the Statue of Liberty is .

0.0031 = 0.31% probability that exactly 7 out of 10 individuals will visit the Statue of Liberty

Step-by-step explanation:

For each individual, there are only two possible outcomes. Either they visit the Statue of the Liberty, or they do not. The probability of an individual visiting the statue is independent of any other individual. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.

The sample probability of an individual traveling to the statue is 25%

This means that

The probability that exactly 7 out of 10 individuals will visit the Statue of Liberty is

This is when . So

0.0031 = 0.31% probability that exactly 7 out of 10 individuals will visit the Statue of Liberty

1gm = 1000mg so 0.1gm = 1000mg

first, convert the weight to kg

160lb * 1kg/2.2lb = 72.7kg

next, the dose

72.7kg * 100mg/kg = 7270mg

the 2 figures are congruent bc if you reflect abcd over the y axis then over the d axis the figures overlap