Answer:
The answer is c.) 0.6 and 0.14
Step-by-step explanation:
Given population has a size of 320.
The proportion of population = 0.6
The mean of population, p = 0.6
The mean of sample, = 0.6
Therefore = 1 -
= 1 - 0.6 = 0.4
The sample size = 12
Therefore the standard deviation of the sample,
The mean and the standard deviation for a sample size of 12 are 0.6 and 0.14 respectively.
The answer is c.) 0.6 and 0.14
Answer:
D) y = -4/5x – 47/10
Step-by-step explanation:
Step 1. Find the <em>midpoint of the segmen</em>t.
The two end points are (-6, -4) and (-2, 1).
The midpoint is at the average of the coordinates.
(xₚ, yₚ) = ((x₁ + x₂)/2, (y₁ + y₂)/2)
(xₚ, yₚ) = ((-6 - 2)/2, (-4 + 1)/2)
(xₚ, yₚ) = (-8/2, -3/2)
(xₚ, yₚ) = (-4, -3/2)
===============
Step 2. Find the <em>slope (m₁) of the segment</em>
m₁ = (y₂ - y₁)/(x₂ - x₁)
m₁ = (1 - (-4))/(-2 - (-6))
m₁ = (1 + 4)/(-2 + 6)
m₁ = 5/4
===============
Step 3. Find the <em>slope (m₂) of the perpendicular bisector </em>
m₂ = -1/m₁
m₂ = -4/5
====================
Step 4. Find the <em>intercept of the perpendicular bisector</em>
y = mx + b
y = -(4/5)x + b
The line passes through (-4, -3/2).
-3/2 = -(4/5)(-4) + b
-3/2 = 16/5 + b Multiply each side by 10
-15 = 32 + 10 b Subtract 32 from each side
-47 = 10b Divide each side by 10
b = -47/10
===============
Step 5. Write the <em>equation for the perpendicular bisector</em>
y = -4/5x – 47/10
The graph shows the midpoint of your segment at (-4, -3/2) and the perpendicular bisector passing through the midpoint and (0, -47/10).
Answer:
N(1)=50 is a minimum
N(15)=4391.7 is a maximum
Step-by-step explanation:
<u>Extrema values of functions </u>
If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.
We are given a function (corrected)
(a)
First, we take its derivative
Solve N'(t)=0
Simplifying
Solving for t
Only t=1 belongs to the valid interval
Taking the second derivative
Which is always positive, so t=1 is a minimum
(b)
N(1)=50 is a minimum
(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15
(d)
N(15)=4391.7 is a maximum