Each student in a gymnastics class tries 20 times to do a cartwheel on the balance beam
1 answer:
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Answer:
a) Null and alternative hypotheses are:
: mu=183 days
: mu>183 days
b) If the true mean is 190 days, Type II error can be made.
Step-by-step explanation:
Let mu be the mean life of the batteries of the company when it is used in a wireless mouse
Null and alternative hypotheses are:
: mu=183 days
: mu>183 days
Type II error happens if we fail to reject the null hypothesis, when actually the alternative hypothesis is true.
That is if we conclude that mean life of the batteries of the company when it is used in a wireless mouse is at most 183 days, but actually mean life is 190 hours, we make a Type II error.
Part 1)
Sam is observing the velocity of a car at different times. After three hours, the velocity of the car is 51 km/h. After five hours, the velocity of the car is 59 km/h.
Part 1 a): Write an equation in two variables in the standard form that can be used to describe the velocity of the car at different times. Show your work and define the variables used
Let
A(3,51) B(5,59)
x------ > represent different times
y------ > represent the velocity of the car
Step 1
Find the slope AB
m=(y2-y1)/(x2-x1)------ > m=(59-51)/(5-3)------ > m=8/2---- > m=4
Step 2
With m=4 and point A(3,51) find the equation of the line
y-y1=m*-(x-x1)------ > y-51=4*(x-3)----- > y=4x-12+51----- > y=4x+39
we know that
The standard form of line equation is Ax + By = C
So
y=4x+39----- > y-4x=39------ > this is the standard form
the answer part 1 a) is
y-4x=39
Part 1 b) How can you graph the equation obtained in Part a) for the first six hours?
To graph the equation obtained in Part a) plot the point A and the point B
and join the points to draw the line
To obtain the velocity for the first six hours, substitute the value of x=6 hour in the equation
for x=6 hour
y-4x=39------ > y-4*6=39------ > y=39+24------ > y=63 km/h
using a graph tool
see the attached figure N 1
Part 2)
g(x)=1+1.5^x
step 1
find the equation of the line of f(x)
let
A(-5,3) B(-3,-1)
m=(-1-3)/(-3+5)----- > m=-4/2---- > m=-2
with m=-2 and point A
y-y1=m*(x-x1)------ > y-3=-2*(x+5)---- > y=-2x-10+3----- > y=-2x-7
so
f(x)=-2x-7
step 2
find the equation of the line of p(x)
let
C(0,2) D(-2,-3)
m=(-3-2)/(-2-0)----- > m=-5/-2---- > m=2.5
with m=2.5 and point C
y-y1=m*(x-x1)------ > y-2=2.5*(x-0)---- > y=2.5x+2
so
p(x)=2.5x+2
Part 2 a) What is the solution to the pair of equations represented by p(x) and f(x)?
We know that
The solution is the intersection of both graphs
Using a graph tool
See the attached figure N 2
The solution is the point (-2,-3)
Part 2 b) Write any two solutions for f(x).
f(x)=-2x-7
for x=0
f(0)=2*0-7---- > f(0)=-7
solution 1 is the point (0,-7)
for x=1
f(1)=2*1-7---- > f(1)=-5
solution 2 is the point (1,-5)
Part 2 c) What is the solution to the equation p(x) = g(x)?
We have
p(x)=2.5x+2
g(x)=1+1.5^x
We know that
The solution is the intersection of both graphs
Using a graph tool
See the attached figure N 3
The solution are the points (0,2) and (7.3,20.2)
Part 3
)
Part A:There are many system of inequalities that can be created such that only contain points D and E in the overlapping shaded regions.
Any system of inequalities which is satisfied by (-4, 2) and (-1, 5) but is not satisfied by (1, 3), (3, 1), (3, -3) and (-3, -3) can serve.
An example of such system of equation is
x < 0
y > 0
The system of equation above represent all the points in the second quadrant of the coordinate system.The area above the x-axis and to the left of the y-axis is shaded.
see the attached figure N 4
Part B:It can be verified that points D and E are solutions to the system of inequalities above by substituting the coordinates of points D and E into the system of equations and see whether they are true.
Substituting D(-4, 2) into the system
we have:
-4 < 0
2 > 0
as can be seen the two inequalities above are true, hence point D is a solution to the set of inequalities.
Also,
substituting E(-1, 5) into the system we have:
-1 < 0
5 > 0
as can be seen the two inequalities above are true, hence point E is a solution to the set of inequalities.
Part C:Given that chicken can only be raised in the area defined by y > 3x - 4.
To identify the farms in which chicken can be raised, we substitute the coordinates of the points A to F into the inequality defining chicken's area.
For point A(1, 3): 3 > 3(1) - 4 ⇒ 3 > 3 - 4 ⇒ 3 > -1 which is true
For point B(3, 1): 1 > 3(3) - 4 ⇒ 1 > 9 - 4 ⇒ 1 > 5 which is false
For point C(3, -3): -3 > 3(3) - 4 ⇒ -3 > 9 - 4 ⇒ -3 > 5 which is false
For point D(-4, 2): 2 > 3(-4) - 4; 2 > -12 - 4 ⇒ 2 > -16 which is true
For point E(-1, 5): 5 > 3(-1) - 4 ⇒ 5 > -3 - 4 ⇒ 5 > -7 which is true
For point F(-3, -3): -3 > 3(-3) - 4 ⇒ -3 > -9 - 4 ⇒ -3 > -13 which is true
Therefore,
the farms in which chicken can be raised are the farms at point A, D, E and F.
