He hold account for 5 years or 5 months. Depending on 3% per year or 3% per month
Okay, so you didn't really put anything for me to know what you wanted to solve, so I'm going to assume this is Scientific Notation:
We have:
10^5 = 100,000 x 4.84 would be: 484,000
10^2 = 100 x 3.9 would be: 390
10^7 = 10,000,000 x 8.91 would be: 89,100,000
Answer:
2b^2
Step-by-step explanation:
according to the question equation is
(a^2 + b^2 ) - (a^2 - b^2)
a^2 + b^2 - a^2 + b^2
plus a square and minus a square gets cancel
b^2 + b^2
since they are like terms they can be added
2b^2
Answer:
<h2>download photomath :»</h2>
Step-by-step explanation:
may scan po don
Answer:
0.9772 = 97.72% probability that a randomly selected firm will earn more than Arc did last year
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be 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.
Suppose the mean income of firms in the same industry as Arc for a year is 90 million dollars with a standard deviation of 7 million dollars.
This means that 
What is the probability that a randomly selected firm will earn more than Arc did last year?
Arc earned 76 million, so this is 1 subtracted by the pvalue of Z when X = 76.



has a pvalue of 0.0228
1 - 0.0228 = 0.9772
0.9772 = 97.72% probability that a randomly selected firm will earn more than Arc did last year