cos (2x) = cos x
2 cos^2 x -1 = cos x using the double angle formula
2 cos ^2 x -cos x -1 =0
factor
(2 cos x+1) ( cos x -1) = 0
using the zero product property
2 cos x+1 =0 cos x -1 =0
2 cos x = -1 cos x =1
cos x = -1/2 cos x=1
taking the arccos of each side
arccos cos x = arccos (-1/2) arccos cos x = arccos 1
x = 120 degrees x=-120 degrees x=0
remember you get 2 values ( 2nd and 3rd quadrant)
these are the principal values
now we need to add 360
x = 120+ 360n x=-120+ 360n x = 0 + 360n where n is an integer
Answer:
40 quarters and 2 dimes
Step-by-step explanation:
I did .25 (the amount of a quarter) and put it into a calculator adding.25 untill I got to 10 and then just 2 dimes bc 10+10 is 20
Answer:
1/6 or 25
Step-by-step explanation:
1/3=10/30
3/10=9/30
1/5=6/30
together is 25/30, which leaves the rest, 5/30 or 1/6 or 25 cookies.
Complete question :
Suppose that of the 300 seniors who graduated from Schwarzchild High School last spring, some have jobs, some are attending college, and some are doing both. The following Venn diagram shows the number of graduates in each category. What is the probability that a randomly selected graduate has a job if he or she is attending college? Give your answer as a decimal precise to two decimal places.
What is the probability that a randomly selected graduate attends college if he or she has a job? Give your answer as a decimal precise to two decimal places.
Answer:
0.56 ; 0.60
Step-by-step explanation:
From The attached Venn diagram :
C = attend college ; J = has a job
P(C) = (35+45)/300 = 80/300 = 8/30
P(J) = (30+45)/300 = 75/300 = 0.25
P(C n J) = 45 /300 = 0.15
1.)
P(J | C) = P(C n J) / P(C)
P(J | C) = 0.15 / (8/30)
P(J | C) = 0.5625 = 0.56
2.)
P(C | J) = P(C n J) / P(J)
P(C | J) = 0.15 / (0.25)
P(C | J) = 0.6 = 0.60