True or False (50 question)
_____ 1. The sum of the probabilities can give one and sometimes exceed one

_____ 2. The alpha value indicates the % of error allowed in the investigation

_____ 3. The critical value is obtained from the formulas applied to each Test

_____ 4. H0 is rejected when the manual value is greater than the critical value

_____ 5. All probability is between zero and 1

_____ 6. The ANOVA Test uses the entire bell

_____ 7. There are 2 types of hypotheses

_____ 8. The null hypothesis may posit that there is no significant difference between the statistic and the parameter or between 2 parameters

_____ 9. When α =.05 then you allow 95% error in the study

____ 10. The critical value is where the rejection zone for H0 begins

. ____ 11. In the binomial distribution r can be greater than n.

____ 12. If H0 ≤ 30 then H1 cannot be determined

____ 13. In the normal distribution the probability is in agreement to the total deviations you find

____ 14. In an incompatible event the probabilities must all give to one

____ 15. In Classical Probability the sample space is always known.

____ 16. In Binomial when n =30, the Pr can give greater than one.
____ 17. All probability is in Pr ≤ 1 and zero
____ 18. In the Binomial Distribution the Pr (0) is part of the probabilities.
____ 19. In Poison Distribution, N is known and its average N is used.
____ 20. Incompatible events are mutually exclusive.
____ 21. The [Pr (Oc) + Pr (Not Oc)] can give greater than one, sometimes.
____ 22. Incompatible events can be seen in classical probability
______23. The variable "eye color" is a qualitative variable, nominal
______24. The "scatter diagram" illustrates the relationship between 2 variables.

______25. In the range of grouped data you subtract the extreme values.

______26. The independent variable is manipulated by the researcher.
______27. In Poison, with lambda = 20, then Pr (X >1) = 1 – [Pr (0, 1)]
______28. In the Binomial distribution, with n=10, the Pr (r > 1) = 1 –[Pr (0, 1)]
______29. The Pr (Oc) + Pr (Not Oc) = 1
______30. In Normal distribution the sum of the Probabilities equals one.
_______31. In the Binomial Distribution r can be greater than n.
_______32. The bar chart is used to illustrate the relationship of nominal qualitative variables
_______33. In Binomial Distribution, with n =25, the sum of all Pr = 1
_______3. 4. In the Poison Distribution, with lambda = 30, the Pr (X = 31) is > 1
_______35. The null and alternate hypotheses are mutually exclusive.
_______36. Let a and b be independent events, Pr(No Oc) = 1 –[Pr(a) Pr (b)].
_______37. Let A and B be Dependent Events, Pr(A)Pr (B/A) = Pr(B)Pr (A/B)
_______38. The Correlation Coefficient can be negative
_______39. The Scatter Plot illustrates the dispersion of the data
_______40. All Pr (Oc) give to one.
_______41. When Z is negative also the probability found
_______42. In Normal Dist when you have Pr(Z ≤ 3) = 1- [Pr (Z=3)]
_____43. The arithmetic average is sought by adding all the data/(n-1)
_____44. The location of the median is found by looking for (n + 1) / 2
_____45. Median of ungrouped data does not use outliers
_____46. In the Normal Dist Z = total of the events
_____47. When Z = 1, the probability = .5 - .3413
_____48. Variance in grouped data is divided by n + 1
_____49. If r > 1 then the variables are directly proportional
_____50. Every probability is a proper fraction

To calculate the required sample size in Excel for estimating the true proportion of satisfied customers within a 3% margin of error with 99% confidence,

In Excel, you can use the NORM.S.INV function to find the critical value corresponding to the desired confidence level. In this case, we use 1 - (1-0.99)/2 to find the z-score for 99% confidence level.

We square this z-score and multiply it by (0.50.5)/(0.030.03), which represents the maximum variance under worst-case scenario and the desired margin of error. The CEILING function is used to round up the result to the nearest whole number since the sample size should be an integer.

you can use the formula =CEILING(MROUND((NORM.S.INV(1-(1-0.99)/2,0)^2)(0.50.5)/(0.03*0.03),1),1). The result will give you the minimum sample size needed for the desired confidence level and margin of error.

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Given the set { 1 , 1 , 1 , 1 , 2 , 3 } . Picking a number at random.The probability of choosing a "1" is 2 / 3 . False

In the given set {1, 1, 1, 1, 2, 3}, there are four occurrences of the number "1" and a total of six elements in the set.

To calculate the probability of choosing a "1", we need to divide the number of favorable outcomes (the occurrences of "1") by the total number of possible outcomes (the total number of elements in the set).

The number of favorable outcomes is 4 (the occurrences of "1"), and the total number of possible outcomes is 6 (the total number of elements in the set).

So, the probability of choosing a "1" is 4/6, which simplifies to 2/3.

Therefore, the statement is true: the probability of choosing a "1" from the given set is 2/3.

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For n e N let an be the number of strings of 0's and l's,

(a) i. a₁ = 2, ii. a₂ = 1, iii. a₃ = 2, iv. a₄ = 3, v. a₅ = 5.

(b) The values of aₙ represent the nth term in the Fibonacci sequence.

(c) The set of strings of length n can be built by appending "1" or "01" to strings of length n-1 and n-2.

(d) The conjecture that the values of aₙ represent the nth term in the Fibonacci sequence is proven using mathematical induction.

(a) To find the values of aₙ, we'll calculate them one by one:

i. a₁: We have two possible strings of length 1 that satisfy the condition: "1" and "0". So a₁ = 2.

ii. a₂: For a string of length 2, the only valid option is "10". So a₂ = 1.

iii. a₃: Now, let's consider strings of length 3. We can build them by appending either "1" or "01" to valid strings of length 2. From the previous step, we know that a₂ = 1. Thus, we have two options: "101" and "100". So a₃ = 2.

iv. a₄: Similarly, for strings of length 4, we can append either "1" or "01" to valid strings of length 3. From the previous step, we know that a₃ = 2. Thus, we have three options: "1010", "1001", and "1000". So a₄ = 3.

v. a₅: Continuing the same pattern, we can append either "1" or "01" to valid strings of length 4. From the previous step, we know that a₄ = 3. Thus, we have five options: "10101", "10100", "10010", "10001", and "10000". So a₅ = 5.

(b) If we look closely at the values of aₙ that we calculated, we notice that they form the Fibonacci sequence: 2, 1, 2, 3, 5, ...

Conjecture: The values of aₙ represent the nth term in the Fibonacci sequence.

(c) To build the set of strings of length n from the sets of strings of length n-1 and n-2, we can append either "1" or "01" to the strings of length n-1 and n-2.

For example, to obtain a string of length 4, we can append "1" or "01" to the strings of length 3. Similarly, to obtain a string of length 5, we can append "1" or "01" to the strings of length 4.

(d) To prove our conjecture that the values of aₙ represent the nth term in the Fibonacci sequence, we can use mathematical induction. We would need to show that a₁ = F₁, a₂ = F₂, and assume that aₖ = Fₖ and aₖ₋₁ = Fₖ₋₁ hold for some k ≥ 2 and prove that aₖ₊₁ = Fₖ₊₁.

Since a₁ = 2 = F₁ and a₂ = 1 = F₂, the base cases hold.

Next, assume that aₖ = Fₖ and aₖ₋₁ = Fₖ₋₁ hold for some k ≥ 2.

From the construction in part (c), we know that to obtain a string of length k+1, we append "1" to a string of length k and append "01" to a string of length k-1.

So, aₖ₊₁ = aₖ + aₖ₋₁ = Fₖ + Fₖ₋₁.

Using the property of the Fibonacci sequence that Fₖ + Fₖ₋₁ = Fₖ₊₁, we can conclude that aₖ₊₁ = Fₖ₊₁.

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The derivative h'(4) is equal to 40.

To find h'(4), we can apply the chain rule. The chain rule states that if we have a composite function, we can find the derivative by taking the derivative of the outer function and multiplying it by the derivative of the inner function.

In this case, h(x) = 5 * 4f(x), so the outer function is multiplying by 5 and the inner function is f(x).

First, we find the derivative of the outer function, which is a constant multiple. The derivative of 5 is 0.

Next, we find the derivative of the inner function, f'(x). We are given that f'(4) = 2.

Now, we can apply the chain rule: h'(x) = 0 * f(x) + 5 * f'(x).

Substituting the given values, we have: h'(4) = 0 * f(4) + 5 * f'(4) = 0 * 5 + 5 * 2 = 0 + 10 = 10.

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Answer:

259 students

Step-by-step explanation:

First, find how many will check out science fiction books

350(0.14)

= 49

Find how many will check out sports books

350(0.12)

= 42

Subtract these amounts from 350

350 - 49 - 42

= 259

So, she should expect 259 students not to check out science fiction and sports books

Answer:

Olivia wrapped 6 gifts and prepared 8 giftbags.

Step-by-step explanation:

Write System of Equations:

3x+1.5y= 30                                                 x+y=14

x+y=  14                                                       y=-x+14

3x+1.5y=30

1.5y=3x+30

1.5y divided by 1.5 is y

-3x+30 divided by y is -2x+20

y=-2x+20

Olivia wrapped 6 gifts and made 8 Giftbags.

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

It demonstrates the equality of the relationship between the expressions printed on the left and right sides. LHS = RHS is a common mathematical formula.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

Given:

Olivia can wrap a single gift in 3 minutes and she can do a giftbag in 1.5 minutes

Olivia prepared a total 14 gifts in 30 minutes.

let number of gifts Olivia wrapped be x.

and, number of giftbags she prepared be y.

So, the equations are:

3x+1.5y= 30 .....(1)

x+y=14

Now, plotting these equations on graph we get intersecting point of two lines which shows x= 6 and y=8.

Hence, Olivia wrapped 6 gifts and made 8 Giftbags.

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Answer:

5

Step-by-step explanation:

2-5=5. Subscribe Single-3 FF on yt

The sample space of all possible pairs of marbles that can be drawn is:

{BB, BR, RB, RR, RR, RR}

It is the total number of possible outcomes from a given set.

We have,

The sample space is the set of all possible outcomes of the experiment.

In this case, we are drawing two marbles from a bag containing 2 blue and 5 red marbles, without replacement (meaning that we do not put the first marble back into the bag before drawing the second one).

The sample space consists of all possible pairs of marbles that can be drawn:

{BB, BR, RB, RR, RR, RR}

where BB means both marbles are blue, BR means the first marble is blue and the second is red, RB means the first marble is red and the second is blue, and RR means both marbles are red.

Note that there are two outcomes corresponding to drawing two red marbles because there are 5 red marbles in the bag.

Thus,

The sample space of all possible pairs of marbles that can be drawn is:

{BB, BR, RB, RR, RR, RR}

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The  complete question:

In an experiment, a bag contains 2 blue marbles and 5 red marbles. Two marbles are drawn from the bag.

List the sample space.

Since the slope of the tangent line on the graph of y=x is 1, any point on the line y=x can be considered as the point where the tangent line is parallel to -5x+2y=4.  The point on the graph of y=x where the tangent line is parallel to the line -5x+2y=4 is (2, 2).

To find the point on the graph of y=x where the tangent line is parallel to the line -5x+2y=4, we need to find the slope of the line -5x+2y=4 and equate it to the slope of the tangent line on the graph of y=x.

The given line -5x+2y=4 can be rewritten as 2y=5x+4, which implies y=(5/2)x+2. Comparing this equation with y=x, we see that the slopes of both lines are equal, which means the tangent line to y=x is parallel to the given line.

Since the slope of the tangent line on the graph of y=x is 1, any point on the line y=x can be considered as the point where the tangent line is parallel to -5x+2y=4. One such point is (2, 2), where x=2 and y=2.

Therefore, the point on the graph of y=x where the tangent line is parallel to -5x+2y=4 is (2, 2).

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option A is the correct statement regarding the exponential distribution.

What is Exponential Distribution?

The exponential distribution is a continuous probability distribution that models the time between events occurring at a constant average rate. It is often used to describe situations where events happen randomly and independently over time, such as the time between phone calls received at a call center or the time between arrivals of customers at a service counter.

The correct answer is:

[tex]\textbf{A. The exponential distribution has the property that its mean equals its variance.}[/tex]

The exponential distribution is characterized by the property that its mean is equal to its variance. This means that the average value of the distribution and the spread of the distribution are the same. In other words, the measure of central tendency and the measure of dispersion are equal in the exponential distribution.

This property holds true for all exponential distributions, regardless of the specific rate parameter or scale parameter. It is a fundamental characteristic of this distribution.

Therefore, option A is the correct statement regarding the exponential distribution.

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Answer:

1.8

Step-by-step explanation:

Answer:

x=22

Step-by-step explanation:

Rule states diagonals = same so 6x-36 = 96 in which x=22 cause 6(22)=132

132-36 = 96

Answer:

C. 50 kilograms

Step-by-step explanation:

For a normal house garden of moderate dimensions 5,000 kg is too much and 50 kg would be enough.

Answer: I'm hoping for it to be C.

Step-by-step explanation: I'll  finish the quiz then l'll let you know ;)

     Edit: GOT IT CORRECT!

      And don't mind that i got three wrong...

The journal entry that Amy Company will make to record the collection of cash from Tory Turnbull, who paid within the discount period, would include a credit to Cash for $160. Therefore, option d is the correct answer.

This is because Tory Turnbull will pay $8,000 - $160 (2% of $8,000) to avail the discount. The Sales Discount account is not involved in the journal entry as the discount was taken by the customer, not given by Amy Company.

Therefore, the correct answer is d. Credit to Cash of $160. This entry reflects the cash received by Amy Company and the reduction in the Accounts Receivable balance for the amount paid by the customer.

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Answer:

B. all of these are true

Step-by-step explanation:

The Square is a rhombus because all the sides of a square are equal in length. Even, the diagonals of both square and rhombus are perpendicular to each other and bisect the opposite angles. Therefore, we can say the square is a rhombus.

Hope this helped!!

Answer:

50.3

Step-by-step explanation:

area for circle formula=πr^2

r=4

π4^2 which is 50.3 rounded

Answer:

50.3

Step-by-step explanation:

A=πr^2

π·4^2≈50.3

Answer:

I am not sure , but i think the answer would be 0.03 m/s .

hope it helps u . ^.^

Answer:

The volume of the rectangular prism is 336

The volume of the triangular prism is 24.875

Add them together, 360.875. You can round if you want

Answer:answer is d

Step-by-step explanation:

I have done the test

The mean of Rt can be calculated as follows:

E[Rt] = E[2Yt - Yt-1] = 2E[Yt] - E[Yt-1] = 2(3) - 3 = 3.

and, the autocovariance function of Rt is given by:

yR(k) = 4y(k) - 2y(k+1) - 2y(k-1) + y(k).

The random variable Rt is defined as 2Yt - Yt-1, where {Yt} is a stationary process with a mean of 3 and autocovariance function y(k). To find the mean and autocovariance function of Rt, we can use the properties of linearity and time shift.

The mean of Rt can be calculated as follows:

E[Rt] = E[2Yt - Yt-1] = 2E[Yt] - E[Yt-1] = 2(3) - 3 = 3.

To find the autocovariance function of Rt, we need to consider the covariance between Rt at time t and Rt at time s for any t and s. Let's denote the autocovariance function of Yt as γ(k). Using the properties of linearity and time shift, we can express the autocovariance function of Rt as follows:

Cov(Rt, Rs) = Cov(2Yt - Yt-1, 2Ys - Ys-1)

= 4Cov(Yt, Ys) - 2Cov(Yt, Ys-1) - 2Cov(Yt-1, Ys) + Cov(Yt-1, Ys-1)

= 4γ(t-s) - 2γ(t-s+1) - 2γ(t-1-s) + γ(t-1-s+1)

= 4γ(t-s) - 2γ(t-s+1) - 2γ(t-s-1) + γ(t-s).

Therefore, the autocovariance function of Rt is given by:

yR(k) = 4y(k) - 2y(k+1) - 2y(k-1) + y(k).

Now, let's determine whether R is stationary. A process is considered stationary if its mean and autocovariance do not depend on time. In this case, the mean of Rt is constant and equal to 3, which satisfies the condition for stationarity. However, the autocovariance function yR(k) of Rt depends on the lag k. Therefore, Rt is not stationary because its autocovariance function varies with time.

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Answer:

Olivia invested $2900 in a savings account that paid 4.5%/yr and $900 in a government bond that paid 5.5%/yr.

Step-by-step explanation:

With the information provided you can write the following equations:

x+y=3800 (1)

0.045x+0.055y=180 (2), where:

x is the amount invested in a savings account

y is the amount invested in a government bond

First, you can solve for x in (1):

x=3800-y (3)

Next, you can replace (3) in (2) and solve for y:

0.045(3800-y)+0.055y=180

171-0.045y+0.055y=180

0.01y=9

y=9/0.01

y=900

Finally, you can replace the value of y in (3) to find x:

x=3800-900

x=2900

According to this, the answer is that Olivia invested $2900 in a savings account that paid 4.5%/yr and $900 in a government bond that paid 5.5%/yr.

The work done by the force field F in moving an object from P to Q is 1459/6.

The work done by a force field in moving an object from point P to point Q can be calculated using the line integral of the force field along the path from P to Q.

Given the force field F(x, y) = x^5i + y^5j and the points P(1, 0) and Q(3, 3), we need to calculate the line integral of F along the path from P to Q.

The line integral of a vector field F along a curve C is given by the formula:

∫C F · dr,

where F is the vector field, dr is the differential vector along the curve, and ∫C represents the line integral over the curve C.

In this case, the path from P to Q can be parameterized by a function r(t) = (x(t), y(t)), where t varies from 0 to 1. We can choose a linear parameterization for simplicity:

x(t) = 1 + 2t,

y(t) = 3t,

where t varies from 0 to 1.

Now, we can calculate the line integral:

∫C F · dr = ∫₀¹ F(x(t), y(t)) · r'(t) dt,

where r'(t) represents the derivative of r(t) with respect to t.

Substituting the values into the formula, we have:

∫₀¹ ( (1 + 2t)^5i + (3t)^5j ) · (2i + 3j) dt.

Simplifying the expression, we get:

∫₀¹ ( (1 + 2t)^5(2) + (3t)^5(3) ) dt.

Now, we can integrate term by term:

= ∫₀¹ ( 2(1 + 2t)^5 + 3^5t^5 ) dt.

= [ (2/6)(1 + 2t)^6 + (3/6)t^6 ] from 0 to 1.

Evaluating the expression at the limits, we have:

= (2/6)(1 + 2(1))^6 + (3/6)(1)^6 - (2/6)(1 + 2(0))^6 - (3/6)(0)^6.

= (2/6)(3^6) + (3/6) - (2/6)(1^6) - (3/6)(0).

= (2/6)(729) + (3/6) - (2/6) - 0.

= (2/6)(729 - 1) + (3/6).

= (2/6)(728) + (3/6).

= 728/3 + 3/6.

= 728/3 + 1/2.

= (1456 + 3)/6.

= 1459/6.

Therefore, the work done by the force field F in moving an object from P to Q is 1459/6.

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Answer:

if y=6x then 3x-6x=21

-3x=21

x=-7

3x - y = 21

y = 6x

We see that,

0 = 42 which is not true.

The system of equations has no solutions.

An equation is a mathematical statement that is made up of two expressions connected by an equal sign.

We have,

System of equations:

Conditions:

1 - If we get x = n (a value)

The system of equations has one solution

2 - If we get x = x or y = y or n = n

The system of equations has infinite solutions.

3 - If we get n ≠ 0

The system of equations has no solutions

Now,

3x - y = 21 _____(1)

y = 6x ______(2)

From (1) we get,

3x = 21 + y

x = (21 + y) / 3 _____(3)

Putting (3) in (2) we get,

y = 6 [(21 + y) / 3]

y = 2 (21 + y)

y = 42 + y

0 = 42

Which is not true.

This means the system of equations has no solutions.

Thus,

The system of equations has no solutions.

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Answer:

Domain:{-9,-4}

Range:{-1,-8,6}

Step-by-step explanation:

I dont really understand what you're trying to ask, it seems like you already have the answer i got.

Answer:

-Se plantarán 50.000 arboles.

-Si se planta un árbol cada 50 m, se plantarán 25.000 arboles.

Step-by-step explanation:

Para encontrar cuántos arboles se plantarán, tienes que dividir la longitud el paseo entre la distancia que habrá entre los árboles:

1250/0,025=50.000

Ahora, para encontrar cuántos arboles se plantarán si se planta uno cada 50m, primero tienes que encontrar cuántos kilómetros hay en 50 m considerando que 1 km es igual a 1000 metros. Para esto, puedes utilizar una regla de tres:

1 km → 1000 m

  x   ←   50 m

x=(50*1)/1000

x= 0,05 km

Ahora puedes dividir la longitud del paseo entre 0,05 para encontrar el número de arboles que se plantarán:

1250/0.05=25.000

De acuerdo a esto, la respuesta es que se plantarán 50.000 arboles.

Si se planta un árbol cada 50 m, se plantarán 25.000 arboles.

Answer:

V=pi x (r)^2 x h

V-volume r-radius h-height

V= pi x (3)^2 x 9

V= pi x 9 x 9

V= pi x 81

V= 254.47 (rounded)

Answer:

C: 108 cans

Step-by-step explanation:

just trust me bro

Answer:

so i'm guessing since 5 times 5 is 25 but there is a 2 on top of the number 25 that means squared meaning your're going to multiply. can i get a thanks and a rate!

Step-by-step explanation:

We can see from the statistical software printout that the factory recently installed new pollution control equipment ("scrubbers") on its smokestacks in hopes of reducing air pollution levels at a nearby national park.

Here, randomly timed measurements of sulfate levels (in micrograms per cubic meter) were taken before (Set C1) and after (Set C2) the installation. We believe that measurements of sulfate levels are normally distributed. A complete conclusion about the effectiveness of these scrubbers based on the statistical software printout shown is:The mean sulfate level for Set C1 is 8.29, and the mean sulfate level for Set C2 is 5.00.

From the Two Sample T for C1 vs C2 table, we find that the t-test statistic is 3.29, and the p-value is 0.0037. Because the p-value (0.0037) is less than the level of significance (α=0.05), we can reject the null hypothesis that the mean sulfate levels are equal before and after the installation of the scrubbers. Therefore, we can conclude that the installation of scrubbers on the factory's smokestacks has had a significant effect in reducing the sulfate levels in the air at the national park.

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Answer: 3

Step-by-step explanation:

Basically all you have to do is subtract 6 from both sides so you will get:

3 = v