What is the probability that either event will occur?
A
B
9
9
P(A or B) = P(A) + P(B) - P(A and B)
P(A or B) = [ ?]
Enter as a decimal rounded to the nearest hundredth.

Answer:

The answer to your problem is, 3 out of 10 or [tex]\frac{3}{10}[/tex]

Step-by-step explanation:

We know that that 7 out of 10 Americans live in an urban city. Lets put 7 out of 10 in a fraction: [tex]\frac{7}{10}[/tex]

Do some simple math:

10 - 7 = 3

So 3 out of 10 or [tex]\frac{3}{10}[/tex] Americans live in a large city.

Thus the answer to your problem is, 3 out of 10 or [tex]\frac{3}{10}[/tex]

Of the options given, cofilin would be expected to increase the rate of actin depolymerization on the minus end of the filament.

The depolymerization of actin filaments is a dynamic process that is regulated by a variety of proteins. Actin filaments grow by the addition of ATP-G-actin subunits to the plus end of the filament and can depolymerize from the minus end. Several proteins regulate actin depolymerization by binding to actin filaments and promoting filament disassembly. Of the options given, cofilin would be expected to increase the rate of actin depolymerization on the minus end of the filament.

Cofilin is a small actin-binding protein that binds to ADP-actin subunits within the filament, inducing a conformational change that destabilizes the filament and promotes depolymerization. Cofilin binds preferentially to the ADP-bound subunits, which are found primarily at the minus end of the filament. By binding to these subunits, cofilin promotes the disassembly of the filament from the minus end, increasing the rate of depolymerization.

In contrast, ATP-G-actin is the monomeric form of actin that is added to the plus end of the filament and promotes filament growth. Cap Z is a capping protein that binds to the plus end of the filament, stabilizing it and preventing depolymerization. Profilin is an actin-binding protein that binds to ATP-G-actin, promoting its polymerization into filaments. Thymosin β4 is a protein that binds to actin monomers, preventing their polymerization into filaments and sequestering them in the cytoplasm.

Therefore, of the options given, cofilin would be expected to increase the rate of actin depolymerization on the minus end of the filament.

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The probability that a randomly selected person has an IQ above 115 is 0.1587.

Someone would need to have an IQ score of at least 130.81 to get into MENSA.

We can expect the middle 95% of the population to have IQ scores between 70.6 and 129.4.

a) Since the IQ test scores follow a normal distribution, the median IQ test score is also equal to the mean IQ test score, which is 100.

b) To find the probability that a randomly selected person has an IQ above 115, we need to calculate the z-score and use a standard normal table or calculator to find the corresponding probability. The z-score is calculated as:

z = (x - μ) / σ = (115 - 100) / 15 = 1

Using a standard normal table or calculator, we can find that the probability of a z-score of 1 or greater is approximately 0.1587. Therefore, the probability that a randomly selected person has an IQ above 115 is 0.1587.

c) To find the IQ score that someone would need to have in order to get into MENSA, we need to find the z-score that corresponds to the 98th percentile and then use the formula to solve for x:

z = invNorm(0.98) = 2.0537

x = μ + zσ = 100 + 2.0537(15) = 130.8055

Therefore, someone would need to have an IQ score of at least 130.81 to get into MENSA.

d) Using the method from problem 2, we need to find the z-scores that correspond to the middle 95% of the distribution. Since the normal distribution is symmetric, we can find the z-scores that correspond to the middle 2.5% and then use the formula to solve for x:

z = invNorm(0.025) = -1.96 and z = invNorm(0.975) = 1.96

x1 = μ + z1σ = 100 + (-1.96)(15) = 70.6

x2 = μ + z2σ = 100 + (1.96)(15) = 129.4

Therefore, we can expect the middle 95% of the population to have IQ scores between 70.6 and 129.4.

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Answer:

Step-by-step explanation:

5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%5.3%uestion

A square with sides measuring 5 millimeters each is drawn within the figure shown. A point within the figure is randomly selected.

What is the approximate probability that the randomly selected point will lie inside the square?

5.3%

8.4%

13.3%

18.1%

uestion

A square with sides measuring 5 millimeters each is drawn within the figure shown. A point within the figure is randomly selected.

What is the approximate probability that the randomly selected point will lie inside the square?

5.3%

8.4%

13.3%

18.1%

uestion

A square with sides measuring 5 millimeters each is drawn within the figure shown. A point within the figure is randomly selected.

What is the approximate probability that the randomly selected point will lie inside the square?

5.3%

8.4%

13.3%

18.1%

uestion

A square with sides measuring 5 millimeters each is drawn within the figure shown. A point within the figure is randomly selected.

What is the approximate probability that the randomly selected point will lie inside the square?

5.3%

8.4%

13.3%

18.1%

uestion

A square with sides measuring 5 millimeters each is drawn within the figure shown. A point within the figure is randomly selected.

What is the approximate probability that the randomly selected point will lie inside the square?

5.3%

8.4%

13.3%

18.1%

uestion

A square with sides measuring 5 millimeters each is drawn within the figure shown. A point within the figure is randomly selected.

What is the approximate probability that the randomly selected point will lie inside the square?

5.3%

8.4%

13.3%

18.1%

uestion

A square with sides measuring 5 millimeters each is drawn within the figure shown. A point within the figure is randomly selected.

What is the approximate probability that the randomly selected point will lie inside the square?

5.3%

8.4%

13.3%

18.1%

True, The matrix corresponding to the relation {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)} is [tex]$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 \end{bmatrix}$[/tex].

The matrix representation of a relation on a set with n elements is an n x n matrix, where the entry in row i and column j is 1 if (i,j) is in the relation, and 0 otherwise. In this case, the set has four elements, so the matrix is 4 x 4.

The relation {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)} includes the pairs (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), and (3, 4), so the corresponding matrix has 1's in the entries (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), and (3, 4), and 0's elsewhere.

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The question is -

Consider the relations on the set {1, 2, 3, 4}.

The matrix corresponding to the relation {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)} is [tex]$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 \end{bmatrix}$[/tex].

State True or False.

The fraction of Cos L is 23/25

First, let's understand what a fraction is. A fraction represents a part of a whole. It has two parts: the numerator and the denominator. The numerator represents the part we are interested in, and the denominator represents the whole.

Now, let's look at our problem. We have a right triangle with sides KL and JL, and we need to find cos L. Cosine is defined as the ratio of the adjacent side to the hypotenuse in a right triangle. So, to find cos L, we need to identify the adjacent and hypotenuse sides.

From the given information, we know that KL is adjacent to angle L, and JL is the hypotenuse. So, we can write:

cos L = KL/JL

Now, we need to simplify this fraction. To simplify a fraction, we need to divide both the numerator and the denominator by their greatest common factor (GCF). The GCF of 23 and 25 is 1, so we cannot simplify the fraction any further.

Therefore, the final answer is:

cos L = KL/JL = 23/25

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Answer:

B

Step by step solution:

A is an obtuse triangle

B is an acute triangle

C is a scalene triangle

D is an equilateral triangle.

The E-M algorithm is used to estimate the parameters mu and Sigma. The implementation in R involves defining the likelihood function, initializing the estimates, and using the E-M steps to update the estimates.

The E-M algorithm for estimating the mean and covariance matrix of a multivariate normal distribution from a set of observations can be performed in the following steps

Initialization: Choose initial values for the mean vector and covariance matrix.

Expectation step: Calculate the posterior probabilities of each observation belonging to each component of the mixture model using Bayes' theorem and the current parameter estimates.

Maximization step: Use the posterior probabilities to update the estimates of the mean vector and covariance matrix for each component of the mixture model.

Repeat steps 2 and 3 until convergence (i.e., until the change in the parameter estimates falls below a specified tolerance level).

In this case, we have a single multivariate normal distribution with unknown mean vector and covariance matrix. Therefore, we can perform the E-M algorithm to estimate these parameters directly.

Here is the R code for implementing the E-M algorithm

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If the function x(t) = cos(2π100t + π/3) is sampled at the minimum sampling rate to avoid aliasing for 10 seconds, 2000 points are generated.

The function x(t) = cos(2π100t + π/3) is sampled at the minimum sampling rate to avoid aliasing for 10 seconds.

To avoid aliasing, we need to use the Nyquist-Shannon Sampling Theorem, which states that the minimum sampling rate should be twice the highest frequency in the signal. In this case, the highest frequency is 100 Hz.

Step 1: Calculate the minimum sampling rate.
Minimum sampling rate = 2 * highest frequency = 2 * 100 Hz = 200 Hz.

Step 2: Calculate the total number of points generated in 10 seconds.
A number of points = sampling rate * time duration = 200 Hz * 10 s = 2000 points.

So, 2000 points are generated when the function x(t) = cos(2π100t + π/3) is sampled at the minimum sampling rate to avoid aliasing for 10 seconds.

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We need to sum at least the first 10 terms of the series to be sure that the remainder is less than 0.0001 in magnitude

We can use the Alternating Series Estimation Theorem to estimate the remainder Rn of the series:

|Rn| ≤ |an+1|

where an+1 is the first neglected term of the series. For this series, the nth term is [tex](-1)^(n+1)/(n^4)[/tex], so the (n+1)th term is [tex](-1)^n+2/((n+1)^4)[/tex].

Thus, we have:

|Rn| ≤ [tex]|(-1)^(n+2)/((n+1)^4)|[/tex]

We want to find the smallest value of n such that |Rn| < 0.0001. This means we need to solve the inequality:

[tex]|(-1)^(n+2)/((n+1)^4)|[/tex] < 0.0001

Simplifying, we get:

[tex]1/((n+1)^4)[/tex] < 0.0001

Taking the fourth root of both sides, we get:

1/(n+1) < 0.1

Solving for n, we get:

n > 9

Therefore, we need to sum at least the first 10 terms of the series to be sure that the remainder is less than 0.0001 in magnitude.

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The total mortgage for the $260,000 purchase is $226,100.00. The Option D is closest.

The down payment of the mortgage arrangement will be:

= 15% of $260,000

= 15% * $260,000

= $39,000

Closing costs:

Credit report: $300.00

Loan origination fee= 1% of $260,000 = $2,600.00

Attorney and notary: $500.00

Documentation stamp:

= 0.50% of $260,000

= $1,300.00

Processing fee: $400.00

Total closing costs will be:

= $300.00 + $2,600.00 + $500.00 + $1,300.00 + $400.00

= $5,100.00

The total mortgage for the $260,000 purchase will equals to:

= Purchase price - Down payment + Closing costs

= $260,000 - $39,000 + $5,100.00

= $226,100.

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The derivative of the vertical position of the pumpkin trajectory with respect to its horizontal position is 2ax, as given by option b.

To find the position of maximum height of the pumpkin, the students need to find the point where the derivative of the vertical position with respect to the horizontal position is equal to zero. Setting 2ax equal to zero and solving for x, we get x=0. This means that the pumpkin reaches its maximum height at x=0, or in other words, at the point where it is launched from the trebuchet.

To find the value of the maximum height, we can substitute x=0 into the original equation for the pumpkin's trajectory. This gives us y(0) = b, which means that the maximum height of the pumpkin is b units.

The derivative of the vertical position of the pumpkin trajectory with respect to its horizontal position is 2ax because the derivative of ax^2 with respect to x is 2ax. This means that the rate of change of the pumpkin's height with respect to its horizontal position is proportional to 2ax. When x is zero, the derivative is also zero, which indicates that the pumpkin has reached its maximum height at that point.

This is because at the maximum height, the rate of change of height with respect to horizontal distance is zero. Finally, we find the value of the maximum height by substituting x=0 into the equation for the pumpkin's trajectory.

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Answer: 82

Step-by-step explanation:

               82

            35   47

       15     20     27

   7      8        12    15

2     5      3        9        6

puzzle two numbers add up to one above

2 numbers under a number add up to make that number

2+5=7

5+3=8 etc

To apply the y-function to the provided cme data, we need to first calculate the specific volume (v) for each data point using the ideal gas law: v = (RT)/P

where R is the gas constant, T is the temperature in Kelvin, and P is the pressure. We will assume that the gas is ideal and use R = 8.314 J/mol K.

Next, we need to calculate the y-values for each data point using the equation:

[tex]Y = (v - v_l) /(v_g - v_l)[/tex]

where [tex]v_l[/tex] and [tex]v_g[/tex] are the specific volumes of the liquid and gas phases, respectively, at the given pressure and temperature. We will use the following values for [tex]v_l[/tex] and [tex]v_g[/tex]:

[tex]v_l = 0.001 (m^3/kg)\\\\v\\_g = 5.0 (m^3/kg)[/tex]

Using the specific volume values and the equation for y, we can calculate the y-values for each data point:

| Pressure (MPa) | Temperature (K) | Specific Volume

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The value of exact solutions of the given system equations graphically is,

⇒ x = (1 ± √ 47i) / 8

We have to given that;

Equations are,

y = x - 3

y = 4x²

Now, Substitute the value of y in (ii);

x - 3 = 4x²

4x² - x + 3 = 0

x = - (- 1) ± √(- 1)² - 4×4×3/8

x = (1 ± √1 - 48) / 8

x = (1 ± √- 47) / 8

Thus, The value of exact solutions of the given system equations graphically is,

⇒ x = (1 ± √ 47i) / 8

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P(T < 50) = 0.3202 * 0.5 + 0.102 * 0.5 = 0.2111

Finally, substituting all these values into Bayes' theorem, we get:

P(T < 50 | T = 50 or 75) = 0

a) To solve this problem, we need to use the properties of the lognormal distribution. Specifically, we know that if T is a lognormal distribution with mean μ and coefficient of variation σ, then ln(T) has a normal distribution with mean ln(μ) and standard deviation σ.

Using this, we can standardize the distribution of ln(T) as follows:

z = (ln(50) - ln(75)) / (0.4 * ln(10))

where ln(10) is the natural logarithm of 10.

We can then use a standard normal distribution table (or a calculator) to find the probability that z is less than or equal to this value.

P(T < 50) = P(ln(T) < ln(50)) = P(z < -0.468) = 0.3202

Therefore, the probability of an earthquake occurring in the next 50 years is approximately 0.3202.

b) Similarly, to find the probability of an earthquake occurring in the next 50 years given that the last earthquake occurred 75 years ago, we standardize the distribution of ln(T) using:

z = (ln(50) - ln(75)) / (0.4 * ln(10))

But this time, we need to adjust the mean by subtracting ln(75) and adding ln(75+75) = ln(150) to account for the time since the last earthquake.

z = (ln(50) - ln(75+75)) / (0.4 * ln(10)) = -1.272

Again, using a standard normal distribution table (or calculator), we find that:

P(T < 50 | T = 75) = P(ln(T) < ln(50) | ln(T) = ln(75)) = P(z < -1.272) = 0.102

Therefore, the probability of an earthquake occurring in the next 50 years given that the last earthquake occurred 75 years ago is approximately 0.102.

c) To find the probability of an earthquake occurring in the next 50 years, given that the last earthquake occurred either 50 or 75 years ago with equal probability, we need to use Bayes' theorem:

P(T < 50 | T = 50 or 75) = P(T = 50 or 75 | T < 50) * P(T < 50) / P(T = 50 or 75)

where P(T = 50 or 75) = 0.5 (since the two cases are equally likely).

To find P(T = 50 or 75 | T < 50), we can use the law of total probability:

P(T < 50) = P(T < 50 | T = 50) * P(T = 50) + P(T < 50 | T = 75) * P(T = 75)

where P(T = 50) = P(T = 75) = 0.5 (since the two cases are equally likely).

From parts a) and b), we know that:

P(T < 50 | T = 50) = 0.3202

P(T < 50 | T = 75) = 0.102

Substituting these values, we get

P(T < 50) = 0.3202 * 0.5 + 0.102 * 0.5 = 0.2111

Finally, substituting all these values into Bayes' theorem, we get:

P(T < 50 | T = 50 or 75) = 0

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The standard deviation may or may not be dependent, as it depends on the nature of the data being analyzed.

When using a dependent samples t-test, the sample consists of two related groups that are being compared to each other.

Therefore, it is not true that there is only one mean or that the sample consists of only one group.

Additionally, the standard deviation may or may not be dependent, as it depends on the nature of the data being analyzed.

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Rounding to 4 decimal places, we expect 9.9600 of the 12 vehicles to use E-ZPass.

Rounding to 4 decimal places, the mode is 10 and the probability associated with the mode is 0.2822.

Rounding to 4 decimal places, the probability of eight or more of the sampled vehicles using E-ZPass is 0.9975.

a. We can use the binomial distribution to model the number of vehicles out of 12 that use E-ZPass. The probability of a vehicle using E-ZPass is 0.83, so the expected number of vehicles out of 12 that use E-ZPass is:

E(X) = np = 12 x 0.83 = 9.96

Rounding to 4 decimal places, we expect 9.9600 of the 12 vehicles to use E-ZPass.

b. The mode of the binomial distribution is the value of X that maximizes the probability mass function (PMF). In this case, the PMF is given by:

P(X = x) = (12 choose x) * 0.83^x * 0.17^(12-x)

We can find the mode by calculating the PMF for each possible value of X and identifying the value(s) with the highest probability. Alternatively, we can use the formula for the mode of the binomial distribution, which is:

mode = floor((n+1)p)

where n is the number of trials and p is the probability of success.

In this case, the mode is:

mode = floor((12+1) x 0.83) = floor(10.08) = 10

The probability associated with the mode is:

P(X = 10) = (12 choose 10) * 0.83^10 * 0.17^2 = 0.2822

Rounding to 4 decimal places, the mode is 10 and the probability associated with the mode is 0.2822.

c. The probability of eight or more of the sampled vehicles using E-ZPass can be found using the binomial distribution:

P(X >= 8) = 1 - P(X < 8) = 1 - P(X <= 7)

Using a binomial calculator or a cumulative binomial table, we can find:

P(X <= 7) = 0.0025

Therefore:

P(X >= 8) = 1 - 0.0025 = 0.9975

Rounding to 4 decimal places, the probability of eight or more of the sampled vehicles using E-ZPass is 0.9975.

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Answer:

146in2

Step-by-step explanation:

i dont like explaining but im in the test to

Answer:

To create a matrix for this linear system, we can arrange the coefficients of the variables and the constants into a matrix as follows:

| 1 3 2 | | x | | 26 |

| 1 -3 4 | x | y | = | 2 |

| 2 1 1 | | z | | 8 |

To solve the system using row reduction, we can perform elementary row operations to transform the matrix into row echelon form or reduced row echelon form. I will use the latter approach for simplicity:

| 1 0 0 | | x | | 6 |

| 0 1 0 | x | y | = | 5 |

| 0 0 1 | | z | | -1 |

Therefore, the solution to the system is x = 6, y = 5, and z = -1.

The probability that x is in the interval (Mx + 20x) is 0.0000.

To calculate the mean (Hy), variance (o^2), and standard deviation (Ox) of the binomial distribution, we use the following formulas:

Hy = np = 5 * 0.12 = 0.6

o^2 = npq = 5 * 0.12 * 0.88 = 0.528

Ox = sqrt(o^2) = sqrt(0.528) = 0.72

These formulas give the same results as the ones given in the section

To calculate the interval (Mx + 20x), we first need to find the values of Mx and Ox:

Mx = Hy + 20 * Ox = 0.6 + 20 * 0.727 = 15.14

Ox = sqrt(o^2) = 0.727

The interval is therefore [15.14 - 0.727, 15.14 + 0.727] = [14.413, 15.867]

To find the probability that x is in this interval, we need to sum the probabilities of the values of x that fall within the interval:

P(14 ≤ x ≤ 15) = p(0) + p(1) + p(2) + p(3) + p(4) = 0.3598 + 0.0009 + 0.0981 + 0.0134 + 0.5277 = 1.0000

Rounding up 14.413 to 15 and rounding down 15.867 to 15, we get the same interval [15, 15] and the probability that x is in this interval is P(15) = p(5) = 0.0000.

Therefore, the probability that x is in the interval (Mx + 20x) is 0.0000.

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The distribution indicates that the most likely outcome of a single draw is to obtain a non-green ball (either red or blue), with a probability of approximately 0.727.

The distribution for the random variable x equals the number of green balls obtained with a single draw can be described as a discrete probability distribution. Since there are a total of 150 balls in the bucket and 41 of them are green, the probability of obtaining a green ball on a single draw is 41/150 or approximately 0.273. Therefore, the probability mass function for this distribution can be written as follows:

P(X = 0) = 109/150 or approximately 0.727
P(X = 1) = 41/150 or approximately 0.273
P(X > 1) = 0

This distribution indicates that the most likely outcome of a single draw is to obtain a non-green ball (either red or blue), with a probability of approximately 0.727. However, there is still a significant chance of obtaining a green ball on a single draw, with a probability of approximately 0.273.

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Answer:

mean = 52

Step-by-step explanation:

In mathematics, the mean is a measure of central tendency that represents the average of a set of numbers. It is also commonly known as the arithmetic mean.

To calculate the mean, you add up all the numbers in a set and then divide by the total number of values in the set.

So in tho case we would add all the values then divide by the amount of values there are.

1) rewrite values:

42  45  58  63

2) add:

42 + 45 + 58 + 63
= 208

3) divide:

Since there arev4 values we will divide by 4.

208 / 4 = 52

Therefore, the mean is 52

Answer:

[tex]\huge\boxed{\sf Mean = 52}[/tex]

Step-by-step explanation:

42, 45, 58, 63

[tex]\displaystyle Mean = \frac{Sum \ of \ data}{No. \ of \ data} \\\\Mean = \frac{42+45+58+63}{4} \\\\Mean = \frac{208}{4} \\\\Mean = 52\\\\\rule[225]{225}{2}[/tex]

Answer:

To calculate the probability that Kaelyn was selected out of 30 people chosen at random from a group of 480 people, we can use the concept of probability.

Given:

Total number of people in the group (N) = 480

Number of people chosen (n) = 30

We want to find the probability of Kaelyn being selected, which is the probability of selecting 1 person out of 480 people, or 1/480.

So, the probability that Kaelyn was selected is:

P(Kaelyn) = 1/480

This is the final answer, as it represents the probability of Kaelyn being selected out of 30 people chosen at random from a group of 480 people.

Step-by-step explanation:

Answer:

To calculate the probability that Kaelyn was selected out of 30 people chosen

Step-by-step explanation:

The art museum had 4,000 visitors that month, as 1,600 visitors during a week represented 40% of the total visitors for the month.

Let's denote the total number of visitors for the month with "x".

According to the problem, the number of visitors during a particular week (1,600) was 40% of the total number of visitors for the month.

We can write this as an equation

1,600 = 0.4x

To solve for "x", we need to isolate it on one side of the equation. We can do this by dividing both sides of the equation by 0.4

1,600 ÷ 0.4 = x

Simplifying the left side

4,000 = x

Therefore, the art museum had a total of 4,000 visitors that month.

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Answer:

The measure of angle B is 80°. B is the correct answer.

Answer:

B. mB = 80°

Step-by-step explanation:

Supplementary angles add to 180 degrees

100 +x = 180

x = 180-100

x = 80 degrees

Answer:

D.) 7

Step-by-step explanation:

5 trees have a height of from 80 ft to 85 ft.

2 trees have a height of from 85 ft to 90 ft.

Answer: D.) 7

Answer:

D) 7 black cherry trees

Step-by-step explanation:

In the bar graph, we can see that starting at 80 feet, there are 5 trees that range from 80-85 feet.

We can also see that there are only 2 trees that range from 85-90 feet.  The question asks for how many trees that have a height of at least 80 feet, so we add 5+2 to get 7 trees.

Hope this helps! :)

The speed of the airplane in still air is 310 miles per hour.

Let's denote the speed of the airplane in still air as "x" (in miles per hour).

When the airplane is flying with a tailwind, its speed relative to the ground increases. We can use the formula:

speed with tailwind = speed in still air + speed of tailwind

To set up an equation:

350 mi/h = x mi/h + 40 mi/h

To simplify, we have:

x mi/h = 350 mi/h - 40 mi/h

x mi/h = 310 mi/h

To learn more about the speed;

https://brainly.com/question/7359669

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answer

56

explain

because I just don't

Yes it should be surface area