A = P(1+r)^(t/5) A = 500000(1+0.05)^(15/5) A = 500000(1.05)^(15/5) A = 500000(1.05)^3 A = 500000*1.157625 A = 578812.5 Telling us that the population will be about 578,812 people in the year 2005
this is really 35 percent
The law of cosines is used for calculating one side of the triangle if the angle opposite to that side is given as well as the other sides. For this problem, we are given angle B. Therefore, the correct answer among the choices given is option 1 where a is equal to 5 and c is equal to 3.
Answer:
The proportion of children in this age range between 70 lbs and 85 lbs is of 0.9306.
Step-by-step explanation:
When the distribution is normal, we use 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 study suggested that children between the ages of 6 and 11 in the US have an average weightof 74 lbs, with a standard deviation of 2.7 lbs.
This means that
What proportion of childrenin this age range between 70 lbs and 85 lbs.
This is the pvalue of Z when X = 85 subtracted by the pvalue of Z when X = 70. So
X = 85
has a pvalue of 1
X = 70
has a pvalue of 0.0694
1 - 0.0694 = 0.9306
The proportion of children in this age range between 70 lbs and 85 lbs is of 0.9306.