This is the all question?
Answer:
x = 1,5 cm
h = 6 cm
C(min) = 135 $
Step-by-step explanation:
Volume of the box is :
V(b) = 13,5 cm³
Aea of the top is equal to area of the base,
Let call " x " side of the base then as it is square area is A₁ = x²
Sides areas are 4 each one equal to x * h (where h is the high of the box)
And volume of the box is 13,5 cm³ = x²*h
Then h = 13,5/x²
Side area is : A₂ = x* 13,5/x²
A(b) = A₁ + A₂
Total area of the box as functon of x is:
A(x) = 2*x² + 4* 13,5 / x
And finally cost of the box is
C(x) = 10*2*x² + 2.50*4*13.5/x
C(x) = 20*x² + 135/x
Taking derivatives on both sides of the equation:
C´(x) = 40*x - 135*/x²
C´(x) = 0 ⇒ 40*x - 135*/x² = 0 ⇒ 40*x³ = 135
x³ = 3.375
x = 1,5 cm
And h = 13,5/x² ⇒ h = 13,5/ (1,5)²
h = 6 cm
C(min) = 20*x² + 135/x
C(min) = 45 + 90
C(min) = 135 $
The two given angles are vertical angles which mean they are the same:
9x + 72 = 4x + 112
Subtract 72 from bot sides:
9x = 4x + 40
Subtract 4x from both sides:
5x = 40
Divide both sides by 5:
X = 8
Answer:
So we reject the null hypothesis and accept the alternate hypothesis that rats learn slower with sound.
Step-by-step explanation:
In this data we have
Mean= u = 18
X= 38
Standard deviation = s= 6
1) We formulate the null and alternate hypothesis as
H0: u = 18 against Ha : u > 18 One tailed test .
2) The significance level alpha = ∝= 0.05 and Z alpha has a value ± 1.645 for one tailed test.
3)The test statistics used is
Z= X- u / s
z= 38-18/6= 3.333
4) The calculated value of z = 3.33 is greater than the z∝ = 1.645
5) So we reject the null hypothesis and accept the alternate hypothesis that rats learn slower with sound.
First we set the criteria for determining the true of value of the variable. That whether the rats learn in less or more than 18 trials.
Then we find the value of z for the given significance value given and the test about to be checked.
Then the test statistic is determined and calculated.
Then both value of z and z alpha re compared. If the test statistics falls in the rejection region reject the null hypothesis and conclude alternate hypothesis is true.
The figure shows that the calulated z value lies outside the given z values
So you are asking how many x's can be written in the form 2*3*number=x
where number is not even
there are 50 odd numbers thereforr there are 50 numbers that satisfy your need