Since the area of the poster doesn't change by putting it in a frame, we presume the question is asking what the area of the framed poster is.
The length of the poster in its frame is ...
(frame width on one side) + (poster length) + (frame width on the other side)
2 in + 32 in + 2 in = 36 in
Likewise, the width of the poster in its frame is ...
2 in + 24 in + 2 in = 28 in
The area of a rectangle 36 in by 28 in is the product of these dimensions:
Area = (36 in)×(28 in) = (36×28) in² = 1008 in²
V=4/3π r^3
(3/4)V=π r^3
3V/4π = r^3
third root√(3V/4π)
<span><span>Hope this helps!
</span>and May the Force Be With You
</span><span>
-Jabba</span>
Answer: The second option
Step-by-step explanation: 14mph times 5 equal 70. 5 hours after 6 am is 11 am. 70 divided by 20 is 3.5. 3.5 hours after 6 am is 9:30. Hence 9:30 to 11
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: