For this case we have:
Let x be the variable that belongs to the real numbers. Then, all reals less than 70 can be expressed as:

The tip of the inequality is directed to the real numbers, since they tell us that they are less than 70, for 70 the inequality remains open.
Answer:

<em>n</em> must be 0, since <em>x </em>ⁿ = 1 for all positive, real <em>x</em> if <em>n</em> = 0. So the answer is A.
Answer:
4.39% theoretical probability of this happening
Step-by-step explanation:
For each coin, there are only two possible outcomes. Either it lands on heads, or it lands on tails. The probability of a coin landing on heads is independent of other coins. So we use the binomial probability distribution 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.
Theoretically, a fair coin
Equally as likely to land on heads or tails, so 
10 coins:
This means that 
What is the theoretical probability of this happening?
This is P(X = 2).


4.39% theoretical probability of this happening
In statistics, a Chi-squared test may be used to determine holiday choice and gender and α (alpha) is the response variable.
<h3>What is the Chi-squared test? </h3>
A statistical technique called the chi-square test is used to compare actual outcomes with predictions.
The goal of this test is to establish if a discrepancy between actual and predicted data is the result of chance or a correlation between the variables you are researching.
Whether there is a statistically significant association between categorical variables is determined by the Chi-square test of independence.
This issue is addressed by a hypothesis test. The chi-square test of association is another name for this assessment.
Hence,in stats would a test looking at gender & holiday preference yes you can do a Chi-squared test and α(alpha) is the response variable.
To learn more about the Chi-squared test refer;
brainly.com/question/14082240
#SPJ1