Find the surface area of the composite figure.
8 cm
6 cm
7 cm
8 cm
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A visit must be of at least 56.33 minutes to put a shopper in the longest 20 percent.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula. In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z=\frac{X-\mu}{\sigma}[/tex]

How long must a visit be to put a shopper in the longest 20 percent?

Here [tex]\mu=41.8 , \sigma=17.3[/tex]

The 100-20=80th percentile, which is X when Z has p-value of 0.8, so Z=0.84

[tex]Z=\frac{X-\mu}{\sigma}[/tex]

[tex]0.84=\frac{X-41.8}{17.3}[/tex]

14.532=X-41.8

X=14.532+41.8

X=56.33

A visit must be of at least 56.33 minutes to put a shopper in the longest 20 percent.

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ANSWER: The third statement appears to be incorrect.

Answer:

7x  + 9/10  ft

Step-by-step explanation:

To find the perimeter of a triangle, add the lengths of the three sides

P = 2x + 1/5  +  4x + 1/2  + x+ 1/ 5

Combine like terms

P = 7x + 2/5 +1/2

Get a common denominator

P = 7x + 2/5(2/2) + 1/2 * (5/5)

   = 7x + 4/10 + 5/10

   = 7x  + 9/10

The number of degrees that the direction of the boat change when it made its first turn is 39°.

From the information given, we would use the cosine function to solve the information.

This will be:

18² = 28² + 25² - 2(28)(25)(cos x)

324 = 784 + 625 - 1400cosx

x = 39.19

Therefore, the value is 39°.

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Answer: ) (2a+7)(3b- 4)

Step-by-step explanation:

Answer:

x = 34

Step-by-step explanation:

The three angles form a straight line and are thus supplementary angles.  The sum of a supplementary angles is always 180, so we can use the equation 180 = x + 12 + 100 + x to find x:

180 = 2x + 112

68 = 2x

x = 34

No, 1775 dollars cannot buy 2 large balloons and 16 small balloons if helium costs $1.60 per cubic foot.

Calculating the Costs of Balloons

It is given that,

Helium costs $1.60 per cubic foot.

Volume of one large balloon = 300 cubic feet

Thus, the price of one large balloon = $1.60 × 300

= $480

Volume of one small balloon = 40 cubic feet

Thus, the price of one small balloon = $1.60 × 40

= $64

How much buying 2 large and 16 small balloons costs?

So, as calculated above,

One large balloon costs $480, and one small balloon costs $64

Therefore, the cost of two large balloons = $480 × 2

= $960

Cost of 16 small balloons = $64 × 16

= $1024

Total cost of balloons = $960 + $ 1024

= $1984

Since, $1984>$1775,

Therefore, $1775 cannot buy all those balloons if helium costs $1.60 per cubic foot.

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Step-by-step explanation:

Chapter 3 of Wheels of Change .

But the bicycle craze had a more lasting impact on women’s clothing than just the use of bloomers. Thanks in large part to cycling, the innovation of rational dress reformers were starting to take effect by the end of the 1890s. Corsets were on their way out, dresses were getting shorter, and women no longer wore the heavy, bulky undergarments that gave them round, unnatural shapes. These changes went a long way toward unburdening women and setting the stage for them to be healthier and more active in the coming century.

Based on the excerpt, women in the 1890s became more independent and their lives improved as a result of

a rise in the popularity of bicycles.

a rise in the popularity of corsets.

cycling becoming less socially acceptable.

clothing becoming more restrictive.

Answer:

to promote safer, more comfortable styles of dress for women

Step-by-step explanation:

to promote safer, more comfortable styles of dress for women

Answer:

1/xyz^3

Step-by-step explanation:

exponents are subtracted when divided

x^0 is 1

negative exponents are positive in the denominator

(4x^0y^-2z^-3) / (4xy-1)

(4xy^-2z^-3) / (4xy^-1)

x-1y^-1z^-3

1/xyz^3

Last option. The correct statement about the graph is g(x) is an odd degree polynomial with a positive leading coefficient.

We have the function as

[tex]f(x) = x^2 - 2x + 1[/tex]

The polynomial that we have here is of an even degree. The highest degree we have here is 2. The coefficient that has the highest degree is 1.

Hence all of the options that have f(x) are incorrect.

The answer to the question is option 4.

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Based on the given variation, y does not vary directly with x and the constant of variation are 8, 3.2 and 1.25 respectively.

y = k × x

where,

k = constant of proportionality

y = -40

x = -5

y = k × x

-40 = k × -5

-40 = -5k

k = -40/-5

k = 8

when,

y = 8 and x = 2.5

y = k × x

8 = k × 2.5

8 = 2.5k

k = 8/2.5

k = 3.2

when,

y = 5 and x = 4

y = k × x

5 = k × 4

5 = 4k

k = 5/4

k = 1.25

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The intercepts are x intercept: (-1/7,0) and y intercept: (0,1/4)

The function is given as:

7x-5=4y-6

Set y = 0 to determine the x-intercept

7x - 5 = -6

Add 5 to both sides

7x = -1

Divide by 7

x = -1/7

Set x = 0 to determine the y-intercept

- 5 = 4y - 6

Add 6 to both sides

4y = 1

Divide by 4

y = 1/4

Hence, the intercepts are x intercept: (-1/7,0) and y intercept: (0,1/4)

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Answer:

? = 38°

Step-by-step explanation:

We can use the property of sine to find the missing angle value.

sin (angle) = opposite / hypotenuse

sin (?) = 16 / 26                                  <----- Insert values

? = sin⁻¹(16 / 26)                                 <----- Rearrange

? = 37.97...                                         <----- Solve

? = 38°                                               <----- Round

Answer:

(っ◔◡◔)っ ♥ the answer is C ♥

Step-by-step explanation:

The height h of the triangle is 5.42 ft

The height of the triangle can be found using the formula below:

Area = 1 / 2 bh

where

Therefore,

A = √s(s - a)(s - b)(s - c)

Hence,

s = 14 + 7 + 11 / 2 = 32 / 2 = 16

A = √16(16 - 11)(16 - 7)(16 - 14)

A = √1440

A = 37.947331922

A = 37.95 ft²

37.95 = 1 / 2 × 14 × h

h = 37.95 / 7

h = 5.42104741743

h = 5.42 ft

Therefore, the height h of the triangle is 5.42 ft

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The solutions are: θ = 90º, 270º, 30º, and 150º

Given,

sin2θ = cosθ

sin2 θ = 2sinθ cosθ

2sinθ cosθ = cosθ

2sinθ cosθ - cosθ = 0

factor

cosθ(2sinθ-1) = 0

cosθ = 0 or 2sinθ - 1 = 0

cosθ = 0 when θ=90º or θ=270º (look on the unit circle)

2sinθ - 1 = 0

2sinθ = 1

sinθ = 1/2

look at the unit circle and see that sinθ=1/2 when θ=30º

sin is positive in Quadrant2 as well as Quadrant 1

180º-30º=150º

θ=150º

answers: θ=90º, 270º, 30º, and 150º

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1) [tex]\overline{AB} \perp \overline{BD}[/tex], [tex]\overline{AB} \cong \overline{DB}[/tex], [tex]\overline{AC} \cong \overline{DE}[/tex] (given)

2) [tex]\angle ABD[/tex] is a right angle (perpendicular lines form right angles)

3) [tex]\triangle ABD[/tex] and [tex]\triangle DEB[/tex] are right triangles (a triangle with a right angle is a right angle)

4) [tex]\triangle ABD \cong \triangle DEB[/tex] (HL)

5) [tex]\angle ACB \cong \angle DEB[/tex] (CPCTC)

The angle of m∠5 is 116 degrees.

When parallel lines are cut by a transversal, angle relationships are formed. This include corresponding angles, alternate angles , vertical angles etc.

Therefore, line m and n are parallel lines cut by the a transversal.

Hence,

m∠2 = 64 degrees

Therefore,

m∠5 = m∠4 (alternate angles)

Therefore,

m∠5  + m∠2 = 180

m∠5  = 180 - 64

m∠5  = 116 degrees.

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The degree measure of the angle labelled 5; m<5 = 116°.

The angle measure of the angle labelled 5 as in the complete task content in the attached image can be determined as follows;

It follows from the complete task content that; that the measure of angle 2 is; 64°.

On this note, it follows from the sum of angle measures on a straight line theorem that; m<4 = 180° - 64° = 116°.

Additionally, it follows from the congruence of alternate angles theorem that <4 and <5 are congruent and hence, their measures are equal.

Therefore, m<5 = 116°.

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The solutions of x² - 3x + 2 ≡ mod 5 does not exist.

We are given the Polynomial;

x² - 3x + 2 ≡ mod 5

Finding the roots of the equation gives;

x² - 2x - x + 2 = 0(mod 5)

x(x - 2) - 1(x - 2) = 0(mod 5)

Thus, RHS = 0(mod 5) = 0

The roots of the equation are;

x = 1 and x = 2

Thus, the solutions of x² - 3x + 2 ≡ mod 5 does not exist.

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The solution to the homogenous differential equation, (11y² - xy)dx + x²dy = 0, is [tex]y = \frac{x}{11 ln (x) + C}[/tex].

A homogeneous differential equation has a homogeneous function as one of its components, if f(λx, λy) = λⁿf(x, y), for any non-zero constant λ, the function is said to be homogenous. The homogeneous differential equation's generic form is of the type f(x, y).dy + g(x, y).dx = 0. The degree of the homogeneous differential equation is the same for the equation's variables x and y.

In the question, we are asked to solve the homogeneous differential equation, (11y² - xy)dx + x²dy = 0.

The given equation can be solved as follows:

Thus, the solution to the homogenous differential equation, (11y² - xy)dx + x²dy = 0, is [tex]y = \frac{x}{11 ln (x) + C}[/tex].

The provided question is incomplete. The complete question is:

"Use the method for solving homogenous equations to solve the following differential equation.

(11y² - xy)dx + x²dy = 0".

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Answer:

The distance between the points (4,3) and (1,-1) is C. 5 units.

Step-by-step explanation:

Answer:

1 +10=11 (since 0 is neither negative or positive the first and second positive integerswith 1 or 0

Considering the area of the rectangle, we have that the length is of 8 units and the width is of 10 units.

The area of a rectangle of length l and width w is given as follows:

A = lw.

In this problem, the area is of 80 square units, while the length and the width are consecutive even integers, hence:

Then:

l(l + 2) = 80

l² + 2l - 80 = 0

(l + 10)(l - 8) = 0.

We need the positive measure, hence:

l -  8= 0 -> l = 8 units.

w = l + 2 = 10 units.

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The probability of winning a lottery by selecting the correct six integers, are

[tex]\begin{aligned}&(a) 1.68 \times 10^{-6} \\&(b) 5.13 \times 10^{-7} \\& (c) 1.91 \times 10^{-7} \\&(d)8.15 \times 10^{-8}\end{aligned}[/tex]

The binomial distribution is a type of probability distribution that expresses the probability that, given a certain set of characteristics or assumptions, a value would take one of two distinct values.

Part (a); positive integers not exceeding 30.

To calculate the probability, use binomial coefficients. Pick six of the six accurate integers and none of the other twenty-four.

[tex]\frac{\left(\begin{array}{c}6 \\6\end{array}\right)\left(\begin{array}{c}24 \\0\end{array}\right)}{\left(\begin{array}{c}30 \\6\end{array}\right)}=\frac{1}{\left(\begin{array}{c}30 \\6\end{array}\right)}=1.68 \times 10^{-6}[/tex]

Part (b); positive integers not exceeding 36.

To calculate the probability, use binomial coefficients. Pick six of the six accurate integers and none of the other thirty.

[tex]\frac{\left(\begin{array}{l}6 \\6\end{array}\right)\left(\begin{array}{c}30 \\0\end{array}\right)}{\left(\begin{array}{c}36 \\6\end{array}\right)}=\frac{1}{\left(\begin{array}{c}36 \\6\end{array}\right)}=5.13 \times 10^{-7}[/tex]

Part (c); positive integers not exceeding 42.

To calculate the probability, use binomial coefficients. Pick six of the six accurate integers and none of the 36 other integers.

[tex]\frac{\left(\begin{array}{l}6 \\6\end{array}\right)\left(\begin{array}{c}36 \\0\end{array}\right)}{\left(\begin{array}{c}42 \\6\end{array}\right)}=\frac{1}{\left(\begin{array}{c}42 \\6\end{array}\right)}=1.91 \times 10^{-7}[/tex]

Part (d); positive integers not exceeding 48.

To calculate the probability, use binomial coefficients. Choose six of the six accurate integers and none of the other 42.

[tex]\frac{\left(\begin{array}{l}6 \\6\end{array}\right)\left(\begin{array}{c}42 \\0\end{array}\right)}{\left(\begin{array}{c}48 \\6\end{array}\right)}=\frac{1}{\left(\begin{array}{c}48 \\6\end{array}\right)}=8.15 \times 10^{-8}[/tex]

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The complete question is-

Find the probability of winning a lottery by selecting the correct six integers, where the order in which these integers are selected does not matter, from the positive integers not exceeding a) 30. b) 36. c) 42. d) 48.

Answer:

Hello,

Step-by-step explanation:

here is the answer in a picture.

Answer:

  yes

Step-by-step explanation:

The point can be demonstrated to lie on the circle by plotting both on a graph. It can also be shown to lie on the circle by showing the distance from center is equal to the radius.

The attachment shows a graph of the circle. Its equation is ...

  (x -h)² +(y -k)² = r² . . . . . . for center (h, k) and radius r

We have (h, k) = (2, 13) and r = 5, so the equation is ...

  (x -2)² +(y -13)² = 25 . . . . graphed equation

The point in question is also graphed, and is shown to lie on the circle.

If the given point satisfies the circle's equation, it lies on the circle.

For (x, y) = (-2, 10), we find ...

  (-2 -2)² +(10 -13)² = (-4)² +(-3)² = 16 +9 = 25 . . . . . the equation is satisfied

Using natural logarithms properties, we will see that:

x = -1.596

Here we can use the property:

[tex]ln(a^b) = b*ln(a)[/tex]

Now we have the equation:

[tex]4^{x + 3 }= 7[/tex]

If we apply the natural logarithm in both sides, we get:

[tex]ln(4^{x + 3 })= ln(7)\\\\(x + 3)*ln(4) = ln(7)\\\\x + 3 = ln(7)/ln(4)\\\\x = ln(7)/ln(4) - 3 = log_4(7) - 3 = -1.596[/tex]

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Answer: (-2, 1)

Step-by-step explanation:

       We will graph the given equations. The point of intersection is the solution to the system of equations. See attached.

       They intersect at the point (-2, 1) giving us an answer of x = -2 and y = 1, but this question asks for the answer as an ordered pair so we leave it as is.

The area covered by the advertisement on the page is [tex]ab \ unit^2[/tex].

The area of a rectangle covers is the space it takes up inside the boundaries of its four sides. The dimensions of a rectangle determine its area. In general, the area of a rectangle is equal to the product of its length and breadth of the rectangle.

The formula of area of rectangle is [tex]length \times breadth[/tex] [tex]unit^2[/tex]

let the given page represents a rectangle.

We take the length of the advertisement is [tex]a[/tex] unit and breadth of the advertisement is [tex]b[/tex] unit.

The area of the advertisement [tex]=a \times b \ unit^2[/tex]

                                                   [tex]=ab \ unit^2[/tex]

Therefore, the area covered by the advertisement on the page is [tex]ab \ unit^2[/tex]

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