Assume you are rolling two fair dice once. The probability of obtaining at least one 3 or one 4 equals

To make a stable structure, the method of joints can be used to calculate the force BD in the truss shown.

This method uses the equations of equilibrium to determine the forces in the members of a truss. To calculate the force BD, one needs to sum the moments about point A and equate them to zero.

The moment is given by M = FxD, where F is the force, and D is the perpendicular distance from the point of interest to the line of action of the force. Therefore, the force BD can be calculated using the equation:

MAB = 0 = (BD)(2) - (AE)(5)

BD = 5AE/2

To find the forces in members CD and CE using the method of sections, one needs to cut through the truss along a line passing through points C and E.

Then, the force in member CD can be found by using the equation of equilibrium, where the sum of the forces in the x-direction and the sum of the forces in the y-direction must be equal to zero. Similarly, the force in member CE can also be calculated using the same method.

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The approximate diameter of the largest pillow Sarita can use is 12 inches.

Sarita has 38 inches of fringe to sew around a circular pillow.

To calculate the approximate diameter of the largest pillow she can use, we can use the formula 2 × (fringe length ÷ 3.14) = diameter. In this case, the diameter would be approximately 12 inches (38 ÷ 3.14 = 12.1).

To solve this problem, we need to divide the length of the fringe (38 inches) by the value of pi (3.14). Dividing 38 by 3.14 gives us 12.1. Since we are looking for the approximate diameter, we can round this number down to 12 inches.

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The percentage  of known edible plants is cultivated or collected by humans for food : 23%

We need to find the percentage  of known edible plants is cultivated or collected by humans for food.

From the infographic, the number of known edible plants are 30,000

and the total number of terrestrial plants known to be edible is about 130000

We know that the percenage of 'x' out of 'n' is given by formula,

p = (x / n) × 100%

here, x = 30,000 and n = 130000

Using above formula of percentage:

p = (30,000/130000) × 100%

p ≈ 23%

the required percentage is: 23%

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The complete question is:

What percentage of known edible plants is cultivated or collected by humans for food? Find that number on the infographic and divide by the total number of terrestrial plants known to be edible. (Round your answer to the nearest whole number.)

Answer:

1.83 pounds of fruit in all

Step-by-step explanation:

1 pound of melon + 0.83 pounds of cherries = 1.83 pounds in all

Answer: He bought 1.83 pounds of fruit

Step-by-step explanation:

Hope this helps :3

Answer:

500 vehicles

Step-by-step explanation:

let t represent number of tolls and v represent number of vehicles.

given that t varies directly with v then the equation relating them is

t = kv ← k is the constant of variation

to find k use the condition t = 129 from v = 20 , then

129 = 20k ( divide both sides by 20 )

6.45 = k

t = 6.45v ← equation of variation

when t = 3225 , then

3225 = 6.45v ( divide both sides by 6.45 )

500 = v

the number of vehicles is 500

The rate of change of the area of a circle when the radius is 4 cm is 4π cm2/sec.

This can be calculated using the formula for the area of a circle (A = πr2) and the chain rule for derivatives. The chain rule states that when the radius (r) changes, the area of a circle (A) is equal to 2πr times the rate of change of the radius (dr/dt).

Therefore, the rate of change of the area of a circle when the radius is 4 cm is equal to 2π(4 cm) × (0.5 cm/sec) = 4π cm2/sec.

Note that if the rate of change of the radius were different, the rate of change of the area would also be different. This formula can be used to calculate the rate of change of the area of a circle at any given radius, as long as the rate of change of the radius is known.

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The length of the arc of circle with radius 13m and with angle of arc [tex]2\pi /3[/tex]would be 13.6 meters.

The part of a circle's circumference known as an arc.It is defined by two endpoints on the circle and the points that lie on the circle between these endpoints.

An arc can be measured by its length, angle, or other properties, and is often used in the calculation of the perimeter and area of a sector of a circle.

Arcs can also be part of other curves, such as ellipses or parabolas, but they are most commonly associated with circles.

To find the length of an arc, you need to know the radius of the circle and the measure of the central angle subtended by the arc.

Arc length = (central angle / 360 degrees) x (2 x pi x radius)

where pi is approximately equal to 3.14.

For example, if the radius is 13m and the central angle is 60 degrees, then the length of the arc would be:

Arc length = (60 / 360) x (2 x 3.14 x 13) = 13.6 meters.

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1. The initial deposit is $10,000 since John deposited this amount to open the savings account.

2. The common ratio can be calculated using the formula for compound interest: A = P(1 + r/n)^(nt)

3. After 8 years, the amount will be $13,685.70 in the savings account.

The formula which will be used to calculated the compound interest is A = P(1 + r/n)^(nt). For this problem, the interest is compounded annually, so n = 1.

The annual interest rate is 4%, or 0.04 as a decimal. Plugging these values into the formula, we get:

A = $10,000*(1 + 0.04/1)^(1*8)

A = $10,000*(1.04)^8

A = $10,000*1.36857

A = $13,685.70

Therefore, after 8 years, there will be $13,685.70 in the savings account.

Full question "John deposited $10,000 to open a new savings account that earned 4 percent annual interest, compounded interest. What's the initial deposit? whats the common ratio? whats the interest rate? how much money is there after 8 years?"

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Answer:

Step-by-step explanation: There are 6 equilateral triangles in a regular hexagon by connecting the opposite vertices of a regular hexagon.

Answer:0.6

Step-by-step explanation: because i got it correct

The value of the angle U in the triangle is 92.10 degrees.

The angle U in the triangle can be found using cosine rule as follows:

Let's use cosine formula to find the angle U

c² = a² + b² - 2ab cos C

Hence,

Therefore,

a = 58.8

b = 38.4

c = 71.4

Hence,

71.4²  = 58.8² +  38.4² - 2(58.8)(38.4) cos U

5097.96 = 3457.44 + 1474.56 - 4515.84 cos U

5097.96 - 4932 =  - 4515.84 cos U

165.96 = - 4515.84 cos U

divide both sides by - 4515.84

cos U = 165.96 / - 4515.84

cos U = - 0.03675063775

U = cos⁻¹ - 0.03675063775

U = 92.1032274244

Therefore,

U = 92.10 degrees

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The best point estimate for the mean GPA for all residents of the local apartment complex would be 1.6.

Given, the mean GPA for the 127 residents of the local apartment complex is 1.6.

The mean or sample mean is used as a point estimate for the population mean or population parameter when the value of the mean of the sample is unbiased and accurately represents the mean of the population.

Here, we are given the mean GPA of 127 residents of the local apartment complex, which is 1.6.

Therefore, the best point estimate for the mean GPA for all residents of the local apartment complex would be 1.6.

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The mean class size at Mountain View School is 4.4 and at Seaside School is 4.2.

The median class size is 5 for both schools.

The range for Mountain View School is 12 and for Seaside School is 8.

The IQR for Mountain View School is 6 and for Seaside School is 3.

To find the measures of center, we can calculate the mean and median for both schools.

Mountain View School:

Mean = (2+1+0+5+8+6+9+8+2+0+1+0+6)/15

Mean = 4.4

Median = 5

Seaside School:

Mean = (0+1+2+5+6+8+8+7+6+5+5+4+4+3+1+0)/15

Mean = 4.2

Median = 5

Part B:

To find the measures of variability, we can calculate the range and interquartile range (IQR) for both schools.

Mountain View School:

Range = largest value - smallest value = 12 - 0

Range = 12

IQR = Q3 - Q1 = 8 - 2 = 6

Seaside School:

Range = largest value - smallest value = 8 - 0

Range = 8

IQR = Q3 - Q1 = 7 - 4 = 3

Part C:

If we are interested in a smaller class size, Seaside School would be the better choice. The median class size is the same at both schools, but Seaside School has a smaller mean, which suggests that on average, class sizes are smaller.

Additionally, the range and IQR are both smaller for Seaside School, indicating less variability in class size.

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g(t) = -100 * [tex](1.05)^t[/tex] +[tex](1.05)^(-5)[/tex] 50 is equation for the function g whose graph is the graph f(t) = -100[tex](1.05)^t[/tex] translated to the right 5 units and up 50 units.

Starting with the function f(t) = -100[tex](1.05)^t[/tex], to translate it 5 units to the right and 50 units up, we can replace t with (t - 5) to shift it to the right, and add 50 to the whole function to shift it up. This gives us:

g(t) = f(t - 5) + 50

g(t) = -100[tex](1.05)^(t - 5)[/tex]+ 50

Simplifying this equation, we can use the properties of exponents to rewrite [tex](1.05)^(t - 5)[/tex] as [tex](1.05)^t[/tex] * [tex](1.05)^(-5)[/tex]:

g(t) = -100[tex](1.05)^t[/tex] *[tex](1.05)^(-5)[/tex]+ 50

g(t) = -100[tex](1.05)^(-5)[/tex] * [tex](1.05)^t[/tex] + 50

So the equation for the function g whose graph is the graph f(t) = -100[tex](1.05)^t[/tex] translated 5 units to the right and 50 units up is:

g(t) = -100[tex](1.05)^(-5)[/tex]* [tex](1.05)^t[/tex]+ 50

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The width of the interval will be reduced to 1.5 if the sample size is increased by a factor of 25.

To find how much the sample size n must be increased if the width of the confidence interval is to be halved, use the following formula:

W = (zα/2 x σ/√n)W/2 = (zα/2 x σ/√n')

Where W is the initial width of the confidence interval,

zα/2 is the critical value of the standard normal distribution that corresponds to the level of significance α and the appropriate two-tailed area,

σ is the population standard deviation,

n is the initial sample size, and

n' is the new sample size needed to halve the width of the interval.

Rearranging the second equation to solve for n', we get:

n' = n(4)

So the sample size needs to be increased by a factor of 4, or 300% if the width of the confidence interval is to be halved.

If the sample size is increased by a factor of 25, the effect it will have on the width of the interval can be found using the following formula:

W' = (zα/2*σ/√25n) = (zα/2*σ/5√n)

The width of the interval will be divided by 5, or reduced to one-fifth of its initial value.

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The slope of the line of best fit in the scatterplot below can be determined using the formula for the slope of a line:
Slope = (y2-y1)/(x2-x1)

In this case, x1 = 3, y1 = 12, x2 = 6, y2 = 19. So, the slope of the line of best fit is (19 - 12) / (6 - 3), which equals 7/3 or 2.33.
The line of best fit is the line that best represents the data on a scatterplot. It is the line that is closest to all the data points in the graph. The slope of the line of best fit helps to indicate whether the data points are moving in a positive or negative direction. If the slope is positive, the data points are increasing, and if the slope is negative, the data points are decreasing.
A positive slope, like the one in this example, means that as x increases, y also increases. This means that in this case, as the x-values (3 and 6) increase, the y-values (12 and 19) also increase.

The slope of the line of best fit can also be used to make predictions about values of y, given values of x. For example, if x = 8, then y = 26, because the slope of the line of best fit is 2.33, and 8*2.33 = 19.8, which is approximately equal to 26.
Overall, the slope of the line of best fit on this scatterplot is 2.33, which is a positive number. This indicates that as the x-values increase, the y-values also increase, and it can also be used to make predictions about y-values, given x-values.

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the worker will have earned $231,584.10 from the period of beginning 1962 through the end 1980.

To solve this problem, we can use the formula for the sum of a geometric series:

S = a(1 - rⁿ)/(1 - r)

where S is the sum of the series, a is the first term, r is the common ratio, and n is the number of terms.

In this case, the first term is $8,400 and the common ratio is 1.05 (since the worker's salary increases by 5% each year). We need to find the sum of the series from 1962 to 1980, which is 19 years. So, the total amount that the worker will have earned from the beginning of 1962 through the end of 1980 is:

S = 8,400(1 - 1.05¹⁹)/(1 - 1.05)

S ≈ $231,584.10

Therefore, the worker will have earned approximately $231,584.10 during this period.

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The sampling method used in the study where the sample is chosen by separating all cars by size, and selecting 10 of each size grouping is known as stratified random sampling.

Stratified random sampling is a method of sampling that involves dividing the population into smaller sub-groups known as strata. In this method, each stratum is composed of elements that have similar characteristics.

The strata could be based on demographic characteristics such as age, sex, education, and so on. Once the strata have been created, the next step is to select a sample from each of the strata to make up the final sample. The selection is done randomly, and the number of elements selected from each stratum is proportional to the size of the stratum. This ensures that the sample is representative of the entire population in terms of the different characteristics of the population.

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The probability that a randomly selected adult weighs between 120 and 165 lbs is approximately 0.8186.

Since the weights of adults are normally distributed with a mean of 140 lbs and a standard deviation of 25 lbs, we can use the standard normal distribution to calculate the probability.

We first need to standardize the values using the formula: z = (x - μ) / σ, where x is the weight, μ is the mean, and σ is the standard deviation.

For x = 120 lbs, z = (120 - 140) / 25 = -0.8, and for x = 165 lbs, z = (165 - 140) / 25 = 1.0. We can then use a calculator to find the probability between -0.8 and 1.0, which is approximately 0.8186.

Thus, the chance of picking an adult at random who weighs between 120 and 165 lbs is roughly 0.8186.

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We can conclude that the end behavior of the function [tex]$h(x)$[/tex] is that it approaches negative infinity as [tex]$x$[/tex] approaches either positive or negative infinity.

What is meant by end behavior?

End behavior refers to the behavior of a function as the input (usually denoted by x) becomes extremely large (approaches positive or negative infinity). It describes the trend of the function as the input approaches infinity or negative infinity, and is determined by the highest-degree term of the function.

To determine the end behavior of the function [tex]$h(x)=-4x^2+11$[/tex], we can use limits. Specifically, we can evaluate the limit of [tex]$h(x)$[/tex] as [tex]$x$[/tex] approaches positive infinity and as [tex]$x$[/tex] approaches negative infinity.

As [tex]$x$[/tex] approaches positive infinity, we have:

[tex]\lim_{x \to \infty} h(x)= \lim_{x \to \infty} (-4x^2+11) = - \infty[/tex]

This tells us that as [tex]$x$[/tex] gets larger and larger, the value of [tex]$h(x)$[/tex] becomes more and more negative, eventually approaching negative infinity.

Similarly, as [tex]$x$[/tex] approaches negative infinity, we have:

[tex]\lim_{x \to -\infty} h(x)= \lim_{x \to -\infty} (-4x^2+11) = - \infty[/tex]

This tells us that as [tex]$x$[/tex] gets more and more negative, the value of [tex]$h(x)$[/tex] becomes more and more negative, also approaching negative infinity.

Therefore, we can conclude that the end behavior of the function [tex]$h(x)$[/tex] is that it approaches negative infinity as [tex]$x$[/tex] approaches either positive or negative infinity.

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A random variable is a variable whose possible values are determined by chance, while a discrete random variable can only take on certain values, and a continuous random variable can take on any value within a given range

There are two types of random variables: discrete random variables and continuous random variables.

The difference between these two types of random variables is as follows:

Discrete random variables: A discrete random variable is one that can only take on a finite or countably infinite number of distinct values. These values are usually represented by integers.

Examples of discrete random variables include the number of students in a class, the number of cars in a parking lot, or the number of heads that come up when a coin is flipped. In other words, a discrete random variable is one that can be counted.

Continuous random variables: A continuous random variable is one that can take on an infinite number of values, and these values can be represented by any real number within a given interval.

Examples of continuous random variables include height, weight, temperature, and time. In other words, a continuous random variable is one that can be measured.

Continuous random variables are usually represented by a function called the probability density function, which describes the likelihood of any given value occurring within the given interval.


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Answer: A'(-1,-4)B'(-5,-8)C'(-5,-4)D'(-4,-2)

Step-by-step explanation:

We have been given a graph of a quadrilateral and we are asked to find the coordinates of our quadrilateral after reflecting it across the x-axis.We can see that coordinates of quadrilateral ABCD are:A (-1,4),B (-5,8),C (-5,4),D (-4,2).

Since we know that rule for reflecting an image across x-axis is . This means that y-coordinates of our given image will be negative.So after reflecting our given quadrilateral across x-axis, coordinates will be:

A'(-1,-4)B'(-5,-8)C'(-5,-4)D'(-4,-2)

Answer:

x [tex]\geq[/tex]2

Step-by-step explanation:

Since the arrow is pointing to the right, we know that it is greater than two. We also know that it could be equal to 2 because the dot is filled in on the number line. So, the answer is x is greater than or equal to 2.

● < P=R " angles opp =sides"

●PQ =QR " Angles opposite = sides"

therefore PQR is an isosceles " 2 opp angles and sides are equal "

Answer:

We can use exponential decay formula to solve this problem:

N(t) = N0 * e^(-kt)

where N(t) is the number of employees with petrol cars at time t, N0 is the initial number of employees with petrol cars (in 2013), k is the decay constant, and e is the base of the natural logarithm.

We need to find N(2019), the number of employees with petrol cars in 2019. We know that N0 = 1536, and we need to find k.

Let's assume that the number of employees with petrol cars decreases by 5% each year. This means that the ratio of the number of employees with petrol cars in two consecutive years is:

0.95

Therefore, we can write:

N(2019) = 1536 * 0.95^(2019-2013)

N(2019) = 1536 * 0.95^6

N(2019) = 1133.76

Rounding to the nearest integer, we get:

N(2019) = 1134

Therefore, we can estimate that about 1134 employees owned a petrol car in 2019, assuming that the number of employees with petrol cars decreases exponentially at a rate of 5% per year.

Answer: i believe its 15.12

Step-by-step explanation:

The relationship between the two confidence intervals will be that the width of i81 will be 3 times the width of i9. So, the correct answer is option D.

In statistics, a confidence interval is a range of values that is used to estimate the unknown population parameter. They are used to express the degree of uncertainty associated with a sample estimate of a population parameter.

Two 99 percent confidence intervals are going to be constructed to estimate the difference in means of two populations, r and w. One confidence interval, i9, is going to be constructed using samples of size 9 from each of r and w.

The other confidence interval, i81, is going to be constructed using samples of size 81 from each of r and w. When all other things remain the same, the width of 181 will be 3 times the width of Ig.

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The numbers in Z84 that are relatively prime to 84 are: 1, 5, 25, 29, 43, 47, 61, 65, 79, and 83.

To find the numbers in Z84 that are relatively prime to 84, we need to find the numbers that do not have any common factors with 84 other than 1.

First, we can factorize 84 as 2^2 * 3 * 7.

Next, we can remove all the factors of 2, 3, and 7 from the set Z84, since any number in Z84 that has any of these factors would have a common factor with 84 other than 1.

So, we are left with the set of numbers {1, 5, 25, 29, 43, 47, 61, 65, 79, 83}.

Therefore, the numbers in Z84 that are relatively prime to 84 are 1, 5, 25, 29, 43, 47, 61, 65, 79, and 83.

As a comma-separated list, the answer would be: 1, 5, 25, 29, 43, 47, 61, 65, 79, 83.

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