Let's get the data out of the problem:
Nv = 18
Ov = 12
Now, we can apply in this equation that relates the new and the old values:
i = Nv / Ov
i = 18 / 12
i = 1.5
Now, we just consider what has passed from 1, and this is 0.5, converting to percentage it is:
p = 0.5 . 100
p = 50%
The increase was 50%.
Hope it helps.
It depends on your Divisor, Product, or Value. There is just not a gaven value for X or X=0
Answer:
For the sum of the digits of Michelle’s father to drop to 1/3 of its’ former value, the second digit must go from 9 to 0, in one year.
Thus, if he is 39 this year, the sum of the digits of his age will be 12. Next year, when the sum of the digits would be 4+0, or 4, then the digits of his age would be 1/3rd of their former value.
Therefore, he is 39, and will be 40 next year
Step-by-step explanation:
Answer:
15.87% probability that a randomly selected individual will be between 185 and 190 pounds
Step-by-step explanation:
Problems of normally distributed samples are solved using 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.
In this problem, we have that:

What is the probability that a randomly selected individual will be between 185 and 190 pounds?
This probability is the pvalue of Z when X = 190 subtracted by the pvalue of Z when X = 185. So
X = 190



has a pvalue of 0.8944
X = 185



has a pvalue of 0.7357
0.8944 - 0.7357 = 0.1587
15.87% probability that a randomly selected individual will be between 185 and 190 pounds
A unit rate where one is in the Denometer, so you would want to divide the bottom by 10.