Classes are distributed between 49.0 and 54.0 minutes find probably that a given class period runs less than 50.75 minutes

Step-by-step explanation:

Assuming that this is a normal distribution, the mean is halfway between 49.0 and 54.0 minutes:  (49.0+54.0)/2 minutes = 51.5 minutes.

We want to find the area under the standard normal curve to the left of 50.75 minutes.  To do this, use the normalcdf( function on a TI-83 Plus calculator:

normalcdf(-1000,50.75,51.5, 1) = 0.23.

Notes:  The left boundary of the area in question is actually -infinity, but using -1000 as this boundary is only minimally different.  50.75 is the right area boundary.  51.5 is the mean.  In the absence of info from you regarding the standard deviation of this distribution, I have arbitrarily chosen to let the standard deviaton be 1.  Thus, the probability that a given class period runs less than 50.75 minutes is 0.23, if the standard deviation is 1.

If the standard deviation were 0.5 instead, then the probability in question would be

normalcdf(-1000,50.75,51.5, 0.5) = 0.31.

the correct answer is semicircle

step-by-step explanation:

the answer is semicircle because the shorter arc joining two points on the circumference of a circle is known as the minor arc, the larger arc is called the major arc, and a semicircle is a half circle, formed by cutting a whole circle along a diameter line.

hope i .

sine

step-by-step explanation: