Answer:
Step-by-step explanation:
We would set up the hypothesis test.
For the null hypothesis,
P = 0.22
For the alternative hypothesis,
P < 0.22
This is a left tailed test
Considering the population proportion, probability of success, p = 0.22
q = probability of failure = 1 - p
q = 1 - 0.22 = 0.78
Considering the sample,
Sample proportion, p = x/n
Where
x = number of success = 230
n = number of samples = 1189
p = 230/1189 = 0.19
We would determine the test statistic which is the z score
z = (p - P)/√pq/n
z = (0.19 - 0.22)/√(0.22 × 0.78)/1189 = - 2.5
Recall, population proportion, p = 0.22
We want the area to the left of 0.22 since the alternative hypothesis is lesser than 0.22. Therefore, from the normal distribution table, the probability of getting a proportion < 0.22 is 0.00621
So p value = 0.00621
Since alpha, 0.01 > than the p value, 0.00621, then we would reject the null hypothesis.
Referring to the above, the null hypothesis will be rejected if the test statistic is - 2.5
Answer:
1
Step-by-step explanation:
1111111111111111111111
Answer:
The imaginary part is 0
Step-by-step explanation:
The number given is:
First, we can expand this power using the binomial theorem:
After that, we can apply De Moivre's theorem to expand each summand:
The final step is to find the common factor of i in the last expansion. Now:
The last part is to multiply these factors and extract the imaginary part. This computation gives:
(It is not necessary to do a lengthy computation: the summands of the imaginary part are the products sin(a)cos(b) and cos(a)sin(b) as they involve exactly one i factor)
A calculator simplifies the imaginary part Im(x⁶) to 0
Answer:
Each person should chip in $0.83
Step-by-step explanation:
Total cost =3000km(66/100)(1.20/L)
=45.46 × 1.20
=$54.55
Cost per person = ($54.55/66)
=$0.83 person
Each person should chip in $0.83
<span>1. X = -2 or 3
2. X = -5 or 3
3. X = -2.5 or 3
4. X = -4 or 2
5. X = 3 or -3
6. X = -4 or 2
I am assuming that you're looking for the intersections between the two equations for each problem. The general approach to each of the given problems is to solve both equations for y (only need to do this with problems 4 through 6 since you've already been given the equations solved for y with problems 1 through 3). After you have two equations solved for y, simply set them equal to each other and then manipulate until you have a quadratic equation of the form:
Ax^2 + Bx + C = 0
After you've gotten your quadratic equation, just find the roots to the equation and you'll know both X values that will result in the same Y value as the equations you've been given for each problem. I'm personally using the quadratic formula for getting the desired roots, but you can also factor manually. So let's do it.
1. y = x+2, y = x^2 - 4
Set the equations equal to each other
x + 2 = x^2 - 4
2 = x^2 - x - 4
0 = x^2 - x - 6
Using the quadratic formula with A=1, B=-1, C=-6, you get the solutions -2 and 3.
2. y = x^2 + 3x - 1, y = x+14
Same thing, set the equations equal to each other.
x^2 + 3x - 1 = x + 14
x^2 + 2x - 1 = 14
x^2 + 2x - 15 = 0
Use the quadratic formula with A=1, B=2, C=-15. Roots are -5 and 3.
3. y = 2x^2 + x - 7, y = 2x + 8
Set the equations equal to each other again.
2x^2 + x - 7 = 2x + 8
2x^2 - x - 7 = 8
2x^2 - x - 15 = 0
Quadratic formula with A=2, B=-1, C=-15, gives you the roots of -2.5 and 3
4. y = x(x + 3), y - x = 8
A little more complicated. Solve the second equation for y
y - x = 8
y = x + 8
Multiply out the 1st equation
y = x(x + 3)
y = x^2 + 3x
Now set the equations equal to each other
x + 8 = x^2 + 3x
8 = x^2 + 2x
0 = x^2 + 2x - 8
And use the quadratic formula with A=1, B=2, C=-8. Roots are -4, 2
5. y = -3x^2 - 2x + 5, y + 2x + 22 = 0
Solve the 2nd equation for y
y + 2x + 22 = 0
y + 22 = -2x
y = -2x - 22
Set equal to 1st equation
-2x - 22 = -3x^2 - 2x + 5
-22 = -3x^2 + 5
0 = -3x^2 + 27
Use the quadratic formula with A=-3, B=0, C=27, giving roots of 3 and -3
6. y + 6 = 2x^2 + x, y + 3x = 10
Solve the 1st equation for y
y + 6 = 2x^2 + x
y = 2x^2 + x - 6
Solve the 2nd equation for y
y + 3x = 10
y = -3x + 10
Set the solved equations equal to each other
2x^2 + x - 6 = -3x + 10
2x^2 + 4x - 6 = + 10
2x^2 + 4x - 16 = 0
Use the quadratic formula with A=2, B=4, C=-16, getting roots of -4 and 2.</span>
