The formula for the future value VV (in dollars) of an investment earning simple interest is V=p+prtV=p+prt, where pp (in dollar
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Answer:
68% of the rates are expected to be betwen 0.718 and 0.762
95% of the rates are expected to be betwen 0.696 and 0.784
99.7% of the rates are expected to be betwen 0.674 and 0.806
Step-by-step explanation:
Check for conditions
For this case in order to use the normal distribution for this case or the 68-95-99.7% rule we need to satisfy 3 conditions:
a) Independence : we assume that the random sample of 400 students each student is independent from the other.
b) 10% condition: We assume that the sample size on this case 400 is less than 10% of the real population size.
c) np= 400*0.74= 296>10
n(1-p) = 400*(1-0.74)=104>10
So then we have all the conditions satisfied.
Solution to the problem
For this case we know that the distribution for the population proportion is given by:
So then:
The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).
68% of the rates are expected to be betwen 0.718 and 0.762
95% of the rates are expected to be betwen 0.696 and 0.784
99.7% of the rates are expected to be betwen 0.674 and 0.806
See the attached figure to better understand the problem
we know that
tan 29°=AC/CB--------> CB=AC/tan 29°
AC=110 ft
CB--------> <span>how far the man is from the helicopter landing pad
</span>so
CB=AC/tan 29°----------> CB=110/ tan 29°----------> CB=198.45 ft
the answer is
the man is 198.45 ft from the helicopter landing pad
we know that
tan 29°=AC/CB--------> CB=AC/tan 29°
AC=110 ft
CB--------> <span>how far the man is from the helicopter landing pad
</span>so
CB=AC/tan 29°----------> CB=110/ tan 29°----------> CB=198.45 ft
the answer is
the man is 198.45 ft from the helicopter landing pad