There is an average of 3.5×10−2 kilograms of dissolved salt in each liter of seawater. The South China Sea contains approximatel
y 9.88×1018 liters of seawater. How many kilograms of dissolved salt are in the South China Sea? Enter your answer, in scientific notation, in the boxes.
1 answer:
Answer:
3.458×10¹⁷ kg
Step-by-step explanation:
(3.5×10⁻² kg/L) × (9.88×10¹⁸ L) = (3.5×9.88) × 10⁽⁻²⁺¹⁸⁾ kg
... = 34.58×10¹⁶ kg
... = 3.458×10¹⁷ kg
_____
<em>Comment on significant digits</em>
It would be appropriate to round the answer to 2 significant digits, matching the number of significant digits in the least-precise factor. Then the answer would be 3.5×10¹⁷ kg.
_____
<em>Comment on scientific notation problems</em>
Your scientific calculator can show results in scientific notation.
You might be interested in
Answer:
0.0314 = 3.14% probability that it takes at most 1 hour of machining time to produce a randomly selected component
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.
Sum of normal variables:
When normal variables are added, the mean is the sum of the means while the standard deviation is the square root of the sum of the variances.
The mean values are 20, 15, and 30 min, respectively:
This means that
The standard deviations are 1, 2, and 1.5 min, respectively.
This means that
What is the probability that it takes at most 1 hour of machining time to produce a randomly selected component?
Mean and standard deviation are in minutes, so this is the pvalue of Z when X = 60.
has a pvalue of 0.0314
0.0314 = 3.14% probability that it takes at most 1 hour of machining time to produce a randomly selected component
