7.62 a^2+b^2=c^2 have a great day
Answer:
0.8041 = 80.41% probability that a given battery will last between 2.3 and 3.6 years
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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.
A certain type of storage battery lasts, on average, 3.0 years with a standard deviation of 0.5 year
This means that 
What is the probability that a given battery will last between 2.3 and 3.6 years?
This is the p-value of Z when X = 3.6 subtracted by the p-value of Z when X = 2.3. So
X = 3.6



has a p-value of 0.8849
X = 2.3



has a p-value of 0.0808
0.8849 - 0.0808 = 0.8041
0.8041 = 80.41% probability that a given battery will last between 2.3 and 3.6 years
For the half-life of a certain substance, the equation that would best represent the scenario is,
At = Ao(0.5^t/n)
where At is the amount at any time t, Ao is the original amount, and n is half-life. Substituting the known values,
At = (220 mg)(0.5^(396/72)
At = 4.86 mg
Thus, after 396 days, there will only be 4.86 mg of Terbium-160.
Answer: the radius is 1/2 times the diameter
Step-by-step explanation:
sorry if wrong
Step-by-step explanation:
divide 45 into the number dominique can type and then subtract 16 minutes.