Which null distribution is used for a hypothesis test of a single population proportion? Select one: O a. t-distribution with df = n - 1 O b. standard normal distribution O c. t-distribution with df = n1 + n2 – 2

Answer:

b

Step-by-step explanation:

oi did the quiz f. 6373737

Where the above is given,

Range

Range = Maximum value - Minimum value

Range = 2.50 - 0.01

= 2.49 inches

Sample Variance

Step 1  - Calculate the mean (average) of the data.

Mean = (0.47 + 0.27 +0.13 + 0.54 + 0.08 + 1.05 + 0.34 +   0.26 + 0.42 + 0.17 + 0.50 + 0.86 + 0.01 + 2.5) / 14

= 1.36 /14

= 0.0971

Step 2  - Calculate the squared deviation from the mean for each data point.

Deviation from mean = Data point - Mean

Squared deviation = (Deviation from mean)²

Sum of squared deviations = (0.3739 + 0.0091 + 0.7307 + 0.1638 + 0.9469 + 0.104 + 0.0145 + 0.0081 + 0.0971 + 0.3037 + 0.0601 + 0.2601 + 1.0679 +   3.1391)

= 7.2799

Sample Variance = Sum of squared deviations / (n - 1)

= 7.2799/ (14 - 1)

= 0.5599

Sample Standard Deviation:

Sample Standard Deviation = √(Sample Variance)

= √0.5599

= 0.7483

Coefficient of Variation:

Coefficient of Variation = (Sample Standard Deviation / Mean) * 100

= (7.2799/ 0.0971) * 100

= 7497.32

Z-Scores for each data point

Z1= (0.47   - 0.0971)/ 0.7212= 0.5318

Z2= (0.27   - 0.0971)/ 0.7212= 0.9983

Z3= (0.13   - 0.0971)/ 0.7212= 0.0456

Z4= (0.54   - 0.0971)/ 0.7212= 0.6189

Z5= (0.08   - 0.0971)/ 0.7212=   -0.0251

Z6= (1.05   - 0.0971)/ 0.7212= 1.2914

Z7= (0.34   - 0.0971)/ 0.7212= 0.4574

Z8= (0.26   - 0.0971)/ 0.7212= 0.2251

Z9= (0.42   - 0.0971)/ 0.7212= 0.4969

Z10= (0.17   - 0.0971)/ 0.7212= 0.1002

Z11= (0.50   - 0.0971)/ 0.7212= 0.6506

Z12= (0.86   - 0.0971)/ 0.7212= 1.0242

Z13= (0.01   - 0.0971)/ 0.7212=   -0.1208

Z14= (2.50   - 0.0971)/ 0.7212= 3.3999

To determine   if there are any outliers, we can compare the absolute values of the Z-Scores to a certain threshold , commonly considered as 2.

If the absolute value of a Z-Score is greater than 2,   it indicates that the corresponding data point is an outlier.

In this case,   Z14 = 3.3999, which is greater than 2 , suggesting that the data point 2.50 inches may be considered an outlier based on the Z-Score criterion.

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Answer:

40,000

Step by step explanation:

842,963

840,000

40,000

Answer:

7

Step-by-step explanation:

the answer I got was 7 hope this helped

Answer:

Open points on 1 and 3 with ray connecting them

Step-by-step explanation:

Because it's right.

Answer:

Step-by-step explanation:

Answer:

y=10

Step-by-step explanation:

[tex]\frac{4}{100}[/tex]=[tex]\frac{y}{250}[/tex]

cross-multiply, 4*250=100y

isolate the variable and solve for y, 1000=100y

divide 100 on both sides, 10=y

The answer is x = 15.

15 or 20.33 are the possible values of the x.

Any mathematical statement that includes numbers, variables, and an arithmetic operation between them is known as an expression or algebraic expression. In the phrase 4m + 5, for instance, the terms 4m and 5 are separated from the variable m by the arithmetic sign +.

We know that sin(x+37)°=cos(90°-(x+37)°) and cos(2x+8)°=sin(90°-(2x+8)°)

So we have sin(x+37)°=cos(2x+8)° becomes sin(x+37)°=sin(82°-2x)

For the above equation to be true, either of the following must hold:

x+37 = 82 - 2x (since the sin function is periodic)

x+37 = 180 - (82-2x)

Solving the first equation for x gives:

3x = 45

x = 15

Solving the second equation for x gives:

3x = 61

x = 20.33 (rounded to two decimal places)

Therefore, the possible values of x are 15 or 20.33 (rounded to two decimal places).

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The probability that all orders are fulfilled on a day chosen at random is approximately 1. Answer: 1.0000 (rounded off to at least four decimal places).

The quantities ordered by customers are independent continuous random variables, and they are uniformly distributed over the interval (0, 9).

The fresh food distributor only has the capacity to ship 477 kg of products daily, and the distributor receives orders from 100 customers daily.

The probability that all orders are fulfilled on a day chosen at random is given by;P(all orders fulfilled) = P(X1 + X2 + ... + X100 < 477)

where X is the quantity ordered by each customer. Since X is a continuous random variable, we can use the probability density function of a uniform distribution to calculate the probability density function of X as;f(x) = 1/9, 0 < x < 9

Hence, the probability that all orders are fulfilled on a day chosen at random is given by;

P(all orders fulfilled) = P(X1 + X2 + ... + X100 < 477)= P[(X1/9) + (X2/9) + ... + (X100/9) < (477/9)]= P[U < (53 + 1/3)], where U ~ Uniform(0, 1)

Now, using the central limit theorem, we can approximate the distribution of U by a normal distribution with mean μ = 1/2 and variance σ^2 = 1/12 such that;Z = (U - μ) / σ ~ N(0, 1)

Hence, P[U < (53 + 1/3)] = P[Z < (53 + 1/3 - μ) / σ]= P[Z < (53 + 1/3 - 1/2) / sqrt(1/12)]≈ P[Z < 9.6067]≈ 1

Thus, the probability that all orders are fulfilled on a day chosen at random is approximately 1. Answer: 1.0000 (rounded off to at least four decimal places).

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Answer:

[tex]\frac{4}{15}[/tex]  (4/15)

Step-by-step explanation:

[tex]\frac{4}{6}=\frac{2}{3}[/tex]

[tex]\frac{2}{5}*\frac{2}{3};[/tex]

1- Multiply the numerators:

[tex]2*2=4[/tex]

2- Multiply the denominators:

[tex]5*3=15[/tex]

3- Thus:

[tex]\frac{2}{5}*\frac{2}{3}= \frac{4}{15}[/tex]

Hope this helps ;)

Answer:

181 feet

Step-by-step explanation:

From the the diagram attached ,

Sinθ = a/b..................... Equation 1

Where θ = angle to the horizontal, a = Height of the tower, b = length of the wire.

make b the subject of the equation

b = a/sinθ................. Equation 2

Given: a = 128 feet, θ = 45°

Substitute these values into equation 2

b = 128/sin45°

b = 128/0.7071

b = 181 feet

b = 181 feet

In Hausdorff-space "X", if A and B are disjoint "compact-subspaces", then there is disjoint "open-sets" U and V such that A is contained in U and B is contained in V, this is because by Hausdorff-Property, the existence of disjoint open neighborhoods for any two "distinct-points".

To prove the existence of disjoint "open-sets" U and V with A⊂U and B⊂V, where A and B are "compact-subspaces" of "Hausdorff-space" X,

Step (1) : A and B are disjoint compact-subspaces, we use Hausdorff property to find "open-sets" Uₐ and [tex]U_{b}[/tex] such that "A⊂Uₐ" and "B⊂[tex]U_{b}[/tex]", and "Uₐ∩[tex]U_{b}[/tex] = ∅". This can be done for every pair of points in A and B, respectively, because X is Hausdorff.

Step (2) : We consider, set U = ⋃ Uₐ, where "union" is taken over all of Uₐ for each-point in A. U is = union of "open-sets", hence open.

Step (3) : We consider set V = ⋃ [tex]U_{b}[/tex], where union is taken over for all [tex]U_{b}[/tex] for "every-point" in B. V is also a union of open-sets and so, open.

Step (4) : We claim that U and V are disjoint. Suppose there exists a point x in U∩V. Then x must be in Uₐ for some point a in A and also in [tex]U_{b}[/tex] for some point b in B. Since A and B are disjoint, a and b are different points. However, this contradicts the fact that Uₐ and [tex]U_{b}[/tex] are disjoint open sets.

Therefore, U and V are disjoint open sets with A⊂U and B⊂V.

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The given question is incomplete, the complete question is

Let A and B be disjoint compact subspaces of a Hausdorff space X. Show that there exist disjoint open sets U and V, with A⊂U and B⊂V.

The solution to the system of inequalities is the region above the x-axis and to the right of the line y = 1, shaded in green.

Given system of inequalities is: x - 2y ≤ 3 ...(1)3x + 4y ≥ 12 ...(2)x ≥ 0 ...(3)y ≥ 1 ...(4)

We graph the lines x - 2y = 3 and 3x + 4y = 12 and shade the appropriate regions.

Let's start with the line x - 2y = 3.

We rewrite this as y = (1/2)x - 3/2 and plot the line as shown below: graph{(1/2)x - 3/2 [-10, 10, -5, 5]}

Now we determine which side of the line we want to shade.

Since the inequality is of the form ≤, we shade below the line y = (1/2)x - 3/2 (including the line itself) as shown below: graph {(1/2)x - 3/2 [-10, 10, -5, 5](-10,-5)--(10,0)}

Next, we graph the line 3x + 4y = 12. We rewrite this as y = (-3/4)x + 3 and plot the line as shown below: graph{(-3/4)x + 3 [-10, 10, -5, 5]}

We determine which side of the line we want to shade. Since the inequality is of the form ≥, we shade above the line y = (-3/4)x + 3 (including the line itself) as shown below: graph{(-3/4)x + 3 [-10, 10, -5, 5](-10,4)--(10,0)}

Finally, we shade the region that satisfies x ≥ 0 and y ≥ 1.

This is the region above the x-axis and to the right of the line y = 1 as shown below: graph{(-3/4)x + 3 [-10, 10, -5, 5](-10,4)--(10,0)(0,1)--(10,1)[above]}

The shaded region is the region that satisfies all three inequalities.

Thus, the solution to the system of inequalities is the region above the x-axis and to the right of the line y = 1, shaded in green.

We graph the lines x - 2y = 3 and 3x + 4y = 12 and shade the appropriate regions.

Let's start with the line x - 2y = 3. We rewrite this as y = (1/2)x - 3/2 and plot the line.

Now we determine which side of the line we want to shade. Since the inequality is of the form ≤, we shade below the line y = (1/2)x - 3/2 (including the line itself).

Next, we graph the line 3x + 4y = 12. We rewrite this as y = (-3/4)x + 3 and plot the line. We determine which side of the line we want to shade.

Since the inequality is of the form ≥, we shade above the line y = (-3/4)x + 3 (including the line itself).

Finally, we shade the region that satisfies x ≥ 0 and y ≥ 1.

This is the region above the x-axis and to the right of the line y = 1. The shaded region is the region that satisfies all three inequalities.

Thus, the solution to the system of inequalities is the region above the x-axis and to the right of the line y = 1, shaded in green.

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Answer:

in right angled triangleBCD

BC=√{DC²-BC²)=√{10²-6²)=8

again in right angled triangle ABC

AB=√(BC²-AC²)

x=√(8²-7²)=3.87

Answer:

2.28871.0005^14

Step-by-step explanation:

Answer:

i got 8067661468.26  

Step-by-step explanation:

The proof of 2ⁿ (-1 + i√3)ⁿ + (-1 - i√3)ⁿ can be expressed as  2ⁿ⁺¹cos(πn/3) is proved below.

To express (-1 + i√3) and (-1 - i√3) in exponential form, we can use Euler's formula, which states that [tex]e^{(i\theta)[/tex] = cos(θ) + isin(θ).

Let's start with (-1 + i√3):

(-1 + i√3) = 2 x (cos(π) + i x sin(π/3))

Now, let's simplify (-1 - i√3):

(-1 - i√3) = 2 (cos(π) - isin(π/3))

Therefore, we have:

(-1 + i√3) = 2  e^(iπ/3)

(-1 - i√3) = 2  e^(-iπ/3)

Now, let's substitute these exponential forms into the expression:

2ⁿ  (-1 + i√3)^n + (-1 - i√3)^n

= 2ⁿ(2  e^(iπ/3))^n + (2  e^(-iπ/3))^n

= 2ⁿ⁺¹  e^(iπn/3) + 2^(n+1) e^(-iπn/3)

Using Euler's formula again, we know that [tex]e^{(i\theta)} + e^{(-i\theta)[/tex] = 2cos(θ).

Therefore, we can rewrite the expression as:

2ⁿ⁺¹ (cos(πn/3) + cos(-πn/3))

=  2ⁿ⁺¹(cos(πn/3) + cos(πn/3))

=  2ⁿ⁺¹ 2  cos(πn/3)

=  2ⁿ⁺¹cos(πn/3)

So, we have shown that  2ⁿ (-1 + i√3)ⁿ + (-1 - i√3)ⁿ can be expressed as  2ⁿ⁺¹cos(πn/3).

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Answer:

12

Step-by-step explanation:

√(x2 - x1)² + (y2 - y1)²

√[4 - (-8)]² + (10 - 10)²

√(12)² + (0)²

√144 + 0

√144

=12

The correct statement regarding r squared is:

r squared equals 0.4268 meaning that Years of Experience explains 42.68 percent of the variability in Salary.

Part A: The correct null and alternate hypotheses are:

H₀: β₁=0;

H₁: β₁≠0.

Part B: The correct p-value for the test in Q1 is 1.74e-13.

Part C: The fitted regression model from this output in Q1 is:

E(Salary) = 31.8387 + 2.8205 x Years of Experience.

Part D: The correct statement regarding r squared is:

r squared equals 0.4268 meaning that Years of Experience explains 42.68 percent of the variability in Salary.

Explanation: The output shows a multiple linear regression model:

Salary=β0+β1x

Years of Experience + ϵ.β0 is the intercept and represents the expected mean salary for an individual with 0 years of experience.

β1 is the slope and represents the expected change in salary due to one year increase in experience.

ϵ is the error term (deviation from the expected salary).

The correct null and alternate hypotheses are:

H₀: β₁=0 (there is no linear relationship between salary and years of experience).

H₁: β₁≠0 (there is a linear relationship between salary and years of experience).

The correct p-value for the test in Q1 is 1.74e-13, which is much smaller than the significance level of 0.05.

Thus, we reject the null hypothesis and conclude that there is sufficient evidence to suggest that there is a linear relationship between salary and years of experience.

The fitted regression model from this output in Q1 is:

E(Salary) = 31.8387 + 2.8205 x Years of Experience.

The coefficient of determination, or R-squared, is a statistical measure that shows how well the regression model fits the observed data.

It is the proportion of the variance in the dependent variable that is explained by the independent variable(s).

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Ancient Egyptian way of computation, division was performed using a method called "repeated subtraction." (a) 26 ÷ 20 = 1 remainder 6.

(b) 55 ÷ 6 = 9 remainder 19.

(c) 71 21 ÷ = 50 remainder 29.

(d) 25 18 ÷ = 7.

(e) 52 ÷ 68 = 0 remainder 52.

(f) 13 36 ÷ = 0 remainder -23.

In the ancient Egyptian way of computation, division was performed using a method called "repeated subtraction." Here's how it would be applied to the given divisions:

(a) 26 ÷ 20:

To divide 26 by 20, we repeatedly subtract 20 from 26 until we cannot subtract anymore. The number of times we subtract is the quotient.

26 - 20 = 6

6 - 20 = -14 (cannot subtract anymore)

Therefore, 26 ÷ 20 = 1 remainder 6.

(b) 55 ÷ 6:

Using the same method, we repeatedly subtract 6 from 55 until we cannot subtract anymore.

55 - 6 = 49

49 - 6 = 43

43 - 6 = 37

37 - 6 = 31

31 - 6 = 25

25 - 6 = 19 (cannot subtract anymore)

Therefore, 55 ÷ 6 = 9 remainder 19.

(c) 71 21 ÷:

To divide 71 21 by a number, we first convert it to a whole number by multiplying the fraction part by the denominator.

71 21 = 71 + (21/100) = 71 + 21/100

Now, we can perform division using repeated subtraction.

71 - 21 = 50

50 - 21 = 29 (cannot subtract anymore)

Therefore, 71 21 ÷ = 50 remainder 29.

(d) 25 18 ÷:

Similar to the previous case, we convert 25 18 to a whole number.

25 18 = 25 + (18/100) = 25 + 18/100

Performing division:

25 - 18 = 7

Therefore, 25 18 ÷ = 7.

(e) 52 ÷ 68:

Since 52 is smaller than 68, the quotient is 0.

Therefore, 52 ÷ 68 = 0 remainder 52.

(f) 13 36 ÷:

Converting to a whole number:

13 36 = 13 + (36/100) = 13 + 36/100

Performing division:

13 - 36 = -23 (cannot subtract anymore)

Therefore, 13 36 ÷ = 0 remainder -23.

Please note that the ancient Egyptian method of division is not as efficient as modern division methods and may not produce exact decimal results.

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Answer:

Given -

In a certain chemical, the ratio of zinc to copper is 4 to 13.

The chemical contains 546 grams of copper.

Prove  -

How many grams of zinc does it chemical contain.

Answer

suppose that scalar multiple of the  zinc and copper be y .

As given

In a certain chemical, the ratio of zinc to copper is 4 to 13.

The chemical contains 546 grams of copper.

Than equation is

13y = 546

13y = 546

y = 42

Zinc contain in the certain chemical = 4y

                                             = 4 42

                                             = 168 grams

Therefore 168 grams of zinc contain in a certain chemical .Step-by-step explanation:

Answer:

-6

Step-by-step explanation:

Step One: The goal is to isolate the x. So first, we would do 4-22, which is -18. The equation is: 3x=-18.

Step Two: Lastly, we need to divide by three to completely isolate the x. This is -6.

a) The probability that the team wins the championship is approximately 0.0141 or 1.41%. b) The probability that the team does not win the championship is approximately 0.9859 or 98.59%.

a) To determine the probability that the team wins the championship, we need to convert the odds against winning into a probability.

The odds against winning are given as 70:1. This means that for every 70 unfavorable outcomes (losing), there is 1 favorable outcome (winning).

To calculate the probability of winning, we divide 1 by the sum of the favorable and unfavorable outcomes:

Probability of winning = 1 / (70 + 1) = 1 / 71 ≈ 0.0141 (or 1.41%)

Therefore, the probability that the team wins the championship is approximately 0.0141 or 1.41%.

b) The probability of not winning the championship is equal to 1 minus the probability of winning:

Probability of not winning = 1 - 0.0141 = 0.9859 (or 98.59%)

Therefore, the probability that the team does not win the championship is approximately 0.9859 or 98.59%.

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Answer:

3 years old and 8 years old

Step-by-step explanation:

8-3=5

3x8=24

plz mark me as brainliest

The values of c₁ and c₂ that satisfy the initial values u(0) = 60 and u'(0) = 56 are:

c₁ = 148 / (3e²)

c₂ = (20 - 148/9)e⁴

The given solution, u(t) = c₁e² + c₂e⁻⁴, indeed satisfies the given ordinary differential equation (ODE) u"(t) + u'(t) - 12u(t) = 0. To find the values of c₁ and c₂ such that the solution satisfies the initial values u(0) = 60 and u'(0) = 56, we substitute these values into the solution.

First, let's find u(0) by substituting t = 0 into the solution:

u(0) = c₁e² + c₂e⁻⁴

Since u(0) = 60, we have:

60 = c₁e² + c₂e⁻⁴    (Equation 2)

Next, let's find u'(0) by differentiating the solution with respect to t and substituting t = 0:

u'(t) = 2c₁e² - 4c₂e⁻⁴

u'(0) = 2c₁e² - 4c₂e⁻⁴

Since u'(0) = 56, we have:

56 = 2c₁e² - 4c₂e⁻⁴    (Equation 3)

Now we have a system of two equations (Equations 2 and 3) with two unknowns (c₁ and c₂). We can solve this system to find the values of c₁ and c₂.

To do that, let's first divide Equation 3 by 2:

28 = c₁e² - 2c₂e⁻⁴

Next, let's multiply Equation 2 by 2:

120 = 2c₁e² + 2c₂e⁻⁴

Adding the two equations, we get:

148 = 3c₁e²

Dividing both sides by 3e², we find:

c₁ = 148 / (3e²)

Substituting this value of c₁ back into Equation 2, we can solve for c₂:

60 = (148 / (3e²))e² + c₂e⁻⁴

60 = 148/3 + c₂e⁻⁴

60 - 148/3 = c₂e⁻⁴

20 - 148/9 = c₂e⁻⁴

c₂ = (20 - 148/9)e⁴

Therefore, the values of c₁ and c₂ that satisfy the initial values u(0) = 60 and u'(0) = 56 are:

c₁ = 148 / (3e²)

c₂ = (20 - 148/9)e⁴

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Therefore, the number of bacteria after 200 minutes can be found by multiplying the initial number of bacteria (450) by the tripling factor (3) raised to the power of 2 (for the full cycles) and multiplying by the remaining fraction of the tripling factor for the partial cycle.

Since the bacteria triples every 3 hours, we can calculate the number of tripling cycles that have occurred in 200 minutes. Since 3 hours is equivalent to 180 minutes, there are 2 full cycles in 200 minutes. To calculate the remaining fraction of the tripling cycle, we divide the remaining time (20 minutes) by the length of a single cycle (180 minutes). The fraction is 20/180, which simplifies to 1/9.

Now, we can calculate the number of bacteria after 200 minutes. We start with the initial number of bacteria, which is 450, and multiply it by the tripling factor (3) raised to the power of the number of full cycles (2). This accounts for the full cycles. Then, we multiply this result by the remaining fraction of the tripling factor (1/9) to account for the partial cycle.

Therefore, the number of bacteria after 200 minutes can be calculated as follows:

Number of bacteria = 450 * (3^2) * (1/9) = 450 * 9 * (1/9) = 450

Hence, after 200 minutes, there will still be 450 bacteria.

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The answer is A, 3.9a + b

Step-by-step explanation:

youre welcome

Answer: b

Step-by-step explanation: When you rotate it around the axis (think of it as a pole) it will become a cylinder with radius 5

Answer:

C

Step-by-step explanation:

4(3n+2b+k)

12n+8b+4k

So it would be C.

---

hope it helps

Answer:

[tex]\frac{b^2 - 4ac}{4a^2}[/tex]

Step-by-step explanation:

Given

See attachment for complete question

Required:

Complete step 6

At step 5, we have:

[tex](x + \frac{b}{2a})^2 = \frac{b^2}{4a^2} - \frac{4ac}{4a^2}[/tex]

Take LCM

[tex](x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}[/tex]

Take square roots of both sides to get step 6

[tex]x + \frac{b}{2a} = \±\sqrt{\frac{b^2 - 4ac}{4a^2}}[/tex]

Hence, the missing radicand is: [tex]\frac{b^2 - 4ac}{4a^2}[/tex]