Construct a 96% confidence interval if a sampling distribution has a mean of 20, standard deviation of 5, and size of 100?
1 answer:
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Answer: 70.909090909091%
Step-by-step explanation: Reading through the problem we have <em>78 is</em>, that's 78 equals, <em>what percent</em>, x/100, <em>of 110</em>, times 110.
It's important to understand that <em>percent</em> means over 100 so what percent would simply mean <em>x/100</em> or any variable but I will be using x.
So we have the equation
So cross canceling the 110 and 100 to 11 and 10, we have
Multiply both sides by 10 to get rid
of the fraction and we have 780 = 11x.
Now divide both sides by 11 and 70.909090909091 = x.
So, 78 is 70.909090909091% of 110.
Work is attached in the image provided.
Answer:
Step-by-step explanation:
This scenario can be modeled using an exponential growth equation.
The exponential growth equations have the following form:
Where P is the population in year t
p is the initial population at t = 0
r is the growth rate
t is the time in years.
In this case we know that the current population is 13,000 and that the growth rate is 11%
So
The equation that models this scenario is:
Answer:
<u>Use tangent:</u>
- BL/BS = tan S
- x / 144 = tan 25°
- x = 144 tan 25°
- x = 67 ft (rounded)
