Evaluate |-6| + 3|4| + |-5| - 0 × 6.
2 answers:
Hey there!
<h2>|-6| +3|4| + |-5| - 0(6) </h2><h2>6+ 3(4) + |-5| - 0(6) </h2><h2>3(4) = 12 </h2><h2>6+12+|-5| -0(6) </h2><h2>6 + 12 = 18 </h2><h2>18 + |-5| - 0(6) </h2><h2>18 + 5 - 0(6) </h2><h2>18 + 5 = 23 </h2><h2>23 - 0(6) </h2><h2>0(6) = 0 </h2><h2>23 - 0 = 23 </h2><h2>Answer: 23 (Option C) </h2><h2>Good luck on your assignment and enjoy your day! </h2><h2>~You might be interested in
Answer:
0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.
The sketch is drawn at the end.
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.
Mean of 0°C and a standard deviation of 1.00°C.
This means that
Find the probability that a randomly selected thermometer reads between −2.23 and −1.69
This is the p-value of Z when X = -1.69 subtracted by the p-value of Z when X = -2.23.
X = -1.69
has a p-value of 0.0455
X = -2.23
has a p-value of 0.0129
0.0455 - 0.0129 = 0.0326
0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.
Sketch:
