Complete Question
Educational Television In a random sample of 200 people, 159 said that they watched educational television. Find the 90% confidence interval of the true proportion of people who watched educational television. Round the answers to at least three decimal places.
Answer:
The 90% confidence interval is
Step-by-step explanation:
From the question we are told that
The sample size is n = 200
The number of people that watched the educational television is 
Generally the sample proportion is mathematically represented as

=> 
From the question we are told the confidence level is 90% , hence the level of significance is
=>
Generally from the normal distribution table the critical value of
is
Generally the margin of error is mathematically represented as
=>
=>
Generally 95% confidence interval is mathematically represented as
=>
=>
Dy/dx = dy/dt * dt/dx
xy = 4
y + x(dy/dx) = 0 by implicit differentiation.
x(dy/dx) = -y
dy/dx = -y/x
<span>dy/dx = dy/dt * dt/dx dy/dt = -2
</span>
<span>-y/x = -2 * dt/dx
</span>
y/(2x) = dt/dx
dt/dx = y/(2x)
dx/dt = 2x/y
When x = -3, xy = 4, y = 4/x = 4/-3 = -4/3
dx/dt = 2*-3/(-4/3) = -6 *-3/4 = 18/4 = 9/2 = 4.5
dx/dt = 4.5
Answer:
r = 9/sin Ф or r = 9 csc Ф
Step-by-step explanation:
The polar form is represented as:
x = r cos Ф and
y = r sin Ф
We are given:
y = 9
Putting value in y = r sin Ф
9 = r sin Ф
=> r = 9/sin Ф
1/sin Ф = csc Ф
or, r = 9 csc Ф
-x = -1. Anything “X” equals to the number One. So do Y = -1 + 1.
3a^3 is the answer
Hope that helps