The random variable x is the number of occurrences of an event over an interval of ten minutes. It can be assumed that the probability of an occurrence is the same in any two time periods of an equal length. It is known that the mean number of occurrences in ten minutes is 5.

The probability that there are 3 or less occurrences is
A) 0.0948
B) 0.2650
C) 0.1016
D) 0.1230

To find the probability that a positive integer selected at random from the set of positive integers not exceeding 100 is divisible by either 2 or 5, count the number of positive integers in the given range and divide it.

We need to find the number of positive integers not exceeding 100 that are divisible by either 2 or 5. We can use the principle of inclusion-exclusion to count these numbers.

The numbers divisible by 2 are: 2, 4, 6, ..., 100. There are 50 such numbers.

The numbers divisible by 5 are: 5, 10, 15, ..., 100. There are 20 such numbers.

However, some numbers (such as 10, 20, 30, etc.) are divisible by both 2 and 5, and we have counted them twice. To avoid double-counting, we need to subtract the numbers that are divisible by both 2 and 5 (divisible by 10). There are 10 such numbers (10, 20, 30, ..., 100).

Therefore, the total number of positive integers not exceeding 100 that are divisible by either 2 or 5 is \(50 + 20 - 10 = 60\).

Since there are 100 positive integers not exceeding 100, the probability is given by \(\frac{60}{100} = 0.6\) or 60%.

Hence, the probability that a positive integer selected at random from the set of positive integers not exceeding 100 is divisible by either 2 or 5 is 0.6 or 60%.

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A vector function r(t) that represents the curve of intersection of the two surfaces, the cone z = x² + y² and the plane z = 2 + y, is r(t) = ⟨t, -t² + 2, -t² + 2⟩.

To find the vector function representing the curve of intersection between the cone z = x² + y² and the plane z = 2 + y, we need to equate the two equations and express x, y, and z in terms of a parameter, t.

By setting x² + y² = 2 + y, we can rewrite it as x² + (y - 1)² = 1, which represents a circle in the xy-plane with a radius of 1 and centered at (0, 1). This allows us to express x and y in terms of t as x = t and y = -t² + 2.

Since the plane equation gives us z = 2 + y, we have z = -t² + 2 as well.

Combining these equations, we obtain the vector function r(t) = ⟨t, -t² + 2, -t² + 2⟩, which represents the curve of intersection.

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The Maclaurin series for [tex]\(f(x) = \sin\left(\frac{\pi x}{2}\right)\)[/tex] is given by:

[tex]\[\sin\left(\frac{\pi x}{2}\right) = \frac{\pi}{2} \left(x - \frac{\left(\pi^2 x^3\right)}{2^3 \cdot 3!} + \frac{\left(\pi^4 x^5\right)}{2^5 \cdot 5!} - \frac{\left(\pi^6 x^7\right)}{2^7 \cdot 7!} + \ldots\right).\][/tex]

The radius of convergence, [tex]\(R\)[/tex] , for this series is infinite since the series converges for all real values of [tex]\(x\).[/tex]

Therefore, the Maclaurin series for [tex]\(f(x) = \sin\left(\frac{\pi x}{2}\right)\)[/tex] is:

[tex]\[\sin\left(\frac{\pi x}{2}\right) = \frac{\pi}{2} \left(x - \frac{\left(\pi^2 x^3\right)}{2^3 \cdot 3!} + \frac{\left(\pi^4 x^5\right)}{2^5 \cdot 5!} - \frac{\left(\pi^6 x^7\right)}{2^7 \cdot 7!} + \ldots\right)\][/tex]

with an associated radius of convergence [tex]\(R = \infty\).[/tex]

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Therefore, the only solutions within the given interval are the values of x for which cos(x) = 0, namely [tex]x = (2k + 1)\pi/2,[/tex] where k is any integer, and 0 < a < π.

To find all solutions of the equation cos(x)sin(x) - 2cos(x) = 0, we can factor out the common term cos(x) from the left-hand side:

cos(x)(sin(x) - 2) = 0

Now, we have two possibilities for the equation to be satisfied:

 cos(x) = 0In this case, x can take values of the form x = (2k + 1)π/2, where k is any integer.

 sin(x) - 2 = 0 Solving this equation for sin(x), we get sin(x) = 2. However, there are no solutions to this equation within the interval 0 < a < π, as the range of sin(x) is -1 to 1.

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The class boundaries for the first class interval are:Lower limit = 32Upper limit = 34Class width = 3Boundaries = 32 - 1.5 = 30.5 and 34 + 1.5 = 35.5. The boundaries for the remaining class intervals can be determined in a similar manner. Therefore, the class boundaries are given below:Temperature (°F)FrequencyClass boundaries32-34130.5-35.535-3735-38.540-4134.5-44.544-4638.5-47.547-4944.5-52.550-5264.5-79.5

The frequency distribution table is given below:Temperature (°F)Frequency32-34135-3738-4041-4344-4647-4950-521The frequency distribution gives a range of values for the temperature in Fahrenheit. In order to answer the questions (a), (b) and (c), the class width, class midpoints, and class boundaries need to be determined.(a) Class WidthThe class width can be determined by subtracting the lower limit of the first class interval from the lower limit of the second class interval. The lower limit of the first class interval is 32, and the lower limit of the second class interval is 35.32 - 35 = -3Therefore, the class width is 3. The answer is 3.(b) Class MidpointsThe class midpoint can be determined by finding the average of the upper and lower limits of the class interval. The class intervals are given in the frequency distribution table. The midpoint of the first class interval is:Lower limit = 32Upper limit = 34Midpoint = (32 + 34) / 2 = 33The midpoint of the second class interval is:Lower limit = 35Upper limit = 37Midpoint = (35 + 37) / 2 = 36. The midpoint of the remaining class intervals can be determined in a similar manner. Therefore, the class midpoints are given below:Temperature (°F)FrequencyMidpoint32-34133.535-37361.537-40393.541-4242.544-4645.547-4951.550-5276(c) Class BoundariesThe class boundaries can be determined by adding and subtracting half of the class width to the lower and upper limits of each class interval. The class width is 3, as determined above. Therefore, the class boundaries for the first class interval are:Lower limit = 32Upper limit = 34Class width = 3Boundaries = 32 - 1.5 = 30.5 and 34 + 1.5 = 35.5. The boundaries for the remaining class intervals can be determined in a similar manner. Therefore, the class boundaries are given below:Temperature (°F)FrequencyClass boundaries32-34130.5-35.535-3735-38.540-4134.5-44.544-4638.5-47.547-4944.5-52.550-5264.5-79.5.

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To find the volume v of the described solid s, a cap of a sphere with radius r and height h, the formula to be used is:v = (π/3)h²(3r - h)First, let's establish the formula for the volume of the sphere. The formula for the volume of a sphere is given as:v = (4/3)πr³

A spherical cap is cut off from a sphere of radius r by a plane situated at a distance h from the center of the sphere. The volume of the spherical cap is given as follows:V = (1/3)πh²(3r - h)The volume of a sphere of radius r is:V = (4/3)πr³Substituting the value of r into the equation for the volume of a spherical cap, we get:v = (π/3)h²(3r - h)Therefore, the volume of the described solid s, a cap of a sphere with radius r and height h, is:v = (π/3)h²(3r - h)The answer is  more than 100 words as it includes the derivation of the formula for the volume of a sphere and the volume of a spherical cap.

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(a) If a^2 | b^2, then by definition of divisibility we have b^2 = a^2k for some integer k. Thus,b^2 - a^2 = a^2(k - 1) = (a√k)(a√k),which implies that a^2 divides b^2 - a^2.

Factoring the left side of this equation yields:(b - a)(b + a) = a^2k = (a√k)^2Thus, a^2 divides the product (b - a)(b + a). Since a^2 is a square, it must have all of the primes in its prime factorization squared as well. Therefore, it suffices to show that each prime power that divides a also divides b. We will assume that p is prime and that pk divides a. Then pk also divides a^2 and b^2, so pk must also divide b. Thus, a | b, as claimed.(b) If a n | b n, then b n = a n k for some integer k. Thus, we can write b = a^k, so a | b, as claimed.

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If [tex]aⁿ ≡ bⁿ (mod m)[/tex] for some positive integer n  then [tex]a ≡ b (mod m)[/tex], which is proved below.

a) Let [tex]a² = b²[/tex]. Then [tex]a² - b² = 0[/tex], or (a-b)(a+b) = 0.

So either a-b = 0, i.e. a=b, or a+b = 0, i.e. a=-b.

In either case, a=b.

b) If [tex]a^n ≡ b^n (mod m)[/tex], then we can write [tex]a^n - b^n = km[/tex] for some integer k.

We know that [tex]a-b | a^n - b^n[/tex], so we can write [tex]a-b | km[/tex].

But a and b are relatively prime, so we can write a-b | k.

Thus there exists some integer j such that k = j(a-b).

Substituting this into our equation above, we get

[tex]a^n - b^n = j(a-b)m[/tex],

or [tex]a^n = b^n + j(a-b)m[/tex]

and so [tex]a-b | b^n[/tex].

But a and b are relatively prime, so we can write a-b | n.

This means that there exists some integer h such that n = h(a-b).

Substituting this into the equation above, we get

[tex]a^n = b^n + j(a-b)n = b^n + j(a-b)h(a-b)[/tex],

or [tex]a^n = b^n + k(a-b)[/tex], where k = jh.

Thus we have shown that if aⁿ ≡ bⁿ (mod m) then a ≡ b (mod m).

Therefore, both the parts are proved.

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a) 0.0062 is the probability that the student completes the exam in less than 45 minutes.

b) 77.4 minutes should be given to complete the exam so 80% of the students will complete the exam in the time given.

a) The probability that a student completes the exam in less than 45 minutes can be calculated using the standard normal distribution. By converting the given values to z-scores, we can use a standard normal distribution table or a calculator to find the probability.

To convert the given time of 45 minutes to a z-score, we use the formula: z = (x - μ) / σ, where x is the given time, μ is the mean, and σ is the standard deviation. Substituting the values, we get z = (45 - 70) / 10 = -2.5.

Using the standard normal distribution table or a calculator, we can find that the probability corresponding to a z-score of -2.5 is approximately 0.0062.

Therefore, the probability that a student completes the exam in less than 45 minutes is approximately 0.0062, or 0.62%.

b) To determine the time needed for 80% of the students to complete the exam, we need to find the corresponding z-score for the cumulative probability of 0.8.

Using the standard normal distribution table or a calculator, we find that the z-score corresponding to a cumulative probability of 0.8 is approximately 0.84.

Using the formula for z-score, we can solve for the time x: z = (x - μ) / σ. Rearranging the formula, we get x = μ + (z * σ). Substituting the values, we get x = 70 + (0.84 * 10) = 77.4.

Therefore, approximately 77.4 minutes should be given to complete the exam so that 80% of the students will complete it within the given time.

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The amount that will have accumulated in 10 years after the last deposit is approximately $13,299.25.

To calculate the accumulated amount, we can use the formula for compound interest:

A = P(1 + r/n)^(nt)

Where:

A = Accumulated amount

P = Principal amount (initial deposit)

r = Annual interest rate (as a decimal)

n = Number of times interest is compounded per year

t = Number of years

In this case, Sklyer has made deposits of $680 at the end of every quarter for 13 years, so the principal amount (P) is $680. The annual interest rate (r) is 5%, which is 0.05 as a decimal. The interest is compounded annually, so the number of times interest is compounded per year (n) is 1. And the number of years (t) for which we need to calculate the accumulated amount is 10.

Plugging these values into the formula, we have:

A = $680(1 + 0.05/1)^(1*10)

  = $680(1 + 0.05)^10

  = $680(1.05)^10

  ≈ $13,299.25

Therefore, the amount that will have accumulated in 10 years after the last deposit is approximately $13,299.25.

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20/21 is the probability that the sample has at least one female.

The total number of students in the club is 4 + 5 = 9.

The sample size is 3. Therefore, the number of ways to choose 3 students out of 9 is: C(9,3) = 84.

There are 5 female students. Therefore, the number of ways to choose 3 students from 5 female students is: C(5,3) = 10.

The probability of selecting at least one female is equal to 1 minus the probability of selecting all male members. The probability of selecting all male members is the number of ways to choose 3 members out of 4 male students divided by the total number of ways to choose 3 members from 9. Therefore, the probability of selecting all male members is: C(4,3) / C(9,3) = 4/84 = 1/21.

So, the probability of selecting at least one female is: P(at least one female) = 1 - P(all male members) = 1 - 1/21 = 20/21.

Therefore, the probability that the sample has at least one female is 20/21.

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Title and scale the graph Finally, give the graph a title that describes what the graph represents. Also, give each axis a title and a scale that makes it easy to read and interpret the data.

To create a scatter plot from the data given in the table with variables `a` and `b`, you can follow the following steps:

Step 1: Organize the dataThe first step in creating a scatter plot is to organize the data in a table. The table given in the question has the data organized already, but it is in a vertical format. We will need to convert it to a horizontal format where each variable has a column. The organized data will be as follows:````| Variable a | Variable b | |------------|------------| | 1 | 12 | | 5 | 8 | | 2 | 10 | | 7 | 5 | | 8 | 4 | | 1 | 10 | | 3 | 8 | | 7 | 10 | | 6 | 5 | | 6 | 6 | | 2 | 11 | | 9 | 4 | | 7 | 4 | | 5 | 5 | | 2 | 12 |```

Step 2: Create a horizontal and vertical axisThe second step is to create two axes, a horizontal x-axis and a vertical y-axis. The x-axis represents the variable a while the y-axis represents variable b. Label each axis to show the variable it represents.

Step 3: Plot the pointsThe third step is to plot each point on the graph. To plot the points, take the value of variable a and mark it on the x-axis. Then take the corresponding value of variable b and mark it on the y-axis. Draw a dot at the point where the two marks intersect. Repeat this process for all the points.

Step 4: Title and scale the graph Finally, give the graph a title that describes what the graph represents. Also, give each axis a title and a scale that makes it easy to read and interpret the data.

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The length of the minor axis for the given ellipse is 8 units. Therefore, the correct option is A: 8.

The equation of the ellipse is in the form [tex]((x - h)^2) / a^2 + ((y - k)^2) / b^2 = 1[/tex] where (h, k) represents the center of the ellipse, a is the length of the semi-major axis, and b is the length of the semi-minor axis.

Comparing the given equation to the standard form, we can determine that the center of the ellipse is (-4, 1), the length of the semi-major axis is 5, and the length of the semi-minor axis is 4.

The length of the minor axis is twice the length of the semi-minor axis, so the length of the minor axis is 2 * 4 = 8.

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If we consider the series given by n = 1/n, we can rewrite it as follows:

n = 1/1 + 1/2 + 1/3 + 1/4 + ...

To determine the value of p for each series, we can compare it to known series forms. In this case, it resembles the harmonic series, which has the form:

1 + 1/2 + 1/3 + 1/4 + ...

The harmonic series is a p-series with p = 1. Therefore, in this case:

P1 = 1

Since the series in question is similar to the harmonic series, we know that if P1 ≤ 1, the series is divergent. Therefore, the series is divergent.

In summary:

P1 = 1 (smaller value)

P2 = N/A (not applicable)

The series is divergent.

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Therefore, we have to take the absolute value of the z-score obtained. Thus, the z-score is z = |1.44| = 1.44.

To determine z such that 8.6% of the standard normal curve lies to the right of z, we can follow the steps below:

Step 1: Draw the standard normal curve and shade the area to the right of z.

Step 2: Look up the area 8.6% in the standard normal table.Step 3: Find the corresponding z-score for the area using the table.

Step 4: Take the absolute value of the z-score obtained since we want the area to the right of z.

Step 1: Draw the standard normal curve and shade the area to the right of z

The standard normal curve is a bell-shaped curve with mean 0 and standard deviation 1. Since we want to find z such that 8.6% of the standard normal curve lies to the right of z, we need to shade the area to the right of z as shown below:

Step 2: Look up the area 8.6% in the standard normal table

The standard normal table gives the area to the left of z.

To find the area to the right of z, we need to subtract the area from 1.

Therefore, we look up the area 1 – 0.086 = 0.914 in the standard normal table.

Step 3: Find the corresponding z-score for the area using the table

The standard normal table gives the z-score corresponding to the area 0.914 as 1.44.

Step 4: Take the absolute value of the z-score obtained since we want the area to the right of z

The area to the right of z is 0.086, which is less than 0.5.

Therefore, we have to take the absolute value of the z-score obtained.

Thus, the z-score is z = |1.44| = 1.44.

Z-score is also known as standard score, it is the number of standard deviations by which an observation or data point is above the mean of the data set. A standard normal distribution is a normal distribution with mean 0 and standard deviation 1.

The area under the curve of a standard normal distribution is equal to 1. The area under the curve of a standard normal distribution to the left of z can be found using the standard normal table.

Similarly, the area under the curve of a standard normal distribution to the right of z can be found by subtracting the area to the left of z from 1.

In this problem, we need to find z such that 8.6% of the standard normal curve lies to the right of z. To find z, we need to perform the following steps.

Step 1: Draw the standard normal curve and shade the area to the right of z.

Step 2: Look up the area 8.6% in the standard normal table.

Step 3: Find the corresponding z-score for the area using the table.

Step 4: Take the absolute value of the z-score obtained since we want the area to the right of z.

The standard normal curve is a bell-shaped curve with mean 0 and standard deviation 1.

Since we want to find z such that 8.6% of the standard normal curve lies to the right of z, we need to shade the area to the right of z.

The standard normal table gives the area to the left of z.

To find the area to the right of z, we need to subtract the area from 1.

Therefore, we look up the area 1 – 0.086 = 0.914 in the standard normal table.

The standard normal table gives the z-score corresponding to the area 0.914 as 1.44.

The area to the right of z is 0.086, which is less than 0.5.

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Therefore, the probability that all 15 pieces are defective is approximately [tex]1.89 * 10^{(-9)[/tex].

To calculate the probability in this scenario, we can use the binomial probability formula.

a) Probability of all defective parts:

Since we want to calculate the probability that all 15 pieces are defective, we use the binomial probability formula:

[tex]P(X = k) = ^nC_k * p^k * (1 - p)^{(n - k)[/tex]

In this case, n = 15 (total number of pieces), k = 15 (number of defective pieces), and p = 0.21 (probability of failure).

Plugging in the values, we get:

[tex]P(X = 15) = ^15C_15 * 0.21^15 * (1 - 0.21)^{(15 - 15)[/tex]

Simplifying the equation:

[tex]P(X = 15) = 1 * 0.21^{15} * 0.79^0[/tex]

= [tex]0.21^{15[/tex]

≈ [tex]1.89 x 10^{(-9)[/tex]

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The amplitude is 7, the period is 12π/11, the horizontal shift is 22π/33 to the right, and the midline is y = 3, where [11π/6(x - 22π/33)] represents the phase shift.

Given the equation y = 7 sin [11π/6(x - 22π/33)] +3 units to the Right

For the given equation, the amplitude is 7, the period is 12π/11, the horizontal shift is 22π/33 to the right, and the midline is y = 3.

To solve for the amplitude, period, horizontal shift and midline for the equation y = 7 sin [11π/6(x - 22π/33)] +3 units to the right, we must look at each term independently.

1. Amplitude: Amplitude is the highest point on a curve's peak and is usually represented by a. y = a sin(bx + c) + d, where the amplitude is a.

The amplitude of the given equation is 7.

2. Period: The period is the length of one cycle, and in trigonometry, one cycle is represented by one complete revolution around the unit circle.

The period of a trig function can be found by the formula T = (2π)/b in y = a sin(bx + c) + d, where the period is T.

We can then get the period of the equation by finding the value of b and using the formula above.

From y = 7 sin [11π/6(x - 22π/33)] +3, we can see that b = 11π/6. T = (2π)/b = (2π)/ (11π/6) = 12π/11.

Therefore, the period of the equation is 12π/11.3.

Horizontal shift: The equation of y = a sin[b(x - h)] + k shows how to move the graph horizontally. It is moved h units to the right if h is positive.

Otherwise, the graph is moved |h| units to the left.

The value of h can be found using the equation, x - h = 0, to get h.

The equation can be modified by rearranging x - h = 0 to get x = h.

So, the horizontal shift for the given equation y = 7 sin [11π/6(x - 22π/33)] +3 units to the right is 22π/33 to the right.

4. Midline: The y-axis is where the midline passes through the center of the sinusoidal wave.

For y = a sin[b(x - h)] + k, the equation of the midline is y = k.

The midline for the given equation is y = 3.

Therefore, the amplitude is 7, the period is 12π/11, the horizontal shift is 22π/33 to the right, and the midline is y = 3, where [11π/6(x - 22π/33)] represents the phase shift.

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The probability density function (PDF) of XI, the lifetime of a certain type of device, is defined as follows:

f(x) = 0, if x ≤ 21

f(x) = 1/21, if x > 21

This means that for any value of x less than or equal to 21, the PDF is zero, indicating that the device cannot have a lifetime less than or equal to 21 months.

For values of x greater than 21, the PDF is 1/21, indicating that the device has a constant probability of 1/21 per month of surviving beyond 21 months.

In other words, the device has a deterministic lifetime of 21 months or less, and after 21 months, it has a constant probability per month of continuing to operate.

It's important to note that this PDF represents a simplified model and may not accurately reflect the actual behavior of the device in real-world scenarios.

It assumes that the device either fails before or exactly at 21 months, or it continues to operate indefinitely with a constant probability of failure per month.

To calculate probabilities or expected values related to the lifetime of the device, additional information or assumptions would be needed, such as the desired time interval or specific events of interest.

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The probability that the length of stay in the ICU is one day or less is approximately 0.0630 to 4 decimal places.

To calculate the probability that the length of stay in the ICU is one day or less, you need to find the cumulative probability up to one day.

Let's assume that the length of stay in the ICU follows a normal distribution with a mean of 4.5 days and a standard deviation of 2.3 days.

Using the formula for standardizing a normal distribution, we get:z = (x - μ) / σwhere x is the length of stay, μ is the mean (4.5), and σ is the standard deviation (2.3).

To find the cumulative probability up to one day, we need to standardize one day as follows:

z = (1 - 4.5) / 2.3 = -1.52

Using a standard normal distribution table or a calculator, we find that the cumulative probability up to z = -1.52 is 0.0630.

Therefore, the probability that the length of stay in the ICU is one day or less is approximately 0.0630 to 4 decimal places.

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Integration by substitution is one of the most useful techniques of integration that is used to solve integrals.

We use integration by substitution when the integrand suggests using it. Whenever there is a complicated expression inside a function or an exponential function in the integrand, we can use the integration by substitution technique to simplify the expression. The method of substitution is used to change the variable in the integrand so that the expression becomes easier to solve.

It is useful for integrals in which the integrand contains an algebraic expression, a logarithmic expression, a trigonometric function, an exponential function, or a combination of these types of functions.In other words, whenever we encounter a function that appears to be a composite function, i.e., a function inside another function, the use of substitution is suggested.

For example, integrands of the form ∫f(g(x))g′(x)dx suggest using the substitution technique. The goal is to replace a complicated expression with a simpler one so that the integral can be evaluated more easily. Substitution can also be used to simplify complex functions into more manageable ones.

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The given statement is false as an assumption of non-parametric tests is that the distribution does not need to be normal.

Question 2The given statement is true as chi-square tests can be used when the data is measured on a nominal scale. Question 3The observed frequencies for a chi-square test can contain fractions or decimal values. Question 4The term expected frequencies refers to the frequencies that are hypothesized for the population being examined. The expected frequencies are computed from the null hypothesis found in the sample data.The chi-square test is a non-parametric test used to determine the significance of how two or more frequencies are different in a particular population. The non-parametric test means that the distribution is not required to be normal. Instead, this test relies on the sample data and frequency counts.The chi-square test can be used for nominal scale data or categorical data. The observed frequencies for a chi-square test can contain fractions or decimal values. However, the expected frequencies are computed from the null hypothesis found in the sample data. The expected frequencies are the frequencies that are hypothesized for the population being examined. Therefore, option D correctly describes the expected frequencies.

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The test statistic is approximately -2.99 using the significance level of 0.05.

To compare the effectiveness of vaccines A and B, we can use a hypothesis test for the difference in proportions. First, we calculate the sample proportions:

p1 = x1 / n1 = 18 / 900 ≈ 0.02

p2 = x2 / n2 = 30 / 600 ≈ 0.05

Where x1 and x2 represent the number of adults who got the virus in each group.

To construct a 95% confidence interval for comparing the two vaccines, we can use the following formula:

CI = (p1 - p2) ± Z * √[(p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2)]

Where Z is the critical value corresponding to a 95% confidence level. For a two-tailed test at a significance level of 0.05, Z is approximately 1.96.

Plugging in the values:

CI = (0.02 - 0.05) ± 1.96 * √[(0.02 * (1 - 0.02) / 900) + (0.05 * (1 - 0.05) / 600)]

Simplifying the equation:

CI = -0.03 ± 1.96 * √[(0.02 * 0.98 / 900) + (0.05 * 0.95 / 600)]

Calculating the values inside the square root:

√[(0.02 * 0.98 / 900) + (0.05 * 0.95 / 600)] ≈ √[0.0000218 + 0.0000792] ≈ √0.000101 ≈ 0.01005

Finally, plugging this value back into the confidence interval equation:

CI = -0.03 ± 1.96 * 0.01005

Calculating the confidence interval:

CI = (-0.0508, -0.0092)

Therefore, the 95% confidence interval for the difference in proportions (p1 - p2) is (-0.0508, -0.0092).

Now, to find the test statistic, we can use the following formula:

Test Statistic = (p1 - p2) / √[(p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2)]

Plugging in the values:

Test Statistic = (0.02 - 0.05) / √[(0.02 * (1 - 0.02) / 900) + (0.05 * (1 - 0.05) / 600)]

Simplifying the equation:

Test Statistic = -0.03 / √[(0.02 * 0.98 / 900) + (0.05 * 0.95 / 600)]

Calculating the values inside the square root:

√[(0.02 * 0.98 / 900) + (0.05 * 0.95 / 600)] ≈ √[0.0000218 + 0.0000792] ≈ √0.000101 ≈ 0.01005

Finally, plugging this value back into the test statistic equation:

Test Statistic = -0.03 / 0.01005 ≈ -2.99

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The given table is: X Y X*Y X*X 4 19 76 16 5 27 135 25 12 17 204 144 17 34 578 289 22 29 638 484

To find the sum of each column:sum X = 4 + 5 + 12 + 17 + 22 = 60   sum Y = 19 + 27 + 17 + 34 + 29 = 126   sum X*Y = 76 + 135 + 204 + 578 + 638 = 1631     sum X*X = 16 + 25 + 144 + 289 + 484 = 958

To find the p-value, we first have to find the value of t using the formula given sample mean = 2,279, $\mu$ = population mean = 1,700, s = sample standard deviation = 560

Hence, the answer to this question is sum X = 60.

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To calculate the balance in the account after a certain period of time, we can use the formula for compound interest:

[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]

Where:

A = Final amount

P = Principal amount (initial deposit)

r = Annual interest rate (in decimal form)

n = Number of times the interest is compounded per year

t = Time in years

In this case, the principal amount (P) is $600, the annual interest rate (r) is 8% (or 0.08 in decimal form), and the interest is compounded monthly, so the number of times compounded per year (n) is 12.

Let's calculate the balance after one year:

[tex]A = 600(1 + \frac{0.08}{12})^{12 \cdot 1}\\\\= 600(1.00666666667)^{12}\\\\\approx 600(1.08328706767)\\\\\approx 649.97[/tex]

Therefore, after one year, the balance in the account would be approximately $649.97.

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In order to find the value of c for which P(z > c) = 0.6454 for the standard normal distribution, we can make use of a z-table which gives us the probabilities for a range of z-values. The area under the normal distribution curve is equal to the probability.

The z-table gives the probability of a value being less than a given z-value. If we need to find the probability of a value being greater than a given z-value, we can subtract the corresponding value from 1. Hence,P(z > c) = 1 - P(z < c)We can use this formula to solve for the value of c.First, we find the z-score that corresponds to a probability of 0.6454 in the table. The closest probability we can find is 0.6452, which corresponds to a z-score of 0.39. This means that P(z < 0.39) = 0.6452.Then, we can find P(z > c) = 1 - P(z < c) = 1 - 0.6452 = 0.3548We need to find the z-score that corresponds to this probability. Looking in the z-table, we find that the closest probability we can find is 0.3547, which corresponds to a z-score of -0.39. This means that P(z > -0.39) = 0.3547.

Therefore, the value of c such that P(z > c) = 0.6454 is c = -0.39.

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The average selling price of a smartphone is estimated to be $318 with a 90% confidence interval.

a. Constructing a 90% confidence interval requires calculating the margin of error, which is obtained by multiplying the critical value (obtained from the t-distribution for the desired confidence level and degrees of freedom) with the standard error.

The standard error is calculated by dividing the population standard deviation by the square root of the sample size. With the given information, the margin of error can be determined, and by adding and subtracting it from the sample mean, the confidence interval can be constructed.

b. To calculate the margin of error, we use the formula: Margin of error = Critical value * Standard error. The critical value for a 90% confidence level and a sample size of 31 can be obtained from the t-distribution table. Multiplying the critical value with the standard error (which is the population standard deviation / square root of the sample size) will give us the margin of error. Adding and subtracting the margin of error to the sample mean will give us the lower and upper limits of the confidence interval, respectively.

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The correct Question is: The average selling price of a smartphone purchased by a random sample of 31 customers was $318, assuming the population standard deviation was $30. a. Construct a 90% confidence interval to estimate the average selling price.

The solution to the problem is as follows:Given that 80% of teenagers aged 12-17 own smartphones. A random sample of 250 teenagers is drawn.

The probability is calculated by using the Central Limit Theorem and the TI-84 calculator, and the answer is rounded to at least four decimal places.PC <0.11)-0 Х $P(X<0.11)To find the probability of less than 75% of the sampled teenagers owned smartphones, convert the percentage to a proportion.75/100 = 0.75

This means that p = 0.75. To find the sample proportion, use the given formula:p = x/nwhere x is the number of teenagers who own smartphones and n is the sample size.Substituting the values into the formula, we get;$$p = \frac{x}{n}$$$$0.8 = \frac{x}{250}$$$$x = 250 × 0.8$$$$x = 200$$Therefore, the sample proportion is 200/250 = 0.8.To find the probability of less than 75% of the sampled teenagers owned smartphones, we use the standard normal distribution formula, which is:Z = (X - μ)/σwhere X is the random variable, μ is the mean, and σ is the standard deviation.

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Answer:

7/30

Step-by-step explanation:

7 out of 30 is 7/30

The marginal density function for the given joint distribution is f(x) = x/3 + x². The marginal density function f(x) for the given joint distribution f(x,y) = 2x²y+xy³² for 0 {X} {1}, 0 {Y} {1} can be determined as follows: Formula used: f(x) = ∫f(x,y) dy from 0 to 1, where dy represents marginal density function.

Given joint distribution: f(x,y) = 2x²y+xy³² for 0 {X} {1}, 0 {Y} {1}

The marginal density function f(x) can be obtained by integrating f(x,y) over all possible values of y. i.e., f(x) = ∫f(x,y) dy from 0 to 1O n

substituting the given joint distribution in the above formula, we get:  f(x) = ∫ (2x²y+xy³²) dy from 0 to 1= 2x² [y²/2] + x [y³/3] from 0 to 1= 2x² (1/2) + x (1/3) - 0On

simplifying the above expression, we get: f(x) = x/3 + x²

Hence, the marginal density function for the given joint distribution is f(x) = x/3 + x².

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The required probability of obtaining a z value between -2.4 to -2.0 is 0.0146.

Given, for a standard normal distribution, the probability of obtaining a z value between -2.4 to -2.0 is.

Now, we have to find the probability of obtaining a z value between -2.4 to -2.0.

To find this, we use the standard normal table which gives the area to the left of the z-score.

So, the required probability can be calculated as shown below:

Let z1 = -2.4 and z2 = -2.0

Then, P(-2.4 < z < -2.0) = P(z < -2.0) - P(z < -2.4)

Now, from the standard normal table, we haveP(z < -2.0) = 0.0228 and P(z < -2.4) = 0.0082

Substituting these values, we get

P(-2.4 < z < -2.0) = 0.0228 - 0.0082= 0.0146

Therefore, the required probability of obtaining a z value between -2.4 to -2.0 is 0.0146.

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Approximately 44.1% of randomly selected females purchased a compact fuel-powered vehicle, while approximately 26.7% of randomly selected customers purchased an electric vehicle.

a) To compute the probability that a randomly selected female purchased a compact-fuel powered vehicle, we divide the number of females who purchased a compact-fuel powered vehicle (45) by the total number of females (102).

The probability is 45/102, which simplifies to approximately 0.441.

b) To compute the probability that a randomly selected customer purchased an electric vehicle, we divide the number of customers who purchased an electric vehicle (55) by the total number of customers (206).

The probability is 55/206, which simplifies to approximately 0.267.

Therefore, the probability that a randomly selected female purchased a compact-fuel powered vehicle is approximately 0.441, and the probability that a randomly selected customer purchased an electric vehicle is approximately 0.267.

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