Answer:
tan(2u)=[4sqrt(21)]/[17]
Step-by-step explanation:
Let u=arcsin(0.4)
tan(2u)=sin(2u)/cos(2u)
tan(2u)=[2sin(u)cos(u)]/[cos^2(u)-sin^2(u)]
If u=arcsin(0.4), then sin(u)=0.4
By the Pythagorean Identity, cos^2(u)+sin^2(u)=1, we have cos^2(u)=1-sin^2(u)=1-(0.4)^2=1-0.16=0.84.
This also implies cos(u)=sqrt(0.84) since cosine is positive.
Plug in values:
tan(2u)=[2(0.4)(sqrt(0.84)]/[0.84-0.16]
tan(2u)=[2(0.4)(sqrt(0.84)]/[0.68]
tan(2u)=[(0.4)(sqrt(0.84)]/[0.34]
tan(2u)=[(40)(sqrt(0.84)]/[34]
tan(2u)=[(20)(sqrt(0.84)]/[17]
Note:
0.84=0.04(21)
So the principal square root of 0.04 is 0.2
Sqrt(0.84)=0.2sqrt(21).
tan(2u)=[(20)(0.2)(sqrt(21)]/[17]
tan(2u)=[(20)(2)sqrt(21)]/[170]
tan(2u)=[(2)(2)sqrt(21)]/[17]
tan(2u)=[4sqrt(21)]/[17]
Answer:
d = 17
Step-by-step explanation:
d = √ (9 − 1) 2 + (15− 0)^2
d = √ (8) ^2 + (15) ^2
d = √ 64 + 225
d = √289
d = 17
The answer is equal to -0.6
Answer:
Probability that a student chosen randomly from the class plays basketball or baseball is
or 0.76
Step-by-step explanation:
Given:
Total number of students in the class = 30
Number of students who plays basket ball = 19
Number of students who plays base ball = 12
Number of students who plays base both the games = 8
To find:
Probability that a student chosen randomly from the class plays basketball or baseball=?
Solution:
---------------(1)
where
P(A) = Probability of choosing a student playing basket ball
P(B) = Probability of choosing a student playing base ball
P(A \cap B) = Probability of choosing a student playing both the games
<u>Finding P(A)</u>
P(A) = 
P(A) =
--------------------------(2)
<u>Finding P(B)</u>
P(B) = 
P(B) =
---------------------------(3)
<u>Finding
</u>
P(A) = 
P(A) =
-----------------------------(4)
Now substituting (2), (3) , (4) in (1), we get


