It is asking you to convert them to different units of measurements.
Answer:
The answer is 90 degrees. Have a good day!
Step-by-step explanation:
<span>238 square units.
The area of a parallelogram is the base multiplied by the height. You can use any of its four sides as the base, so pick the one that is easiest to deal with. Examining the parallelogram, you'll notice that line segments AB and CD are both parallel to the x axis which makes it extremely easy to calculate the height which is 12 - (-5) = 17.
The length of AB is 13 - (-1) = 14. So the area of the parallelogram is 14 * 17 = 238</span>
Answer:
22.29% probability that both of them scored above a 1520
Step-by-step explanation:
Problems of normally distributed samples are 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.
In this problem, we have that:

The first step to solve the question is find the probability that a student has of scoring above 1520, which is 1 subtracted by the pvalue of Z when X = 1520.
So



has a pvalue of 0.5279
1 - 0.5279 = 0.4721
Each students has a 0.4721 probability of scoring above 1520.
What is the probability that both of them scored above a 1520?
Each students has a 0.4721 probability of scoring above 1520. So

22.29% probability that both of them scored above a 1520
Answer:
The odd function is written f (x) = (x) and rounds the odd numbers to the nearest integer.
The graph of f (x) = (x) includes the point (17/4), (7/2), (15/6)
Step-by-step explanation:
The odd function seeks to round odd numbers, with fractions of first odd numbers and then even by adding 3 to them.