Answer:
7/25
Step-by-step explanation:
θ lies in quadrant ii
so 2θ lies in quadrant iv
csc θ=5/3
sin θ=3/5 (sin θ=1/csc θ)
[cos(α+β)=cosαcosβ-sinαsinβ]
cos (2θ)=cos(θ+θ)=cos θ cos θ-sin θ sin θ=cos² θ-sin ²θ=1-sin²θ-sin²θ=1-2sin²θ
=1-2 (3/5)²
=1-2(9/25)
=1-18/25
=(25-18)/25
=7/25
Answer:
7
Step-by-step explanation:
First expand 2/3 (9x-6) = 4 (x+2)+ 2 By doing 2/3 (9x - 6)/3
Which gives you 2/3 (9x - 6)
Then you factor it 2/3 (9x - 6) 6 (3x - 2) get 6 (3x - 2) / 3
then you divide 6/3 with 2
Getting 2 (3x - 2)
The use the distributive law
2 x 3x - 2 x 2
simplify the equation
6x - 4
expand 4 (x +2) + 2: 4x + 10
which then equals 6x-4=4x+10
add 4 to both sides
6x-4+4=4x+10+4
simplify
6x=4x+14
subtract 4x from both sides
simplify again
2x=14
divide both sides by 2
2x/2 = 14/2
simplify one last time and get the answer
x=7
I think it’s true because happy body happy mine
Let's define the following variables first.
A = number of tickets sold for adults
C = number of tickets sold for children
From the question, we can say that or form the following equations:
1. A + C = 790 tickets
2. $7A + $4C = $4, 390
The first equation can also be written as A = 790 - C. We can use this equation and replace "A" in the second equation.
From that, we can solve "C" by solving the equation formed above.
Therefore, 380 tickets for children were sold.
Since there are 790 tickets in total that are sold and 380 tickets for children were sold, we can say that 410 tickets for adult was sold.
The question is asking for the lower bound of the 95% two tailed Confidence interval of the normally distributed population.
95% C.I. is given by 200 + or - 1.96(25) = 200 + or - 49 = (151, 249)
Therefore, the minimum weight of the middle 95% of players is 151 pounds.
