Round the decimal to the nearest hundredths 983.625

The standard deviation for the number of times a 3 will be rolled in 360 dice rolls is approximately 17.32, calculated using the binomial distribution's standard deviation formula.

The question asks for the standard deviation of the number of times a 3 will be rolled when a die is rolled 360 times. Since each face of a six-sided die has an equal probability of landing upright, the chance of rolling a 3 is 1/6. To calculate the standard deviation, we first find the expected value (mean) of the number of times a 3 will come up, which is 360 rolls multiplied by the probability of 1/6, resulting in an expected value of 60. Then we use the binomial distribution formula, where the standard deviation (σ) is given by the square root of the product of the number of trials (n), the probability of success (p), and the probability of failure (q), which is σ = √(npq). In this case, it is σ = √(360*(1/6)*(5/6)) which simplifies to σ = √(300) or approximately 17.32.

Answer:

Perimeter of one end zone = 126.66 yards.

Step-by-step explanation:

Perimeter of the football field between the pylons = Perimeter of  ABCD in the figure = [tex]306\frac{2}{3}[/tex]

Now in figure we can see, If we remove the length of line segment AB and CD, we will left with sides of end zones AD and BC.

AD + BC = 306.67 - 200 ( each increment by 10 , therefore 10 x 10 + 10 x 10 = 200) = 106.67 yards

In figure, we can see AD = PQ = BC = half of 106.67 yards

So, PQ + AD = 106.67

Now, For one end zone which is of triangular shape, Perimeter = PQ + AD + AP + QD = 106.67 + 10 + 10 = 126.67 yards

Note:- In figure, we can see AP = QD = 10 (∵ increment by 10 yards)

Thus, the perimeter of one end zone is 126.67 yards.

The perimeter of one end zone is 126.67 yards.

To calculate this, we first need to know the total perimeter of the football field. The numbers on the field indicate 10-yard increments, so we can count 10 yards between each pair of pylons. There are 10 pairs of pylons on the field, so the total perimeter of the field is 10 * 10 = 100 yards.

The football field is 120 yards long, including the end zones. This means that each end zone is 10 yards long. Therefore, the perimeter of one end zone is 2 * 10 = 20 yards.

However, the end zones are not straight lines. They are curved, with a radius of 10 yards. This means that the perimeter of one end zone is not just 20 yards. We need to use the formula for the circumference of a circle to find the perimeter of the curved part of the end zone.

The formula for the circumference of a circle is C = 2πr, where C is the circumference, r is the radius, and π is a mathematical constant approximately equal to 3.14159.

Plugging in the values for the radius of the end zone, we get C = 2π(10) ≈ 62.83 yards.

Therefore, the perimeter of one end zone is 20 + 62.83 = 82.83 yards.

Answer:

y=(x–5)^2

Step-by-step explanation:

62 +62 = 124

360 - 124 = 176

176/2 = 88

BED = 88 degrees

To determine the dimensions of the rectangle, set up a quadratic equation using the given area and the relationship between the length and width, then solve for the width and consequently the length.

To find the dimensions of the rectangle with an area of 55 yd2 and a length that is 4 yards less than three times the width, we need to set up and solve a quadratic equation. Let the width be x yards. According to the problem, the length is 3x - 4 yards. The area of a rectangle is calculated by multiplying the length by the width, so we have the equation:

x(3x - 4) = 55

Distribute x across the expression:

3x2 - 4x = 55

To solve for x, we need to move all terms to one side of the equation to set it equal to zero:

3x2 - 4x - 55 = 0

This is a quadratic equation that we can solve by factoring, completing the square, or using the quadratic formula. It appears that this equation is not easily factorable; thus, the quadratic formula might be the best approach:

x = ∛{-b ± √{b2 - 4ac}}/2a

In this equation, a = 3, b = -4, and c = -55:

x = ∛{4 ± √{(-4)2 - 4(3)(-55)}}/(2·3)

Simplifying under the square root and solving for x, we get the width of the rectangle, and subsequently, we get the length by substituting x back into 3x - 4.

Answer:

points

Step-by-step explanation:

The angular speed of Earth is [tex]\fbox{\begin\\\ \bf 7.268\times 10^{-5}\ \text{radian per second}\\\end{minispace}}[/tex].

Further explanation:

It is given that the radius of Earth is [tex]6.37\times10^{6}\ \text{meters}[/tex] and it rotates once in every  [tex]24[/tex] hours.

The angular speed of earth is nearly constant and Earth takes one day to complete one rotation that means the time interval for earth to complete one revolution is [tex]1\ \text{day}[/tex].

The total angular displacement covered by the earth to complete one revolution is [tex]2\pi\text{radian}[/tex].

One day is equivalent to [tex]24\ \text{hours}[/tex], [tex]1\ \text{hour}[/tex] is equivalent to [tex]60\ \text{minutes}[/tex] and [tex]1\ \text{minute}[/tex] is equivalent to [tex]60\ \text{seconds}[/tex].

The total number of seconds in [tex]1\ \text{day}[/tex] is calculated as follows:

[tex]\fbox{\begin\\\ \begin{aligned}1\ \text{day}&=24\ \text{hours}\\&=24\times60\ \text{minutes}\\&=1440\ \text{minutes}\\&=1440\times60\ \text{seconds}\\&=86400\ \text{seconds}\end{aligned}\\\end{minispace}}[/tex]    

Therefore the time taken to complete one revolution is [tex]86400\ \text{seconds}[/tex].

The angular velocity is defined as the rate of change of angular displacement. So the formula for angular velocity is, as follows:

[tex]\fbox{\begin\\\ \omega=\dfrac{\theta}{t}\\\end{minispace}}[/tex]

In the above equation [tex]\omega[/tex] represents angular velocity, [tex]\theta[/tex] represents angular displacement and [tex]t[/tex] represents time.

To obtain the value of angular velocity of Earth substitute [tex]2\pi[/tex] for [tex]\theta[/tex] and [tex]86400[/tex]  for [tex]t[/tex] in the above equation.

[tex]\begin{aligned}\omega&=\dfrac{2\pi}{86400}\\&=\dfrac{2\times3.14}{86400}\\&=\dfrac{6.28}{86400}\\&=7.268\times10^{-5}\end{aligned}[/tex]

Therefore, the value of [tex]\omega[/tex] is [tex]7.268\times10^{-5}[/tex].

Thus, the angular speed of the Earth is [tex]\fbox{\begin\\\ \bf 7.268\times 10^{-5}\ \text{radian per second}\\\end{minispace}}[/tex].

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Answer details:

Grade: High school

Subject: Mathematics

Chapter: Angular motion

Keyword: Angular velocity, angular displacement, time, one revolution, radian/second, Earth, 86400 seconds, 2pi, speed, minutes, omega, theta.

The difference in value is due to the decimal place values. A digit in the hundredths place indeed has ten times the value of the same digit in the thousands place. This happens due to the way decimal place values work.

The digit in the hundredths place and the digit in the thousands place refer to decimal place values. When we mention that a digit in the hundredths place has ten times the value of the same digit in the thousands place, it's because of the way decimal place values function. For example, if we take a number 0.1203, the digit 2 is in the hundredths place. Now suppose you have another number, 0.003, where 3 is in the thousands place. Even if the digit remains same i.e., 2 or 3, their value will continue to change based on their place. Hence, one should know the value of the same digit can change depending on its position in the number.

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Answer:

C. sqrt of 11

Step-by-step explanation:

Answer:

bouncing a check

Step-by-step explanation:

Bouncing a check occurs when you write a check for more money than you have in your account.

A bank charges a fee any time you bounce a check.  

The store that you write a bounced check to will also charge you a fee.

Answer:

-4< X ≥ 7

Step-by-step explanation:

The sign of < indicates greater than, so that would express numbers greater than -4 and the sign ≥means smaller or less valuable or equal to, so with that you represent equal or less than 7, with that expression you can represent all the real numbers that are greater than -4 and less or equal to 7.