what would the radiusof a hemisphere be if the volume is 140000pi

The probability that between 124 and 125​, inclusive are on time is approximately 0.0655.

Given:

The probability of a flight arriving on time is 0.88

Number of flights selected randomly = 145

Let X be the number of flights arriving on time.

(a) P(exactly 128 flights are on time)

Using the normal approximation to the binomial distribution, we have:

Mean, µ = np = 145 × 0.88 = 127.6

Standard deviation, σ = sqrt(np(1-p)) = sqrt(145 × 0.88 × 0.12) = 3.238

P(X = 128) can be approximated using the standard normal distribution:

z = (128 - µ) / σ = (128 - 127.6) / 3.238 = 0.1234

P(X = 128) ≈ P(z = 0.1234) = 0.4511

Therefore, the probability that exactly 128 flights are on time is approximately 0.4511.

(b) P(at least 128 flights are on time)

P(X ≥ 128) can be approximated as:

z = (128 - µ) / σ = (128 - 127.6) / 3.238 = 0.1234

P(X ≥ 128) ≈ P(z ≥ 0.1234) = 0.4515

Therefore, the probability that at least 128 flights are on time is approximately 0.4515.

(c) P(fewer than 124 flights are on time)

P(X < 124) can be approximated as:

z = (124 - µ) / σ = (124 - 127.6) / 3.238 = -1.1154

P(X < 124) ≈ P(z < -1.1154) = 0.1326

Therefore, the probability that fewer than 124 flights are on time is approximately 0.1326.

(d) P(between 124 and 125​, inclusive are on time)

P(124 ≤ X ≤ 125) can be approximated as:

z1 = (124 - µ) / σ = (124 - 127.6) / 3.238 = -1.1154

z2 = (125 - µ) / σ = (125 - 127.6) / 3.238 = -0.7388

P(124 ≤ X ≤ 125) ≈ P(-1.1154 ≤ z ≤ -0.7388) = P(z ≤ -0.7388) - P(z < -1.1154)

P(124 ≤ X ≤ 125) ≈ 0.1981 - 0.1326 = 0.0655

Therefore, the probability that between 124 and 125​, inclusive are on time is approximately 0.0655.

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Answer:

The answer is D.) A random sample may represent the population.

Step-by-step explanation:

Collecting samples from anyone at random increases the chances of representing the overall people because that's where they're getting it from. But there is still a chance it will not represent everyone or the general population.

Answer:B) A random sample may represent the population is your best answer.

Step-by-step explanation:

this is what i got for this one. have a good day everyone. :)

Mean- 6

Median- 6

Mode- 6 and 8

The intervals over which  it is increasing or decreasing is:

Increasing on: ([tex]-\infty[/tex], -1)

Decreasing on: (-1, [tex]\infty[/tex])

The definitions for increasing and decreasing intervals are given below.

The function is :

[tex]f(x)=-x^2-2x+3[/tex]

We have to find the interval of function is decrease/increase .

Now, We have to first differentiate with respect to x , then:

f'(x) = - 2x + 2

This derivative is never 0 for real x.

In order to determine the intervals over which  it is increasing or decreasing.

Increasing on: ([tex]-\infty[/tex], -1)

Decreasing on: (-1, [tex]\infty[/tex])

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The equation of the parabola is given as follows:

y = -28/19600(x - 140)² + 28.

The distances from the center for a height of 13 meters are given as follows:

37.53 m and 242.47 m.

The equation of a parabola of vertex (h,k) is given by the equation presented as follows:

y = a(x - h)² + k.

In which a is the leading coefficient.

It has a span of 280 meters, hence the x-coordinate of the vertex is given as follows:

x = 280/2

x = 140 -> h = 140.

The maximum height is of 28 meters, hence the y-coordinate of the vertex is given as follows:

y = 28 -> k = 28.

Hence the equation is:

y = a(x - 140)² + 28.

When x = 0, y = 0, hence the leading coefficient a is obtained as follows:

19600a = -28

a = -28/19600

Hence:

y = -28/19600(x - 140)² + 28.

The distance from the center at which the height is 13 meters is obtained as follows:

13 = -28/19600(x - 140)² + 28.

28/19600(x - 140)² = 15

(x - 140)² = 15 x 19600/28

(x - 140)² = 10500.

Hence the distances are obtained as follows:

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The 80% confidence interval for the mean waste recycled per person per day for the population of Montana is given as follows:

(1.9, 2.3).

The t-distribution is used when the standard deviation for the population is not known, and the bounds of the confidence interval are given according to the following rule:

[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]

In which the variables of the equation are presented as follows:

The critical value, using a t-distribution calculator, for a two-tailed 80% confidence interval, with 18 - 1 = 17 df, is t = 1.33.

The parameters are given as follows:

[tex]\overline{x} = 2.1, s = 0.74, n = 18[/tex]

The lower bound of the interval is given as follows:

2.1 - 1.33 x 0.74/sqrt(18) = 1.9.

The upper bound of the interval is given as follows:

2.1 + 1.33 x 0.74/sqrt(18) = 2.3.

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The equation of the plane passing through point P with normal vector 2i + 6j + 7k is x + 3y + 4z = 48.

A: True.

The equation of a plane in 3D space is given by Ax + By + Cz = D, where A, B, C are the components of the normal vector and D is the distance from the origin to the plane along the direction of the normal vector.

In this case, the normal vector is 2i + 6j + 7k, so A = 2, B = 6, and C = 7. To find D, we can substitute the coordinates of the given point P into the equation of the plane:

2(1) + 6(3) + 7(4) = D

2 + 18 + 28 = D

D = 48

Therefore, the equation of the plane passing through point P with normal vector 2i + 6j + 7k is x + 3y + 4z = 48.

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a) The quadratic equation x² + 7x + 10 = 0 has two distinct real roots
b) The quadratic equation 4x² - 3x + 4 = 0 has two complex (non-real) roots.

The discriminant of a quadratic equation of the form ax² + bx + c = 0 is given by the expression b² - 4ac. The value of the discriminant can help us determine the nature of the roots of the quadratic equation.

Specifically:

If the discriminant is positive, then the quadratic equation has two distinct real roots.

If the discriminant is zero, then the quadratic equation has one real root (also known as a double root or a repeated root).

If the discriminant is negative, then the quadratic equation has two complex (non-real) roots.

Using this information, we can determine the number of real solutions for each of the given quadratic equations without actually solving them:\

a) x² + 7x + 10 = 0

Here, a = 1, b = 7, and c = 10.

Therefore, the discriminant is:

b² - 4ac = 7² - 4(1)(10) = 49 - 40 = 9

Since the discriminant is positive, this quadratic equation has two distinct real roots.

b) 4x² - 3x + 4 = 0

Here, a = 4, b = -3, and c = 4.

Therefore, the discriminant is:

b² - 4ac = (-3)² - 4(4)(4) = 9 - 64 = -55

Since the discriminant is negative, this quadratic equation has two complex (non-real) roots.

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We have shown that (10101) times (10011) in GF(2^5) with [tex](x^5 + x^4 + x^3 + x^2+ 1)[/tex] as the modulus is equal to (101111) in binary or [tex]x^4 + x^2 + x + 1[/tex] in polynomial form.

To multiply (10101) by (10011) in GF [tex](2^5)[/tex] with [tex](x^5 + x^4 + x^3 + x^2+ 1)[/tex] as the modulus, we first need to write these polynomials as binary numbers:

[tex](10101) = 1x^4 + 0x^3 + 1x^2 + 0x + 1 = 16 + 4 + 1 = (21)_10 = (10101)_2[/tex]

[tex](10011) = 1x^4 + 0x^3 + 0x^2 + 1x + 1 = 16 + 2 + 1 = (19)_10 = (10011)_2[/tex]

We will use long multiplication to multiply these polynomials in GF[tex](2^5)[/tex], as shown below:

    1 0 1 0 1   <-- (10101)

  x 1 0 0 1 1   <-- (10011)

  ------------

    1 0 1 0 1   <-- Step 1: Multiply by 1

1 0 1 0 1      <-- Step 2: Multiply by x and shift left

------------

1 0 0 1 0 1    <-- Step 3: Add steps 1 and 2

1 0 0 1 0 <-- Step 4: Multiply by x and shift left

1 0 1 1 1 1 <-- Step 5: Add steps 3 and 4

Now, we have the product (101111)_2, which corresponds to the polynomial [tex]1x^4 + 0x^3 + 1x^2 + 1x + 1 = x^4 + x^2 + x + 1[/tex] in GF[tex](2^5)[/tex] with [tex](x^5 + x^4 + x^3 + x^2+ 1)[/tex] as the modulus. We can verify that this polynomial is indeed in GF(2^5) with modulus [tex](x^5 + x^4 + x^3 + x^2+ 1)[/tex] by noting that all of its coefficients are either 0 or 1, and none of its terms have degree greater than 4. Additionally, we can check that it satisfies the modulus:

[tex]x^4 + x^2 + x + 1 = (x^4 + x^3 + x^2 + x) + (x^3 + 1)[/tex]

[tex]= x(x^3 + x^2 + x + 1) + (x^3 + 1)[/tex]

[tex]= x(x^3 + x^2 + x + 1) + (x^3 + x^2 + x + 1)[/tex]

(since [tex]x^3 + x^2 + x + 1 = 0[/tex] in GF[tex](2^5))[/tex]

[tex]= (x+1)(x^3 + x^2 + x + 1)[/tex]

Therefore, we have shown that (10101) times (10011) in GF(2^5) with [tex](x^5 + x^4 + x^3 + x^2+ 1)[/tex] as the modulus is equal to (101111) in binary or [tex]x^4 + x^2 + x + 1[/tex] in polynomial form.

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The surface area of the triangular base prism is 174 ft².

The prism above is a triangular prism. Therefore, let's find the surface area of the triangular prism as follows:

The prism has two triangular faces and three rectangular faces.

Therefore,

area of the triangle = 1 / 2 bh

where

Therefore,

area of the triangle = 1 / 2 × 6 × 4

area of the triangle = 24 / 2

area of the triangle = 12 ft²

Therefore,

area of the rectangle = l × w

where

Hence,

area of the rectangle =  8 × 5 = 50 ft²

Surface area of the triangular prism = 12(2) + 3(50)

Surface area of the triangular prism = 24 + 150

Surface area of the triangular prism = 174 ft²

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The statement that is true about the 2 functions, in which the relationship between the x and y-values in the table of values for both functions is a linear relationship is that The y-intercept of the graph of A is equal to the y-intercept of the graph of B

The equation representing the relationship in function A in point-slope form is therefore;

y - 12 = 6·(x - 2)

y - 12 = 6·x - 12

y = 6·x - 12 + 12 = 6·x

The equation in slope-intercept form, y = m·x + c, where c is the y-intercept is therefore; y = 6·x

The true statement is therefore; The y-intercept of the graph of A is less than the y-intercept of the graph of B

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There was a significant reduction in peoples over estimation of the line length, t = XXX with df = XXX, p < 0.01 two-tailed, with M = 10%, n = 25, s = 15%, Cohen's d = XXX , M = 10%, 95% CI [XXX, XXX].

8a. A. was
8b. B. t = XXX with df = XXX
8c. A. < 0.01 two-tailed
8d. C. M = 10%, n = 25, s = 15%, Cohen's d = XXX , M = 10%, 95% CI [XXX, XXX].

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Answer:

$7200

step by step Explanation:

Cameron borrowed $18,000 at an interest rate of 10% for a period of 4 years. To calculate the interest, we can use the simple interest formula: I = P * r * t, where I is the interest, P is the principal amount, r is the interest rate, and t is the time period.

Plugging in the values, we get I = 18,000 * 0.10 * 4 = $7,200. Therefore, Cameron paid a total of $7,200 in interest over the 4-year period.

By the principle of mathematical induction, we can conclude that n! > 2ⁿ for all n ∈ Z≥4.

The art of demonstrating a claim, theorem, or formula that is regarded as true for each and every natural number n is known as proof.

We can prove by mathematical induction that n! > 2ⁿ for all n ∈ Z≥4.

First, we will prove the base case n = 4:

4! = 4 x 3 x 2 x 1 = 24

2⁴ = 16

Since 24 > 16, the base case is true.

Next, we assume that the inequality is true for some arbitrary k ≥ 4:

k! > [tex]2^k[/tex]

To complete the induction step, we must prove that the inequality is also true for k + 1:

(k+1)! = (k+1) x k!

(k+1)! > (k+1) x [tex]2^k[/tex]    (by the induction hypothesis)

(k+1)! > 2 x [tex]2^k[/tex]

(k+1)! > [tex]2^{(k+1)[/tex]

Since the inequality is true for k+1, this completes the induction step.

Therefore, by the principle of mathematical induction, we can conclude that n! > 2ⁿ for all n ∈ Z≥4.

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If you invest $2,000 for 3 years at an interest rate of 6%, compounded every 6 months. The value of your investment at the end of the period is $2,397.39.

The interest rate is 6% and it is compounded every 6 months, so the period is 6 months. To calculate the value of the investment at the end of the period, we need to use the formula for compound interest:

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

Where:
A = the final amount.
P = the principal amount (initial investment)
r = the annual interest rate (6%).
n = the number of times the interest is compounded per year (2, since it's compounded every 6 months).
t = the time period in years (3)

Plugging in the numbers, we get:

A = 2,000(1 + 0.06/2)^(2*3)
A = 2,000(1 + 0.03)^6
A = 2,000(1.03)^6
A = $2,397.39

Therefore, the value of your investment at the end of the period is $2,397.39.

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There are 64 functions from set A to set B.

To determine how many functions there are from A = {1, 2, 3} to B = {a, b, c, d}, you can use the following step-by-step explanation:

1. Understand that a function maps each element of set A to exactly one element in set B.
2. Notice that set A has 3 elements, and set B has 4 elements.
3. For each element in set A, there are 4 choices in set B it can be mapped to.
4. Therefore, the total number of functions is equal to the product of the number of choices for each element in set A, which is 4 × 4 × 4 = 64.

So, there are 64 functions from A = {1, 2, 3} to B = {a, b, c, d}.

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Answer:

The answer is m = 2 .

Step-by-step explanation:

14 - 3m = 4m

14 = 4m + 3m

14 = 7m

14/7 = m

2 = m

m = 2

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The solution to the differential equation is y(t) = 3cosh(3t) + 2sin(4t).

To solve this differential equation using Laplace transform, we first apply the transform to both sides of the equation:

L[day + 2dy/dt + y] = L[sinh(3t) - 5cosh(3t)]

Using the properties of Laplace transform and the derivative property, we get:

sY(s) - y(0) + 2[sY(s) - y(0)]/dt + Y(s) = 3/(s^2 - 9) - 5s/(s^2 - 9)

Substituting the initial conditions y(0) = -2 and y'(0) = 5, and simplifying the expression, we get:

Y(s) = (3s - 19)/(s^3 - 2s^2 - 3s + 18)

Now, we need to find the inverse Laplace transform of Y(s) to obtain the solution y(t). This can be done using partial fraction decomposition, which gives:

Y(s) = -1/(s - 3) + 4/(s + 2) + 2/(s - 3)^2

Taking the inverse Laplace transform of each term using the Laplace transform table, we get:

y(t) = -e^(3t) + 4e^(-2t) + 2te^(3t)

Therefore, the solution to the differential equation is y(t) = -e^(3t) + 4e^(-2t) + 2te^(3t).

To solve this differential equation using Laplace transform, we first apply the transform to both sides of the equation:

L[day/dt^2 - 4y - 5y] = L[e^3 + sin(4t)]

Using the properties of Laplace transform, we get:

s^2Y(s) - sy(0) - y'(0) - 4Y(s) - 5Y(s) = 3/(s - 3) + 4/(s^2 + 16)

Substituting the initial conditions y(0) = 0 and y'(0) = 0, and simplifying the expression, we get:

s^2Y(s) - 9Y(s) = 3/(s - 3) + 4/(s^2 + 16)

Using partial fraction decomposition, we get:

Y(s) = (3s - 9)/(s^2 - 9) + (4s)/(s^2 + 16)

Taking the inverse Laplace transform of each term using the Laplace transform table, we get:

y(t) = 3cosh(3t) + 2sin(4t)

Therefore, the solution to the differential equation is y(t) = 3cosh(3t) + 2sin(4t).

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a) The probability that X is within three standard deviations of the mean is approximately 1.

b) the probability that the drone will land out of the targeted area exactly 4 times is 0.00052.

c) The probability that the drone will land out of the targeted area at most 4 times is 0.1029

d) The expected value of X is 9.6.

e) The meaning of the expected value in the context of the story is average landing performance of the drone based on the given probability of success.

f) The variance of X is 0.7319.

g) The probability that it missed the target at most 2 times is 3.121.

h) The probability that it missed the target at least 2 times is 0.7319.

I) The probability that X is within three standard deviations of the mean is 1.3856.

The binomial distribution is a discrete probability distribution that describes the number of successes in a fixed number of independent trials with a constant probability of success.

The characteristics of a binomial random variable include the number of trials (n), the probability of success (p), the number of successes (x), and the mean and variance of the distribution.

Here we have

Binomial distribution We are testing the landing performance of a new automated drone. The drone lands on the targeted area 80% of the time. We test the drone 12 times.

a. X is a binomial random variable because we have a fixed number of independent trials and each landing has only two possible outcomes (landing on the targeted area or landing outside of it) with a constant probability of success (0.8).

The characteristics of the binomial distribution are:

The number of trials is fixed (n=12)

Each trial has only two possible outcomes (success or failure)

The probability of success (p) is constant for each trial

The trials are independent of each other

b. P(X = 4) = (12 choose 4) × (0.8)⁴ × (0.2)⁸ = 0.00052

c. P(X< = 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= 0.0687 + 0.2060 + 0.3020 + 0.2670 + 0.1854 = 0.1029

d. E(X) = np = 120.8 = 9.6

e. The expected value of X represents the average number of successful landings (in the targeted area) we would expect to see in a sample of 12 landings.

In the context of the story, it tells us the average landing performance of the drone based on the given probability of success.

f. Var(X) = np(1-p) = 120.80.2 = 1.92

g. P(X<=2 | X<=4) = P(X<=2)/P(X<=4)

= (P(X=0) + P(X=1) + P(X=2))/(P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4))

= 0.3217/0.1029 = 3.121

h. P(X>=2 | X<=4) = 1 - P(X<2 | X<=4) = 1 - P(X<=1 | X<=4) = 1 - (P(X=0) + P(X=1))/(P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)) = 1 - 0.2747/0.1029 = 0.7319

i. The standard deviation of a binomial distribution is √(np(1-p)). So, the standard deviation of X is √(120.80.2) = 1.3856. Three standard deviations above and below the mean would be 3*1.3856 = 4.1568.

Therefore,

The probability that X is within three standard deviations of the mean is approximately 1.

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Answer:

B) 6

Step-by-step explanation:

Firstly, we need to know what the question is asking for.

"How many miles does she need to run each hour to complete the run" is asking for a speed in miles per hour.

miles / hour = speed in mph

18 miles / 3 hours = 18/3 mph

18/3 simplifies to 6

Lisa needs to run 6 mph

Answer:

Its the first one

Step-by-step explanation:

correct answer

To begin, let's recall that synthetic division is a method used to divide a polynomial by a linear factor (i.e. a binomial of the form x-a, where a is a constant). The result of synthetic division is the quotient of the division, which is a polynomial of one degree less than the original polynomial.

In this case, we are given that x is a solution of a third-degree polynomial equation. This means that the polynomial can be factored as (x-r)(ax^2+bx+c), where r is the given solution and a, b, and c are constants that we need to determine.

To use synthetic division, we will divide the polynomial by x-r, where r is the given solution. The result of the division will give us the coefficients of the quadratic factor ax^2+bx+c.

Here's an example of how to do this using synthetic division:

Suppose we are given the polynomial P(x) = x^3 + 2x^2 - 5x - 6 and we know that x=2 is a solution.

1. Write the polynomial in descending order of powers of x:

P(x) = x^3 + 2x^2 - 5x - 6

2. Set up the synthetic division table with the given solution r=2:

2 | 1  2  -5  -6

3. Bring down the leading coefficient:

2 | 1  2  -5  -6
  ---
   1

4. Multiply the divisor (2) by the result in the first row, and write the product in the second row:

2 | 1  2  -5  -6
  ---
   1  2

5. Add the second row to the next coefficient in the first row, and write the sum in the third row:

2 | 1  2  -5  -6
  ---
   1  2 -3

6. Multiply the divisor by the result in the third row, and write the product in the fourth row:

2 | 1  2  -5  -6
  ---
   1  2 -3
       4

7. Add the fourth row to the next coefficient in the first row, and write the sum in the fifth row:

2 | 1  2  -5  -6
  ---
   1  2 -3
       4 -2

The final row gives us the coefficients of the quadratic factor: ax^2+bx+c = x^2 + 2x - 3. Therefore, the factorization of P(x) is

P(x) = (x-2)(x^2+2x-3).

To find the real solutions of the equation, we can use the quadratic formula or factor the quadratic further:

x^2 + 2x - 3 = (x+3)(x-1).

Therefore, the real solutions of the equation are x=2, x=-3, and x=1.

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The statement is true, if u > x and v > y then ged(u, v) > ged(x,y).

First, let's define ged(u,v) as the greatest common divisor of u and v.

Assuming that u > x and v > y, we can express u and v as:

u = x + m
v = y + n

where m and n are positive natural numbers.

Now, let's assume that ged(x,y) = d, where d is a positive natural number that divides both x and y.

Therefore, we can express x and y as:

x = dp
y = dq

where p and q are positive natural numbers.

Now, we can express u and v in terms of d as well:

u = dp + m
v = dq + n

Since m and n are positive natural numbers, it follows that ged(u,v) is a positive natural number as well.

Now, we need to show that ged(u,v) > d.

Assume the contrary, i.e. ged(u,v) ≤ d.

This means that there exists a positive natural number k that divides both u and v, and k ≤ d.

Since k divides both u and v, it must also divide their difference:

u - v = (d * p + m) - (d * q + n) = d * (p - q) + (m - n)

Therefore, k must also divide (m - n).

But since m and n are positive natural numbers, we have:

|m - n| < max(m,n) ≤ max(u,v)

Therefore, k cannot divide both (m - n) and max(u,v), which contradicts the assumption that k divides both u and v.

Therefore, our initial assumption that ged(u,v) ≤ d must be false, which means that ged(u,v) > d.

Therefore, the statement is true.

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We can state the critical values for the rejection rule as follows:

test statistic ≤ -1.645 (left-tailed test)

test statistic ≥ 1.645 (right-tailed test)

The sample mean can be calculated by adding up all the weekly pays and dividing by the sample size:

sample mean = (687.73 + 543.15 + ... + 703.06) / 50 = 638.55 (rounded to two decimal places)

To test whether the average weekly earnings for workers who have received a high school diploma is significantly greater than average weekly earnings for workers who have not received a high school diploma, we can perform a two-sample t-test assuming equal variances. The null hypothesis is that there is no difference in the means of the two groups, and the alternative hypothesis is that the mean for the high school diploma group is greater than the mean for the non-high school diploma group.

Using a calculator or software, we can calculate the test statistic and p-value. Assuming a two-tailed test and a significance level of 0.05, the critical values for the rejection rule are -1.96 and 1.96.

test statistic = 3.196 (rounded to three decimal places)

p-value = 0.0012 (rounded to four decimal places)

Since the p-value (0.0012) is less than the significance level (0.05), we reject the null hypothesis and conclude that the average weekly earnings for workers who have received a high school diploma is significantly greater than average weekly earnings for workers who have not received a high school diploma.

For a one-tailed test with α = 0.05, the critical value would be 1.645. The rejection rule would be: if the test statistic is greater than 1.645, reject the null hypothesis. Therefore, we can state the critical values for the rejection rule as follows:

test statistic ≤ -1.645 (left-tailed test)

test statistic ≥ 1.645 (right-tailed test)

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The number of ways there are for her to select the starting line-up is 68,640 ways.

1. Center: There are 4 players who can play this position, so there are 4 choices.
2. Right Forward: Since one player has been selected as Center, there are now 13 players remaining. So, there are 13 choices for this position.
3. Left Forward: After selecting the Center and Right Forward, 12 players remain, resulting in 12 choices for this position.
4. Right Guard: With three players already chosen, there are 11 players left to choose from, giving us 11 choices.
5. Left Guard: Finally, after selecting players for the other four positions, 10 players remain, providing 10 choices for this position.

Now, we can calculate the total number of ways to select the starting line-up using the counting principle by multiplying the number of choices for each position:

4 (Center) × 13 (Right Forward) × 12 (Left Forward) × 11 (Right Guard) × 10 (Left Guard) = 68,640 ways

So, there are 68,640 ways for the coach to select the starting line-up.

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The circle intersects the x-axis at the points (-5, 0) and (5, 0).

We have,

Using the Pythagorean theorem to find the radius of the circle.

So,

r = √(0-3)² + (0-4)²

r = √(9+16)

  = √25

  = 5

The equation of the circle is x² + y² = 5² = 25.

To find the points where the circle intersects the x-axis,

We substitute y = 0 in the equation of the circle and solve for x:

x² + 0² = 25

x² = 25

x = ±5

Therefore,

The circle intersects the x-axis at the points (-5, 0) and (5, 0).

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The possible values of the coefficient (a) and an exponent (b) are CThe missing coefficient (a) can be any multiple of 3. The missing exponent (b) must be 5.

The two monomials are given as

ax³ 6xᵇ

Such that we have the LCM to be

LCM = 18x⁵

Since the coefficient of the LCM is 18, then the following is possible

a * 6 = multiples of 18

Divide both sides by 6

a = multiples of 3

Next, we have

LCM of x³ * xᵇ = x⁵

So, we have

b = 5 (bigger exponent)

Hence, the true statement is (c)

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Answer:

The answer to your problem is, [tex]7\frac{1}{9}[/tex]

Step-by-step explanation:

Calculation process:

= [tex]\frac{16}{3}[/tex] ÷ [tex]\frac{3}{4}[/tex]

= [tex]\frac{16}{3}[/tex] × [tex]\frac{4}{3}[/tex]

= [tex]\frac{16*4}{3*3}[/tex]

= [tex]\frac{64}{9}[/tex] = [tex]7\frac{1}{9}[/tex]

Thus the answer to your problem is, [tex]7\frac{1}{9}[/tex]

Answer:

b

Step-by-step explanation:

trust ne