what does your work show about the behavior of the standard deviation of a binomial distribution as the probability of a success gets closer to zero?

A fixed ratio schedule is a type of reinforcement schedule in which a response is reinforced after a specific number of responses. In this case, the response of putting money in the machine is reinforced after every 1 response.

If each time you put a dollar in a vending machine you get a snack, then your response is reinforced according to a fixed ratio schedule.

Variable interval, fixed interval, and variable ratio schedules are all different types of reinforcement schedules that are based on different criteria. A variable interval schedule is a type of reinforcement schedule in which a response is reinforced after a random interval of time. A fixed interval schedule is a type of reinforcement schedule in which a response is reinforced after a fixed interval of time. A variable ratio schedule is a type of reinforcement schedule in which a response is reinforced after a random number of responses.

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Answer:

E. 2,850 ft²

Step-by-step explanation:

Answer:

OptionE)

Step-by-step explanation:

Area of a circle = [tex]\pi r^{2}[/tex]

[tex]3.14X30X30[/tex]

= [tex]2826[/tex]

Hence the closest is 2,850

To obtain a smaller margin of error, choose a smaller confidence level.

Margin of error is defined as the degree of the sampling errors in statistics. It can be calculated using the formula below.

MOE = z x (SD / √n)

where MOE = margin of error

z = found by using a z-score table

SD = sample standard deviation

n = sample size

Based on the formula, to reduce the margin of error, the z-score, z, must be small. In order to have a smaller value of z, the probability, p, should also be small. And to obtain a small probability, choose a smaller confidence level.

Hence, to reduce the margin of error, use a lower confidence level.

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The equation of the line that represents the graph will be y = x.

A connection between a number of variables results in a linear model when a graph is displayed. The variable will have a degree of one.

The linear equation is given as,

y = mx + c

Where m is the slope of the line and c is the y-intercept of the line.

From the graph, the two points are (0, 0) and (1, 1).

The line is passing through the origin that is (0, 0), then we have

0 = m (0) + c

c = 0

Then the equation is given as,

y = mx

The equation also passes through (1, 1), then we have

1 = m (1)

m = 1

Then the equation is written as,

y = x

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No, it is not possible to draw an quilateral triangle for which the circumecenter, incenter, centroid, and orthocenter are not all the same point.

An equilateral triangle, which is frequently referred to as a "regular" triangle, is a triangle with three equal-length sides.

For the stated question-

Thus, the statement given by the friend is incorrect.

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The value of the given expression, 3^4 + 18 ÷ 3 +2, is 89

From the question, we are to determine the value of the given expression

The given expression is

3^4 + 18 ÷ 3 +2

First, we will write the expression properly

The given expression written properly is

3⁴ + 18 ÷ 3 +2

Using PEMDAS

P - Parentheses

E - Exponent

M - Multiplication

D - Division

A - Addition

S - Subtraction

Then,

3⁴ + 18 ÷ 3 +2

81 + 18 ÷ 3 + 2

81 + 6 + 2

= 89

Hence, the value is 89

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37.96 sq units is the area enclosed by the graph [tex]y=\frac{20}{1+x^2}[/tex] and y=2

To calculate the area enclosed by 2 graphs we have to firstly find the intersection point of both graph

to find the intersection point we will equate given equation

=>    [tex]\frac{20}{1+x^2}[/tex] = 2

=>  [tex]1+x^2[/tex] = 10

=> [tex]x^2[/tex] = 9

=> x= 3,-3

so we have to find area of graph from x=-3 and x=3

so the formula for equation is given by:

[tex]\int\limits^b_a {\frac{20}{1+x^2} -2} \, dx[/tex]

where a=-3 and b=3

=>[tex]\int\limits^a_b {\frac{20}{1+x^2} } \, dx - \int\limits^a_b {2} \, dx[/tex]                      

[tex]\int\limits {\frac{1}{1+x^2} } \, dx = tan^-^1x[/tex] +c

=> 20 [tex]\left \{ {{x=3} \atop {x=-3}} \right.[/tex]  [tex]tan^-^1x[/tex] - [tex]\left \{ {{x=3} \atop {x=-3}} \right.[/tex]  2x

=> 20 (  [tex]tan^-^1 3 - tan^-^1 (-3)[/tex]  - 2*(3+3)

=>40 * 1.249 -12

=> 49.96 -12

=> 37.96

so the area enclosed by the given graphs is 37.96 sq. units

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The number sentence needed to find the number of boxes of plain oatmeal is given as follows:

6 x 7 = 42.

From the image given at the end of the answer, we have that:

Then the expression that gives the total number of boxes of plain oatmeal can be given by the addition of six terms seven, as follows:

7 + 7 + 7 + 7 + 7 + 7.

Consecutive additions are nothing more than multiplication expressions, hence the number sentence needed to find the number of boxes of plain oatmeal is given as follows:

6 x 7 = 42.

(this is because the product of 6 and 7 has a result of 42, that is, 7 + 7 + 7 + 7 + 7 + 7 = 42).

The problem is given by the image shown at the end of the answer.

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The length of side AC of the triangle ABC using cosine rule is; 10.252 cm

The cosine formula to find the side lengths of the triangle is given by:

b = √[a² + c² – 2ac cos B]

Where a, b and c are the sides of the triangle.

The Cosine Rule is useful in any triangle that we are trying to relate all three sides to one angle. Thus, when we want to find the length of a side, we need to know the other two sides and the opposite angle.

Considering triangle ABC, we have;

a = 8cm

c = 11 cm

B = 63°

Thus;

b = √[8² + 11² – 2(8 * 11) cos 63]

b = √(185 - 79.9)

b = 10.252 cm

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Jennie has 20 points.

Explanation:

to Ada's points is 4 to 7. If Ada has 15 more points than Jennie, how many points does Jennie have?

To solve this problem, we can use proportions. We know that the ratio of Jennie's points to Ada's points is 4 to 7, so we can set up the following proportion:

4/7 = Jennie's points / Ada's points

We also know that Ada has 15 more points than Jennie, so we can add 15 to both sides of the proportion to represent this:

4/7 = Jennie's points / (Jennie's points + 15)

Next, we can cross-multiply to solve for Jennie's points:

4 * (Jennie's points + 15) = 7 * Jennie's points

4 * Jennie's points + 60 = 7 * Jennie's points

We can then subtract 4 * Jennie's points from both sides to isolate the variable:

60 = 3 * Jennie's points

Finally, we can divide both sides by 3 to find the value of Jennie's points:

Jennie's points = 60/3 = 20

Thus, Jennie has 20 points.

Answer:

86.5

We add them together, which is 173, and since there are two terms, divide by 2

173/2=86.5

Answer:

86.5

127 + 46 / 2 = 86.5

I could be wrong but I believe this is correct.

Step-by-step explanation:

x/2 = 3×x/6

sin(x/2) = sin(3x/6)

cos(x/2) = cos(3x/6)

tan(x/2) = tan(3x/6) = sin(3x/6)/cos(3x/6)

looking up triple angle identities for trigonometric functions we find :

cos(3x) = 4cos³(x) - 3cos(x)

so,

cos(x/2) =

cos(3x/6) = 4cos³(x/6) - 3cos(x/6) = 4(3/5)³ - 3×3/5 =

= 4×27/125 - 9/5 = 108/125 - 9×25/125 =

= 108/125 - 225/125 = -117/125 =

= -0.936

sin(3x) = 3×sin(x) - 4×sin³(x)

so,

sin(x/2) =

sin(3×x/6) = 3×sin(x/6) - 4×sin³(x/6)

also remember,

sin²(x) + cos²(x) = 1

therefore,

sin²(x/6) + cos²(x/6) = 1

sin(x/6) = sqrt(1 - cos²(x/6)) = sqrt(1 - (3/5)²) =

= sqrt(1 - 9/25) = sqrt(25/25 - 9/25) =

= sqrt(16/25) = 4/5 = 0.8

so, again

sin(x/2) =

sin(3×x/6) = 3×sin(x/6) - 4×sin³(x/6) =

= 3×4/5 - 4×(4/5)³ =

= 12/5 - 4×64/125 = 12×25/125 - 256/125 =

= 300/125 - 256/125 = 44/125 =

= 3×0.8 - 4×0.8³ = 0.352

tan(x/2) = sin(x/2)/cos(x/2) =

= 44/125 / -117/125 = -44/117 =

= 0.352 / -0.936 =

= -0.376068376...

Media tonelada son 500kg, con esa cantidad Rosa logra hacer 100 cajas de 5kg. En una semana ha vendido 125 kilos, es decir 25 cajas. Aún le quedan 375kg.

Para calcular los kilos de ropa que tiene Rosa debemos tener en cuenta que una tonelada equivale a 1000kg. De acuerdo con esto, media tonelada sería equivalente a 500kg.

1000kg / 2 = 500kg

Para saber cuántas cajas de 5kg hace con media tonelada debemos realizar la siguiente operación:

500 / 5 = 100

Para identificar cuántos kilos ha vendido Rosa, debemos identificar a cuántos kilos equivale 1/4 de media tonelada, para ello realizamos la siguiente operación.

500kg / 4 = 125kg

Para identificar a cuántas cajas equivale lo vendido por Rosa, debemos realizar la siguiente operación:

125kg / 5kg = 25 cajas

Para identificar cuántos kilos le faltan por vender, debemos realizar la siguiente operación:

500kg - 125kg = 375kg.

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The product of the value of each outcome and the likelihood that it will occur results in the expected value. The likelihood of each outcome occuring for the ticket wheel is inversely correlated to the number of points on the wheel for that outcome.

When N is the number of slots on the wheel for 15 tickets, for instance, the probability of winning 15 tickets is N/20.

So, the following is the expected value of the number of tickets won:

[tex](15 × \frac{N}{20} ) + (25 × \frac{N}{20} ) + (30 × \frac{N}{20} ) + (40 × \frac{N}{20} ) + (10 × \frac{2}{20} ) + (100 × \frac{6}{20} ) + (500 \times \frac{3}{20} )[/tex]

Simplifying this expression, we get:

[tex] \frac{(15N + 25N + 30N + 40N + 20 + 600 + 1500)}{20} [/tex]

Further simplifying, we get:

[tex]\frac{(100N + 2120)}{20} [/tex]

This expression can be further simplified by dividing the numerator and denominator by 5 to get:

[tex] \frac{(20N + 424) }{4} [/tex]

The response to the first query is that this expression is equal to the expected value of the quantity of tickets won. Therefore, (d) 102 tickets is the right response.

The percentage of data that falls within two standard deviations of the mean for all normal curves is what is being questioned in the second question. The correct response is (b) 95% because this is roughly 95% according to the Empirical Rule.

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When the variables r = 3 and y = -6, the value of the expression is -27.

In Mathematics, an expression is sometimes referred to as an equation and it can be defined as a mathematical equation which is typically used for illustrating the relationship that exist between two (2) or more variables, parameters, and numerical quantities (number).

Based on the information provided, the mathematical expression is given by;

Mathematical expression = √x⁴ - y²

Substituting the given parameters into the mathematical expression, we have;

Mathematical expression = √(3)⁴ - (-6)²

Mathematical expression = √81 - 36

Mathematical expression = 9 - 36

Mathematical expression = -27.

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Complete Question:

Type the correct answer in the box. use numerals instead of words. Enter the number of the topic sentence.

Consider this expression. √x⁴ - y². When r = 3 and y = -6, the value of the expression is____.

Answer:

graph B

Step-by-step explanation:

if x ≥ 0

x + 3 > 7

x > 7 - 3

x >4

if x < 0

-x + 3 > 7

-x > 7-3

-x > 4

x < -4

final solution

x < -4 ∨ x > 4

The value of the side x for the given similar triangle will be 12 units.

The three-sided shape known as a triangle is sometimes used to allude to it. Every triangle has three sides and three angles, some of which might be the same.

The area of a triangle is defined as (1/2) base height, and the sum of all three angles within a triangle will be 180°.

As per the given both triangle,

The one angle of both triangles is 90° and one angle is the common angle.

Since the two angles are the same thus the third angle will be the same thus by the AAA postulate both triangles will be similar.

By a similar triangle rule,

20/x = 25/(√(25² - 20²))

x = 20/25 (15)

x = 12

Hence "The value of side x will be 12 units by AAA postulate".

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The value of SU is 4.7 units

An isosceles trapezoid can be defined as a trapezoid in which non-parallel sides and base angles are of the same measure.

Given is an isosceles trapezoid,

In an isosceles trapezoid, the triangles Δ SWV and Δ TWU must be congruent. Then, the corresponding sides are congruent too:

TW ≅ SW

UW ≅ VW

From the given information,

UW = 3

SU = SW+UW

Finally, using the values of SW and UW:

SU = 1.7+3 = 4.7

Hence, the value of SU is 4.7 units.

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Answer:

4.7

Step-by-step explanation:

Which function is growing on the (-, ) interval? O A. g(x) = "-4(2^x)" O B. f(x)= "-3x"+7 O C. h(x)= 2x—1 O D. j(x)= x2 + 8x+1

Try searching for Which function is rising on the range (-, ) instead. O A. g(x) = -4(2^x) O B. f(x) = -3x + 7, O c. h(x) = 2x - 1, and O D. j(x) = x2 + 8x + 1.

Which function is increasing on the interval (- ∞ ∞?

Image result for Which function is increasing on the interval (-∞, ∞)? O A. g(x) = -4(2^x) O B. f(x) = -3x + 7 O c. h(x) = 2^x - 1 O D. j(x) = x² + 8x + 1

Since, x and y are arbitrary values, therefore, f (x) < f (y) whenever x < y. Therefore, the interval (-∞, ∞) is a strictly increasing interval for f(x) = 3x + 5

The interval is increasing if the value of the function f(x) increases with an increase in the value of x and it is decreasing if f(x) decreases with a decrease in x.

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Line u and line v are perpendicular lines by the relation of their slopes.

Slope of a line is the ratio of the difference of the y coordinates of two points to the x coordinates of those points. Slope is commonly denoted by the letter 'm'.

If a line passes through the points (x₁, y₁) and (x₂, y₂), then slope of the line,

m = (y₂ - y₁) / (x₂ - x₁)

Slopes of two lines, m₁ and m₂, are equal if they are parallel ⇒ m₁ = m₂.

Slopes of two lines, if the lines are perpendicular have the relation,

m₁ × m₂ = -1.

First let's find the slopes of the given lines.

Let m₁ be the slope of line u and m₂ be the slope of line v.

Line u passes through (7, 2) and (10, 10).

m₁ = (10 - 2) / (10 - 7) = 8/3

Line v passes through (10, 8) and (2, 11).

m₂ = (11 - 8) / (2 - 10) = 3/(-8)

Slopes are not equal, so they are not parallel.

m₁ × m₂ = [tex]\frac{8}{3}[/tex] × [tex]\frac{3}{-8}[/tex] = -1.

So the lines are perpendicular.

Hence the given lines u and v are perpendicular.

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Answer:

48p²+4p-4

Step-by-step explanation:

(8p-2)(6p+2)

=48p²+16p-12p-4

=48p²+4p-4

Answer: x = 0.1396263404

Step-by-step explanation:

The expression x+16 = 4x-15

(3 x (-1)) * degrees =-0.0523598776

6 x degrees) + (5 x degrees) =0.191986218

For #1

For a penny of diameter 19mm to be of the same apparent size as the moon it should be held at a distance of about 2.1m away from the eye.

Therefore, the answer is 2.1m.

Distance of moon from earth is 384,400km, that is 384,400,000m. Also 3,474.8km that is 3,474,800m.

For penny to have same apparent size as the moon, it should subtend equal angle as moon to the viewer. See the attached figure for reference. Thus ΔABC is similar to ΔEDB by AA criterion as ∠A is common and ∠BAC = ∠DEB as they are corresponding angles.

Let distance between penny and observer be x. Then

x/ ED = 384,400,000m/ AB

x/ 19mm = 384,400,000m/ 3,474,800m

x/ 19mm = 961000/ 8687

x = 961000/ 8687 × 19

x ≈ 2101.88mm

x = 2.1m

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well, from 1990 to 1998 is 8 years, and we know the amount went from $4800 to $6300, let's check for the rate of growth.

[tex]\qquad \textit{Amount for Exponential Growth} \\\\ A=P(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$6300\\ P=\textit{initial amount}\dotfill &\$4800\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{years}\dotfill &8\\ \end{cases} \\\\\\ 6300=4800(1 + \frac{r}{100})^{8} \implies \cfrac{6300}{4800}=(1 + \frac{r}{100})^8\implies \cfrac{21}{16}=(1 + \frac{r}{100})^8[/tex]

[tex]\sqrt[8]{\cfrac{21}{16}}=1 + \cfrac{r}{100}\implies \sqrt[8]{\cfrac{21}{16}}=\cfrac{100+r}{100} \\\\\\ 100\sqrt[8]{\cfrac{21}{16}}=100+r\implies 100\sqrt[8]{\cfrac{21}{16}}-100=r\implies \stackrel{\%}{3.46}\approx r[/tex]

now, with an initial amount of $4800, up to 2007, namely 17 years later, how much will that be with a 3.46% rate?

[tex]\qquad \textit{Amount for Exponential Growth} \\\\ A=P(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &4800\\ r=rate\to 3.46\%\to \frac{3.46}{100}\dotfill &0.0346\\ t=years\dotfill &17\\ \end{cases} \\\\\\ A=4800(1 + 0.0346)^{17} \implies A=4800(1.0346)^{17}\implies A \approx 8558.02[/tex]

Answer:

Step-by-step explanation:

11 = 5 * 2 * 22 + 2 * 5

11 = 110 + 10

11 = 120

11 = 5 * 4 * 22 + 2 * 3

11 = 440 + 6

11 = 446

Please let me know if you have any questions.

Answer:Cannot be factored

Step-by-step explanation:

Answer:

[tex]x(x+2)(x-4)[/tex]

Step-by-step explanation:

Given:

[tex]x^3-2x^2-8x[/tex]

Factor out the common term x:

[tex]\implies x(x^2-2x-8)[/tex]

Factor (x² - 2x - 8) by rewriting -2x as (-4x + 2x):

[tex]\implies x \left[ x^2-4x+2x-8 \right][/tex]

Factor the first two terms and the last two terms separately:

[tex]\implies x \left[x(x-4)+2(x-4) \right][/tex]

Factor out the common term (x - 4):

[tex]\implies x \left[ (x+2)(x-4) \right][/tex]

Therefore:

[tex]x^3-2x^2-8x=x(x+2)(x-4)[/tex]

The values of the variables and the measures of the angles, found using the congruency of ΔAOI and ΔIEA are;

w = 2        [tex]{}[/tex]   ∠EAI = 45°

x = 4   [tex]{}[/tex]         ∠AOI = 128°

y = 7 [tex]{}[/tex]           ∠OIA = 45°

z = 20 [tex]{}[/tex]        ∠IEA = 128°

Two triangles are congruent if two angles and a non included side of one triangle are congruent to two angles and a non included side of another triangle.

The relationship between the triangles ΔAOI and ΔIEA is ΔAOI ≅ ΔIEA

Therefore; ∠IEA is congruent to ∠AOI, by Corresponding Parts of Congruent Triangles are Congruent, CPCTC

∠IEA ≅ ∠AOI

Therefore; ∠IEA = ∠AOI by the definition of congruency

(6·z + 8)° = (7·z - 12)°

7·z - 6·z = (8 + 12)° = 20°

z = 20°

∠AOI = (7·z - 12)°

Therefore;

∠AOI = (7 × 20 - 12)° = 128°

∠IEA = 128°

Similarly; AO ≅ EI by CPCTC

AO = EI definition of congruency

5·w - 6 = 2·w

5·w - 2·w = 6

3·w = 6

w = 6 ÷ 3 = 2

w = 2

∠OIA ≅ ∠EAI by CPCTC

Therefore; ∠OIA = ∠EAI definition of congruency

(7·x + 17)° = 3·(4·x - 1)° substitution property

(7·x + 17)° = 3·(4·x - 1)° = (12·x - 3)°

12·x - 7·x = 17° + 3° = 20°

5·x = 20°

x = 20° ÷ 5 = 4°

x = 4°

∠EAI = 3·(4·x - 1)°

∠OIA = (7·x + 17)°

Therefore;

∠EAI = 3·(4×4 - 1)° = 45°

∠OIA = (7×4 + 17)° = 45°

y° + (7·z - 12)° + 3·(4·x - 1)° = 180° (angle sum property of a triangle)

∠AOI = (7·z - 12)° = 128°

∠EAI = 3·(4·x - 1)° = 45°

Therefore;

y° + 128° + 45° = 180°

y° = 180° - (128° + 45°) = 7°

y° = 7°

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The triangles that are congruent to (a) ΔCDE, and (b) ΔCDF include ΔBFG, ΔADG, ΔBDE, and  ΔABH, ΔBHCΔ, ΔGHE respectively

We should know that when two or more objects in geometry have the same shape and size, or if one has the same shape and size as the mirror image of the other, they are congruent.

From the object drawn the congruent triangles to ΔCDE include the following

ΔBDF

ΔABG

ΔBDE

ΔABI

ΔACH

ΔBCJ

ΔIHJ

All the triangle has side, side, right angle

The triangles that are congruent to ΔCDF include the following

ΔABG

ΔBHC

HGE

These triangles have SIDE, SIDE, SIDE (SSS)

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