What is the probability of flipping at least one head? Round the answer to the nearhundredth of a percent.

Answer:

The 2nd box is the 1st box is not

Step-by-step explation:

it just is thi sway

Answer:

nooe!

Step-by-step explanation:

the first set ads up to 133/120 whike the second ads up to 2 so they aren't :)

Answer:

x= -2

-------

7

Step-by-step explanation:

<1. combine like terms.>

<2.divide both sides of the equation by same terms.>

<3.Simplify.>: then Answer while come x= -2

------

7

Answer: \(7.39466666\)

Step-by-step explanation:

Setting \(x=2.2\) and \(n=3\),

\(e^{2.2} \approx \sum^{3}_{k=0} \frac{2.2^{k}}{k!}=\frac{2.2^0}{0!}+\frac{2.2^1}{1!}+\frac{2.2^2}{2!}+\frac{2.2^3}{3!} \approx 7.39466666\)

Answer:

7.39466667 (8 d.p.)

Step-by-step explanation:

We can approximate the value of eˣ for any x using the following sum formula:

\(\boxed{e^{x} \approx \displaystyle \sum^n_{k=0}\dfrac{x^k}{k!}}\)

To approximate \(e^{2.2}\) with n = 3, substitute x = 2.2 and n = 3 into the given sum formula:

\(e^{2.2} \approx \displaystyle \sum^3_{k=0}\dfrac{2.2^k}{k!}\)

To calculate the sum, substitute k with each value from 0 to 3 and add the results together:

\(\begin{aligned}e^{2.2} &\approx \displaystyle \sum^3_{k=0}\dfrac{2.2^k}{k!}\\\\&= \dfrac{2.2^0}{0!}+\dfrac{2.2^1}{1!}+\dfrac{2.2^2}{2!}+\dfrac{2.2^3}{3!}\\\\&= \dfrac{1}{1}+\dfrac{2.2}{1}+\dfrac{4.84}{2}+\dfrac{10.648}{6}\\\\&= 1+2.2+2.42+1.774666666...\\\\&=7.39466667\; \sf (8\;d.p.)\end{aligned}\)

Therefore, the approximate value of \(e^{2.2}\) is:

\(\large \textsf{$e^{2.2}$}=\boxed{7.39466667}\; \sf (8\;d.p.)}\)

Note: To obtain a more accurate approximate value, increase the value of n.

Answer:

Step-by-step explanation:

The growth rate of the investment is represented by the constant multiplier in the exponential term of the formula y = 20000(1.035)x. In this case, the constant multiplier is 1.035, which represents the annual growth rate of the investment account.

To calculate the growth rate as a percentage, we can subtract 1 from the constant multiplier, and then multiply the result by 100.

So, the growth rate of the investment is:

1.035 - 1 = 0.035

0.035 * 100 = 3.5%

Therefore, the growth rate of the investment is 3.5%.

Answer:

Refer to the step-by-step explanation, please follow along very carefully. Answers are encased in two boxes.

Step-by-step explanation:

Given the following function, find it's derivative using the definition of derivatives. Evaluate the function when θ=1, 11, and 3/11

\(p(\theta)=\sqrt{11\theta}\)

\(\hrulefill\)

The definition of derivatives states that the derivative of a function at a specific point measures the rate of change of the function at that point. It is defined as the limit of the difference quotient as the change in the input variable approaches zero.

\(f'(x) = \lim_{{h \to 0}} \dfrac{{f(x+h) - f(x)}}{{h}}\)\(\hrulefill\)

To apply the definition of derivatives to this problem, follow these step-by-step instructions:

Step 1: Identify the function: Determine the function for which you want to find the derivative. In out case the function is denoted as p(θ).

\(p(\theta)=\sqrt{11\theta}\)

Step 2: Write the difference quotient: Using the definition of derivatives, write down the difference quotient. The general form of the difference quotient is (f(x+h) - f(x))/h, where "x" is the point at which you want to find the derivative, and "h" represents a small change in the input variable. In our case:

\(p'(\theta) = \lim_{{h \to 0}} \dfrac{{p(\theta+h) - p(\theta)}}{{h}}\\\\\\\Longrightarrow p'(\theta)= \lim_{h \to 0} \dfrac{\sqrt{11(\theta + h)} - \sqrt{11\theta} }{h}\)

Step 3: Take the limit:

We need to rationalize the numerator. Rewriting using radical rules.

\(p'(\theta)= \lim_{h \to 0} \dfrac{\sqrt{11(\theta + h)} - \sqrt{11\theta} }{h} \\\\\\\Longrightarrow p'(\theta)= \lim_{h \to 0} \dfrac{\sqrt{11\theta + 11h} - \sqrt{11\theta} }{h}\\\\\\\Longrightarrow p'(\theta)= \lim_{h \to 0} \dfrac{\sqrt{11}\sqrt{\theta+h} - \sqrt{11}\sqrt{\theta} }{h}\)

Now multiply by the conjugate.

\(p'(\theta)= \lim_{h \to 0} \dfrac{\sqrt{11}\sqrt{\theta+h} - \sqrt{11}\sqrt{\theta} }{h} \cdot \dfrac{\sqrt{11}\sqrt{\theta+h} + \sqrt{11}\sqrt{\theta} }{\sqrt{11}\sqrt{\theta+h} + \sqrt{11}\sqrt{\theta} } \\\\\\\Longrightarrow p'(\theta)= \lim_{h \to 0} \dfrac{(\sqrt{11}\sqrt{\theta+h} - \sqrt{11}\sqrt{\theta} )(\sqrt{11}\sqrt{\theta+h} + \sqrt{11}\sqrt{\theta} )}{h(\sqrt{11}\sqrt{\theta+h} + \sqrt{11}\sqrt{\theta} )} \\\\\\\)

\(\Longrightarrow p'(\theta)= \lim_{h \to 0} \dfrac{11h}{h(\sqrt{11}\sqrt{\theta+h} + \sqrt{11}\sqrt{\theta} )}\\\\\\\Longrightarrow p'(\theta)= \lim_{h \to 0} \dfrac{11}{\sqrt{11}\sqrt{\theta+h} + \sqrt{11}\sqrt{\theta} }\)

Step 4: Simplify the expression: Evaluate the limit by substituting the value of h=0 into the difference quotient. Simplify the expression as much as possible.

\(p'(\theta)= \lim_{h \to 0} \dfrac{11}{\sqrt{11}\sqrt{\theta+h} + \sqrt{11}\sqrt{\theta} }\\\\\\\Longrightarrow p'(\theta)= \dfrac{11}{\sqrt{11}\sqrt{\theta+(0)} + \sqrt{11}\sqrt{\theta} }\\\\\\\Longrightarrow p'(\theta)= \dfrac{11}{\sqrt{11}\sqrt{\theta} + \sqrt{11}\sqrt{\theta} }\\\\\\\Longrightarrow p'(\theta)= \dfrac{11}{2\sqrt{11}\sqrt{\theta} }\\\\\\\Longrightarrow p'(\theta)= \dfrac{11}{2\sqrt{11\theta} }\\\\\\\Longrightarrow p'(\theta)= \dfrac{11}{2\sqrt{11\theta} }\)

\(\therefore \boxed{\boxed{p'(\theta)= \dfrac{\sqrt{11} }{2\sqrt{\theta} }}}\)

Thus, we have found the derivative on the function using the definition.

It's important to note that in practice, finding derivatives using the definition can be a tedious process, especially for more complex functions. However, the definition lays the foundation for understanding the concept of derivatives and its applications. In practice, there are various rules and techniques, such as the power rule, product rule, and chain rule, that can be applied to find derivatives more efficiently.\(\hrulefill\)

Now evaluating the function at the given points.  

\(p'(\theta)= \dfrac{\sqrt{11} }{2\sqrt{\theta}}; \ p'(1)=??, \ p'(11)=??, \ p'(\frac{3}{11} )=??\)

When θ=1:

\(p'(\theta)= \dfrac{\sqrt{11} }{2\sqrt{\theta}}\\\\\\\Longrightarrow p'(1)= \dfrac{\sqrt{11} }{2\sqrt{1}}\\\\\\\therefore \boxed{\boxed{p'(1)= \dfrac{\sqrt{11} }{2}}}\)

When θ=11:

\(p'(\theta)= \dfrac{\sqrt{11} }{2\sqrt{\theta}}\\\\\\\Longrightarrow p'(11)= \dfrac{\sqrt{11} }{2\sqrt{11}}\\\\\\\therefore \boxed{\boxed{p'(11)= \dfrac{1}{2}}}\)

When θ=3/11:

\(p'(\theta)= \dfrac{\sqrt{11} }{2\sqrt{\theta}}\\\\\\\Longrightarrow p'(\frac{3}{11} )= \dfrac{\sqrt{11} }{2\sqrt{\frac{3}{11} }}\\\\\\\therefore \boxed{\boxed{p'(\frac{3}{11} )= \dfrac{11\sqrt{3} }{6}}}\)

Thus, all parts are solved.

The roots of given equation x^2 - 2x +3 are 1+4i , 1-4i

What is Quadratic Equation ?

Quadratic equation can be defined as the equation in which it is in the form of ax^2 + bx + c = 0

where c is a constant.

Given equations,

x^2 - 2x +3 = 0

so, we know that

the roots of a quadratic equation

= (- b + (√ b^2 - 4ac ))  / 2a , (- b - (√ b^2 - 4ac ))  / 2a

so,

here a = 1 b = -2 c = 3

by substituting the given values,

we get,

=  2+ (√ 4 - 4*1*3 ) / 2  ,  2- (√ 4 - 4*1*3 ) / 2

= 2+ (√-8)  / 2 , 2- (√-8)  / 2

= 2+8i / 2 , 2-8i / 2

= 1 + 4i , 1- 4i

Hence, The roots of given equation x^2 - 2x +3 are 1+4i , 1-4i

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Distance = √(17) Because 1 + 16 = 17

Answer:

35 units

Step-by-step explanation:

Perimeter of rectangle ABCD =2(L + W)

L = AD or BC (we only need to calculate one of the two)

W = AB or DC. (we only need to calculate one of the two).

Distance between A(-2, 6) and D(-8, -4):

\( AD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

Let,

\( A(-2, 6) = (x_1, y_1) \)

\( D(-8, -4) = (x_2, y_2) \)

\( AD = \sqrt{(-8 -(-2))^2 + (-4 - 6)^2} \)

\( AD = \sqrt{(-6)^2 + (-10)^2} \)

\( AD = \sqrt{36 + 100} = \sqrt{136} \)

\( AD = 11.7 \) (nearest tenth)

Distance between A(-2, 6) and B(3, 3):

\( AD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

Let,

\( A(-2, 6) = (x_1, y_1) \)

\( B(3, 3) = (x_2, y_2) \)

\( AB = \sqrt{(3 -(-2))^2 + (3 - 6)^2} \)

\( AB = \sqrt{(5)^2 + (-3)^2} \)

\( AB = \sqrt{25 + 9} = \sqrt{34} \)

\( AB = 5.8 \) (nearest tenth)

Perimeter of rectangle ABCD =2(AD + AB)

= 2(11.7 + 5.8)

= 2(17.5)

= 35 units

Step-by-step explanation:

The equation for this function is y=2x

So 32*2=64

36/2=18

2*x=2x

y/2=y/2

2x*2=4x

(x+3)*2=2x+6

Answer: 200 miles

Step-by-step explanation:

Answer:

20

Step-by-step explanation:

the first step is plugging in -2 for b and 4 for a.

4(-2) + 7(4)

-8 + 28 = 20

give brainliest please!

hope this helps :)

The apparent midpoint of AB is (3, 3) (option b).

To find the apparent midpoint of AB, we need to determine the coordinates that represent the midpoint of the line segment AB.

The given triangle ABC is placed on a grid. Since the coordinates are not provided for points A and B, we cannot directly calculate the midpoint using their coordinates. Therefore, we'll have to rely on the visual representation provided.

Looking at the grid, we can see that the line segment AB is a diagonal of the square formed by the grid lines. The square has sides of length 3 units, as it extends from (1, 1) to (4, 4).

The midpoint of a line segment is the point that divides the segment into two equal parts. Since the square has sides of length 3, the midpoint of AB should be at the halfway point between (1, 1) and (4, 4).

To calculate the coordinates of the midpoint, we take the average of the x-coordinates and the average of the y-coordinates.

The x-coordinate of the midpoint is (1 + 4) / 2 = 5 / 2 = 2.5.

The y-coordinate of the midpoint is (1 + 4) / 2 = 5 / 2 = 2.5.

Therefore, the apparent midpoint of AB is (2.5, 2.5).

However, none of the given options match the calculated midpoint. It's possible that there is an error or discrepancy in the given options. Based on the calculations, the correct apparent midpoint of AB should be (2.5, 2.5). Thus, the correct option is a.

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Answer:

0.3

Step-by-step explanation:

 2.1

-1.8

---------

0.3

Answer:

39 sq inches

Step-by-step explanation:

The probability that next year there will be no snowfall in that town on January 1 is,

P (no snowfall) = 0.770

The term probability refers to the likelihood of an event occurring. Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.

We have to given that;

Based on a weather record, the probability of snowfall in a certain town in New York on January 1 is 0.230.

Now, WE get;

the probability that next year there will be no snowfall in that town on January 1 is,

P (no snowfall) = 1 - P (snowfall)

P (no snowfall) = 1 - 0.230

P (no snowfall) = 0.770

Thus, The probability that next year there will be no snowfall in that town on January 1 is,

P (no snowfall) = 0.770

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Answer:

Step-by-step explanation:

That is cheating no help for you sorry

The inequality expression that corresponds to the solution of the inequality graph is x² + 3x - 3 < 0.

option B.

The inequality expression that corresponds to the solution of the inequality graph is determined by simplifying the equations as follows;

The solution of the graph,

x > -4 and x < 1

The first equation with "≥" is ruled out because the graph doesn't have a thick dot.

Let's simplify the second expression;

x² + 3x - 3 < 0

solve using quadratic formula;

x > -3.79 or x < 0.79

The third expression is ruled out since its solution will be complex.

For the last expression;

x² + 3x - 3 > 0

x < -3.79 or x > 0.79

Thus, the correct inequality expression is x² + 3x - 3 < 0.

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Answer:

You have 7,167

Step-by-step explanation:

If this was to be converted into a whole number, it would be:

7,167

Answer:

1

Step-by-step explanation:

If C were to be 0, y would have to be 1 bc 7+5=12, and c is already zero, so you need 1 more, so y is 1

Answer:

$0.64 per cup

Step-by-step explanation:

There are 16 cups in 1 gallon, so the number of cups in 2 gallons is:

1 gallon: 16 cups

2 gallon = 2 x 1 gallon = 2 x 16 cups = 32 cups

So we need to find the price of each cup:

1 cup = ($20.48 / 32 cups) = $0.64 per cup.

Answer:

11 + 10i

Step-by-step explanation:

Just treat i like any other variable, and combine like terms. Hope that helps!

The number that is equal to four and five sixths is A. four and eighty three hundredths with the three repeating

A fraction is simply a piece of a whole. The number is represented mathematically as a quotient where the numerator and denominator are split. In a simple fraction, the numerator as well as the denominator are both integers.

The number that is equal to 4 5/6 will be illustrated thus:

= 4 5/6

= 4 + 0.8333.

= 4.8333

In this vase, the 3 is repeating. Therefore, the correct option is A.

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The total surface area of solid is,

S = 220 m²

And,  The volume of the prism is, 200 m³

We have to given that;

A solid prism is shown in figure.

Since, The surface area of a prism is,

S = (2 × Base Area) + (Base perimeter × height)

Where, "S" is the surface area of the prism.

Hence, We get;

base area = 5 x 10 = 50 m²

height = 4 m

Base Perimeter = 2 (5 + 10) = 30

Hence, We get;

S = (2 x 50) + (30 x 4)

S = 100 + 120

S = 220 m²

Since, A prism is a solid shape that is bound on all its sides by plane faces. The volume of a prism is expressed as;

V = base area × height.

Now, For given figure,

Volume of the prism = base area × height

base area = 5 x 10 = 50 m²

height = 4 m

Hence, Volume = 50 × 4 m³

= 200 m³

Thus, The volume of the prism is, 200 m³

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Answer:

17.3 cm

Step-by-step explanation:

20² = 10² + x²

400 = 100 + x²

300 = x²

x = 17.3 cm

The selling price of the item at 10% marlkup is $55

From the question, we have the following parameters that can be used in our computation:

Using the above as a guide, we have the following:

Selling price = Cost To Store * (1 + Markup)

substitute the known values in the above equation, so, we have the following representation

Selling price = 50 * (1 + 10%)

Evaluate

Selling price = 55

Hence the selling price of the item is $55

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Answer:

-4.9x²+75x=-4.9x+50x+38

75x=50x+38

25x=38

x=1.52

1.52 -4.9(1.52)²+75(1.52)=102.67904≈102.7 meters

Double check: -4.9(1.52)²+50(1.52)+38=102.67904≈102.7. Yes.

102.7 meter.

Step-by-step explanation:

The various answers to the question are:

a)

A number of extension lines needed to accommodate $90 in calls immediately:

Use the calculation for busy k servers.

\($$P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$$\)

The probability that 2 servers are busy:

The likelihood that 2 servers will be busy may be calculated using the formula below.

\(P_{2}=\frac{\frac{\left(\frac{20}{12}\right)^{2}}{2 !}}{\sum_{i=0}^{2} \frac{\left(\frac{20}{12}\right)^{t}}{i !}}$$\approx 0.3425$\)

Hence, two lines are insufficient.

The probability that 3 servers are busy:

Assuming 3 lines, the likelihood that 3 servers are busy may be calculated using the formula below.

\(P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{2} \frac{\left(\frac{\lambda}{\mu}\right)^{i}}{i !}}$ \\\\$P_{3}=\frac{\frac{\left(\frac{20}{12}\right)^{3}}{3 !}}{\sum_{i=0}^{3} \frac{\left(\frac{20}{12}\right)^{1}}{i !}}$$\approx 0.1598$\)

Thus, three lines are insufficient.

The probability that 4 servers are busy:

Assuming 4 lines, the likelihood that 4 of 4 servers are busy may be calculated using the formula below.

\(P_{j}=\frac{\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}}{\sum_{i=0}^{k} \frac{\left(\frac{\lambda}{\mu}\right)^{t}}{i !}}$ \\\\$P_{4}=\frac{\frac{\left(\frac{20}{12}\right)^{4}}{4 !}}{\sum_{i=0}^{4} \frac{\left(\frac{20}{12}\right)^{7}}{i !}}$\)

Generally, the equation for is  mathematically given as

To answer 90% of calls instantly, the organization needs four extension lines.

b)

The probability that a call will receive a busy signal if four extensions lines are used is,

\(P_{4}=\frac{\left(\frac{20}{12}\right)^{4}}{\sum_{i=0}^{4} \frac{\left(\frac{20}{12}\right)^{1}}{i !}} $\approx 0.0624$\)

Therefore, the average number of extension lines that will be busy is Four

c)

In conclusion, the Percentage of busy calls for a phone system with two extensions:

The likelihood that 2 servers will be busy may be calculated using the formula below.

\(P_{j}=\frac{\left(\frac{\lambda}{\mu}\right)^{j}}{j !}$$\\\\$P_{2}=\frac{\left(\frac{20}{12}\right)^{2}}{\sum_{i=0}^{2 !} \frac{\left(\frac{20}{12}\right)^{t}}{i !}}$$\approx 0.3425$\)

For the existing phone system with two extension lines, 34.25 % of calls get a busy signal.

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Step-by-step explanation:

x - y = 16 ---eq(1)

x - y = 16 ---eq(2)

From eq(1)

x - y = 16

x = 16 + y ---eq(3)

Substituting the value of eq(3) into eq(2)

(16 + y) - y = 16

16 = 16

We can't get a proper value for this pair of linear Equations

But, Since both the equations are the same, the lines formed by them will be coincident and therefore they will have infinitely many solutions

Answer:

m = 5

Step-by-step explanation:

                   ( x , y )

here x is m and y is 5, lets put it in the equation to solve:

5x-y=20

5(m) - 5 = 20

5m = 20 + 5

5m = 25

m = 25/5

m = 5