## Problem 1037

The graph of a normal curve is given. Use the graph to identify the value of µ and σ.

Solution:-

The mean of the distribution µ is the point where the curve reaches the maximum value.

Determine the mean of the graph.

µ = 50

The points at x = µ – σ  and x = µ + σ are the inflection points on the normal curve. The inflection points are the points on the curve where the curvature of the graph changes.

The curvature of the accompanying graph change at points x = 30 and x = 70.

Consider the equation for the right inflection point, x = µ + σ.

µ + σ = x

µ + σ – µ  = x – µ

σ = x – µ

Find σ.

σ  = x  – µ

= 70  – 20

Therefore , the mean of the distribution is 50 and the standard deviation is 20.

## Problem 1035

Solve the inequality symbolically. Express the solution set in interval notation.

-7(z-8)≥4(9-2z)

Solution:-

To solve the inequality, isolate the variable on one side the inequality using the properties if inequalities.

First, use the distributive property to eliminate the parentheses on both sides of the equation.

-7(z-8) ≥ 4(9-2z)

-7z + 56 ≥ 36 – 8z (Simplify the right side.)

Add 8z to both sides so that all expressions containing a variable are on the left side.

-7z +56 ≥ 36 – 8z

z + 56 ≥ 36 (Add 8z to both sides)

To isolate z, subtract 56 form both sides.

z + 56 ≥ 36

z ≥ 56 ≥ 36

z ≥ -20 (Subtract)

Thus, the solution is the set of all real numbers greater than or equal to -20.

In interval notation, a square bracket is used to show that a number is part of the interval and a parenthesis is used to indicate that a number is not part of the interval.

Write the solution set is interval notation.

[-20,∞)

## Problem 1034

Express the following in interval notation.

{x|7≤x<11}

Solution:-

To express this set using interval notation, first identify the left endpoint. The left endpoint of the set {x|7≤x<11}  is 7.

Next, identify the right endpoint. The right endpoint of the set{x|7≤x<11} is 11.

Now, identify if a left bracket,[, or a left parenthesis,(, should be used for the left endpoint. A bracket is used if the left endpoint is included in the set. A parenthesis is used if the left endpoint is not included in the interval.

The ≤ sign indicates that the left endpoint is included in the interval so that a left bracket should be used.

Now, identify if a right bracket,], or a right parenthesis,), should be used for the right endpoint. A bracket is used if the right endpoint is included in the interval. A parenthesis is used if the right endpoint is not included in the interval.

The > sing indicates that the right endpoint is not included in the interval so the a right parenthesis should be used.

The set in interval notation is [7,11).

## Problem 1033

Express in interval notation.

x < 0

Solution:-

The set x < 0 represents those numbers that are less than 0.

A simplified notation, called interval notation, is used for writing intervals.

The symbol  < indicates that 0 is not included in the set; therefore, it would not be included in this interval.

The symbol -∞ is used to show that the interval included all real numbers less than 0. It is used in the position of the left endpoint. A parenthesis is always included next to the infinity symbol in interval notation.

Using interval notation, the set x < 0 is written as (-∞,0)

## Problem 1028

Determine the domain of the function.

y  =

Solution:-

The numerator can be any number. Values of x for which the expression is undefined are values that make the denominator equal to 0.

Set the denominator equal to zero to find these values.

x2 + 2x -8 = 0

Factor the denominator.

(x + 4 )(x-2) = 0

Solve the equation to obtain the values of x that are exchange in the given expression.

x + 4 = 0 or x- 2 = 0

x = -4 or x = 2

The domain of the original function is {x | x ≠ – 4 and x ≠ 2}

## Problem 1027

Solve the following equation for r.

F =

Solution:-

First rewrite the equation with the expression containing r on the left side .

= F

To rewrite the formula with only the r on the left side, multiply by 3, then simplify.

3* = F*3

Divide both sides by πh to isolate the expression containing r on the left side.

Use the square root property to solve for r.

r= ±

## Problem 1026

Solve for the specified variable.

w2   + y2 = z2 for y

Solution:-

First isolate the term y2 on the left side of the equation.

w2   + y2 = z2

y2 = z2 – w2

Use the square root property and simplify. Let k be a nonnegative number. Then the solution to the equation x2 = k are given by x = ±.

y2 = z2 – w2

y = ±.

## Problem 1025

A cylindrical aluminum can is being constructed to have a height h of 3 inches. If the can is to have a volume of 24 cubic inches, approximate tis radius r.

Solution:-

Recall that the volume of a cylinder is given by the formula V = πr2h, where V represents the volume, h the height, and r the radius of the can.

To find the radius of the can, substitute V = 24 and h = 3 into the equation V = πr2h and solve for r.

V = πr2h

24 = πr2*3

Solve for r

r2 =

r2 =

r=

Therefore, the radius of the can is about 1.6 inches.

## Problem 1024

A farmer decide to enclose a rectangular garden, using the side of  a barn as one side of the rectangle. What is the maximum area that the framer can enclose with 120 ft of fence? What should the dimension of the garden be to given this area?

Solution:-

Picture the bran and the garden as in the following figure.

A formula for the use of the fencing is 120 = 2W + L.

A formula for the area of the garden A = LW.

Now, you must express A as a function of L or W.

Find A as a function of W. To do this, you must solve 120 = 2W + L for L. This gives

L = 120 – 2W.

Substituting for L in the equation for area, given A = (120-2W)W.

To find the maximum function value, you must complete the square.

A = 120W – 2W2

A = -2(W2 – 60W)

A = -2(W2 – 60W + 900) + 1800

A = -2(W – 30)2 + 1800

This tells you that the maximum function value, or the maximum area, is 1800 sq ft.

The maximum area is obtained when the width is 30 ft and the length is 60 ft.

## Problem 1023

Use the graph of the quadratic function f to write its formula as

f(x) = a(x – h)2 +k.

Solution:-

First determine the value of a, h, and k in f(x) = a(x – h)2+k.

Recall that the coordinates of the vertex x and y correspond to the values of h and k.

Identify the vertex of the given parabola.

Vertex = (-1, 4)

The vertex is (-1,4). Thus h = -1 and k = 4. Now substitute values of h and k in f(x) = a(x-h)2 + k.

f(x) = x(x+1)2 + 4

To find a, substitute the coordinates of a point on the graph in the equation and solve for a. Selection any point on the graph of f other than the vertex.

Consider the point(0, -2). The point (0,-2)lies on the graph, so f(0) = -2.

f(x) = a(x  + 1)2 + 4

let x = 0 and f(0) = -2.

-2 = a(0+1)2 + 4

-2 = a + 4

-6 = a

Finally, substitute a into the equation

f(x) = a(x + 1)2+ 4.

f(x) = -6(x + 1)2 + 4