Answer:
In February, 423 daytime minutes is used
Step-by-step explanation:
Let the base plan charges be x
And cost per daytime minute be y
In December,
x + 510y = 92.25------------------(1)
In January,
x + 397y = 77.56---------------------(2)
Subtracting eq(2) from eq(1)
x + 510y = 92.25
x + 397y = 77.56
-------------------------------
0 + 113y = 14.69
-------------------------------
y = \frac{14.69}{113}
y = 0.13----------------------------------(3)
Substituting (3) in (1)
x + 510(0.13) = 92.25
x + 66.3 = 92.25
x = 92.25 - 66.3
x = 25.95
So In February
base plan + (daytime minute)(cost per daytime minute) = 80.9
25.95 + (daytime minute)(0.13) = 80.9
(daytime minute)(0.13) = 80.9 - 25.95
(daytime minute)(0.13) = 54.95
(daytime minute) =
daytime minutes = 422.69
daytime minute 
Answer:
Hello your question is incomplete attached below is the complete question
Ix = 0 Ux = 
Iz = 0 Uz = 
Iy = 5 Uy = 10
Step-by-step explanation:
Ix = 0 Ux =
Iz = 0 Uz =
Iy = 5 Uy = 10
attached below is the detailed solution
To find t<span>he relative maximum value of the function we need to find where the function has its first derivative equal to 0.
Its first derivative is -7*(2x)/(x^2+5)^2
</span>7*(2x)/(x^2+5)^2 =0 the numerator needs to be eqaul to 0
2x=0
x=0
g(0) = 7/5
The <span>relative maximum value is at the point (0, 7/5).</span>
The tangent of a given circle is perpendicular to the radius a the point called point if tangency. The radius of the circle is perpendicular to the tangent at the point of tangency. This property is especially useful in cases where the radius that connects to the point of tangency forms a part of right angle because the pythagorean theorem and trigonometry apply to right angles.
Answer:
a) Narrower
b) Narrower
c) Wider
Step-by-step explanation:
We are given the following in the question:
Proportion of coworker who received flu vaccine = 32%
98% confidence interval: (0.231, 0.409)
Confidence interval:
a) Sample size had been 600 instead of 150
If we increase the sample size, thus the standard error of the interval decreases.
Since the standard error decreases, the confidence interval become narrower.
b) Confidence level had been 90% instead of 98%
As the confidence level increases, the confidence interval becomes narrower. This is due to a smaller value of z-statistic at 90% confidence level.
c) Confidence level had been 99% instead of 98%
As the confidence level increases, the width of the confidence interval increases and the confidence interval become wider. This is because of a larger value of z-statistic at 99% confidence interval.
