If A= pi/6 then what is B?

Answers

Answer 1

Answer: $\bf \textit{the graph starts at }0\textit{ and goes to }\pi \n\nA=(\pi )/(6)\textit{ now, B is away from }\pi \textit{ 1A}\n\n\textit{since A and B have a same y-value,}\n\textit{and the graph is symmetric, thus}\n\nB=\pi -(\pi )/(6)$

Answer 2

Answer: the other person is correct

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One table at a bake sale has 75 oatmeal cookies. Another table has 60 lemon cupcakes. Which table allows for more rectangular arrangements when all the cookies and cupcakes are displayed?

Answers

75 = 3·5², so has 6 divisors. 6 rectangles are possible if you make the distinction between 1×75 and 75×1.

60 = 2²·3·5, so has 12 divisors. 12 rectangles are possible under the same conditions.

The cupcake table can be arranged more ways.

_____

When 1 is added to each exponent of the prime factors, the product of those sums is the number of divisors. For 75: (1+1)(1+2) = 6; for 60: (1+2)(1+1)(1+1) = 12.

Arrangement is simply the order, which items are displayed or presented.

The table of 60 lemon cupcakes allow more rectangular arrangements

The given parameters are:

$\mathbf{Oatmeal = 75}$

$\mathbf{Lemon = 60}$

The rectangular arrangement (R) is calculated as follows:

$\mathbf{R = (Area)/(n)}$

Where n represents the number of items, and Area represents the area of the rectangular table

For the oatmeal, we have:

$\mathbf{R_1 = (Area)/(75)}$

$\mathbf{R_1 = 0.0133Area}$

For the lemon, we have:

$\mathbf{R_2 = (Area)/(60)}$

$\mathbf{R_2 = 0.0167Area}$

By comparison, 0.0167Area is greater than 0.0133Area

Hence, the table of 60 lemon cupcakes allow more rectangular arrangement

Answers

It’s c i think or can be d but my best bet is c !!

If you change the sigh of a point y-coordinate, how will the location of the
point change?

Answers

Answer:

It will reflect across the x axis

Hope this helps!

Let be a continuous random variable that follows a normal distribution with a mean of and a standard deviation of . Find the value of so that the area under the normal curve to the right of is . Round your answer to two decimal places.

Answers

Complete Question

Let x be a continuous random variable that follows a normal distribution with a mean of 550 and a standard deviation of 75.

Find the value of x so that the area under the normal curve to the left of x is .0250.

Find the value of x so that the area under the normal curve to the right ot x is .9345.

Answer:

$x = 403$

$x = 436.75$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 550$

The standard deviation is $\sigma = 75$

Generally the value of x so that the area under the normal curve to the left of x is 0.0250 is mathematically represented as

$P( X < x) = P( (x - \mu )/( \sigma) < (x - 550 )/(75 ) ) = 0.0250$

$(X -\mu)/(\sigma ) = Z (The \ standardized \ value\ of \ X )$

$P( X < x) = P( Z < z ) = 0.0250$

Generally the critical value of 0.0250 to the left is

$z = -1.96$

=> $(x- 550 )/(75) = -1.96$

=> $x = [-1.96 * 75 ]+ 550$

=> $x = 403$

Generally the value of x so that the area under the normal curve to the right of x is 0.9345 is mathematically represented as

$P( X < x) = P( (x - \mu )/( \sigma) < (x - 550 )/(75 ) ) = 0.9345$

$(X -\mu)/(\sigma ) = Z (The \ standardized \ value\ of \ X )$

$P( X < x) = P( Z < z ) = 0.9345$

Generally the critical value of 0.9345 to the right is

$z = -1.51$

=> $(x- 550 )/(75) = -1.51$

=> $x = [-1.51 * 75 ]+ 550$

=> $x = 436.75$

A box is formed by cutting squares from the four corners of a sheet of paper and folding up the sides.Suppose the paper is 7"-wide by 9"-long.
a. Estimate the maximum volume for this box?
b. What cutout length produces the maximum volume?

Answers

To answer this question it is necessary to find the volume of the box as a function of "x", and apply the concepts of a maximum of a function.

The solution is:

a) V (max) = 36.6 in³

b) x = 1.3 in

The volume of a cube is:

V(c) = w×L×h ( in³)

In this case, cutting the length "x" from each side, means:

wide of the box ( w - 2×x ) equal to ( 7 - 2×x )

Length of the box ( L - 2×x ) equal to ( 9 - 2×x )

The height is x

Then the volume of the box, as a function of x is:

V(x) = ( 7 - 2×x ) × ( 9 -2×x ) × x

V(x) = ( 63 - 14×x - 18×x + 4×x²)×x

V(x) = 4×x³ - 32×x² + 63×x

Tacking derivatives, on both sides of the equation

V´(x) = 12×x² - 64 ×x + 63

If V´(x) = 0 then 12×x² - 64 ×x + 63 = 0

This expression is a second-degree equation, solving for x

x₁,₂ = [ 64 ± √ (64)² - 4×12*63

x₁ = ( 64 + 32.74 )/ 24

x₁ = 4.03 this value will bring us an unfeasible solution, since it is not possible to cut 2×4 in from a piece of paper of 7 in ( therefore we dismiss that value)

x₂ = ( 64 - 32.74)/24

x₂ = 1.30 in

The maximum volume of the box is:

V(max) = ( 7 - 2.60) × ( 9 - 2.60)×1.3

V(max) = 4.4 × 6.4 × 1.3

V(max) = 36.60 in³

To chek for maximum value of V when x = 1.3

we find the second derivative of V V´´, and substitute the value of x = 1.3, if the relation is smaller than 0, we have a maximum value of V

V´´(x) = 24×x - 64 for x = 1.3

V´´(x) = 24× 1.3 - 64 ⇒ V´´(x) < 0

Then the value x = 1.3 will bring maximum value for V

Related Link: brainly.com/question/13581879

Final answer:

The maximum volume of the box that can be formed is approximately 17.1875 cubic inches. The cutout length that would result in this maximum volume is approximately 1.25 inches.

Explanation:

To solve this problem, we will use optimization in calculus. Let's denote the length of the square cutout as 'x'. When you cut out an x by x square from each corner and fold up the sides, the box will have dimensions:

Length: 9 inches (the original length) - 2x (the removed parts)
Width: 7 inches (the original width) - 2x
Height: x inches (the height is the cutout's length)

So the volume V of the box can be given by the equation: V = x(9-2x)(7-2x). We want to maximize this volume.

To find the maximum, differentiate V with respect to x, equate to zero and solve for x. V' = (9-2x)(7-2x) + x(-2)(7-2x) + x(9-2x)(-2) = 0. We obtain x=1.25 inches, but we need to verify whether this value gives us a maximum. Second differentiation, V'' = -12 is less than zero for these dimensions so the V is maximum.

a. So, when we solve, the maximum volume will be approximately 17.1875 cubic inches.

b. The cutout length that would produce the maximum volume is therefore about 1.25 inches.