We have that
sin ∅=-0.7660
the sin ∅ is negative
so
∅ belong to the III or IV quadrant
but
180°< ∅ < 270°
hence
∅ belong to the III quadrant
sin ∅=-0.7660
sin² ∅+cos² ∅=1
cos² ∅=1-sin² ∅------> cos² ∅=1-(0.7660)²-----> 0.4132
cos ∅=√0.4132-----> cos ∅=0.6428
the value of cos ∅ is negative-------> III quadrant
cos ∅=-0.6428
tan ∅=sin ∅/ cos ∅----> tan ∅=-0.7660/-0.6428----> tan ∅=1.1917
the answer is
tan ∅ is 1.1917
Whats the question?
Sorry but can't answer if theres no question
A=h(b+c)
A/h=b+c
b=(A/h)-c
Answer: b=(A/h)-c
Answer:
Hey!
Step-by-step explanation:
Your answer would be 1/8
Have a nice day!♥
And ty XD
Answer:


Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the grade points avergae of a population, and for this case we know the following properties
Where
and
The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).
So we can find the z score for the value of X=3.44 in order to see how many deviations above or belowe we are from the mean like this:

So the value of 3.44 is 2 deviations above from the mean, so then we know that the percentage between two deviations from the mean is 95% and on each tail we need to have (100-95)/2 = 2.5% , because the distribution is symmetrical, so based on this we can conclude that:
