The sequence is all odd numbers. the next will be 15, 17, 19, 21, etc...
Answer:
D. One
Step-by-step explanation:
A parallelogram will have two sets of equal sides. Since we have that fact, all we need to know are the angles.
A quadrilateral will have a total of 360° in its angles. If only one angle is equal to 90°, then we can do <u>the math</u> to figure out that all the others must be 90° as well. A rectangle is a parallelogram with the opposite sides being parallel and all the angles being 90°. This then answers the question.
"<u>The Math</u>"
-> Keep in mind there will be four angles
-> Once we subtract the first angle of 90°, we will have three angles left. Hence why we divide by 3.
360° - 90° = 270°
270° / 3 = 90° ✓
Answer:
a) NORM.S.INV(0.975)
Step-by-step explanation:
1) Some definitions
The standard normal distribution is a particular case of the normal distribution. The parameters for this distribution are: the mean is zero and the standard deviation of one. The random variable for this distribution is called Z score or Z value.
NORM.S.INV Excel function "is used to find out or to calculate the inverse normal cumulative distribution for a given probability value"
The function returns the inverse of the standard normal cumulative distribution(a z value). Since uses the normal standard distribution by default the mean is zero and the standard deviation is one.
2) Solution for the problem
Based on this definition and analyzing the question :"Which of the following functions computes a value such that 2.5% of the area under the standard normal distribution lies in the upper tail defined by this value?".
We are looking for a Z value that accumulates 0.975 or 0.975% of the area on the left and by properties since the total area below the curve of any probability distribution is 1, then the area to the right of this value would be 0.025 or 2.5%.
So for this case the correct function to use is: NORM.S.INV(0.975)
And the result after use this function is 1.96. And we can check the answer if we look the picture attached.