a bag contains 3 red marbles, 8 blue marbles and 5 green marbles. If two marbles are drawn out of the bag, what is the probability, to the nearest 1000th. that both marbles drawn will be blue ​

Answer:

50

Step-by-step explanation:

Answer:

the answer is 200

Step-by-step explanation:

Answer:

A rectangle!!

Step-by-step explanation:

When you cut a square horizontally or vertically, you get two smaller triangles that are half the area of the square. When you split the rectangle again, you get a square 1/4 the size of the regular square.

Answer:

95

Step-by-step explanation:

66-66=0

0-29=-29

66+29=95

The range of the temperatures during this period is 95.

Given the following data:

To determine the range of the temperatures during this period:

Range is simply calculated by taking the difference between the highest number and the lowest number in a sample.

Mathematically, range is given by the formula;

[tex]Range = highest \;number -lowest \;number[/tex]

Substituting the given parameters into the formula, we have;

[tex]Range = 66-(-29)\\\\Range =66+29[/tex]

Range = 95

Therefore, the range of the temperatures during this period is 95.

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The equation p-20=-30 can be rewritten as an addition problem by adding 20 to both sides of the equation. The resulting addition equation is p = -30 + 20.

The equation p-20=-30 can be written as an addition problem by moving -20 to the other side of the equation. This is done by adding 20 to both sides of the equation, which results in the equation p = -30 + 20. In other words, you can add -30 and 20 together to find the value of p. This is how you write the equation p-20=-30 as an addition problem. So the equation becomes p= -30+20

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Answer:

A- square

Step-by-step explanation:

A square should have 4 right angles.

Answer:

A) Confidence Interval will become narrower. B) Confidence Interval will become narrower. C) Confidence Interval will become broader.

Step-by-step explanation:

Confidence Interval is the probable range around sample statistic, in which the population parameter is expected to lie.

Confidence Level shows the average percentage level of confidence interval, expected to contain population parameter. Lower confidence level implies narrower Confidence Interval

Bigger sample size reduces margin error (sample statistic, population parameter difference). Parameter-statistic proximity implies: narrower confidence interval around statistic, expected to contain parameter.

Standard Deviation is a measure of dispersion, spread. So, higher standard deviation implies more spread & broader confidence interval.

Answer:

1 / 15

Step-by-step explanation:

1/3 / 5 =

1/ 15

Final answer:

Each person will get 1/15 of a pound of peanuts when a 1/3 pound bag is shared evenly among 5 people.

Explanation:

To figure out how much each of the 5 people should get from a 1/3 pound bag of peanuts, we need to divide the total weight of the peanuts by the number of people. This gives us:

1/3 pound ÷ 5 = 1/15 pound per person.

This means each person will get 1/15 of a pound of peanuts. In other calculations such as the candy survey, determining percent uncertainty, or unit conversions as in Michaela's party scenario, a similar process of division or unit conversion is applied to find the answer.

Answer:

10% probability that a given class period runs between 51.25 and 51.75 minutes.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability of finding a value X between c and d, d greater than c, is given by the following formula:

[tex]P(c \leq X \leq d) = \frac{d-c}{b-a}[/tex]

Uniformly distributed between 49 and 54 minutes

This means that [tex]b = 54, a = 49[/tex]

Find the probability that a given class period runs between 51.25 and 51.75 minutes.

[tex]P(51.25 \leq X \leq 51.75) = \frac{51.75 - 51.25}{54 - 49} = 0.1[/tex]

10% probability that a given class period runs between 51.25 and 51.75 minutes.

Answer:

Step-by-step explanation:

Hello!

You have the output:

Two-Sample T-Test and Cl

Sample N Mean StDev SE Mean

1(M) 112 7.38 6.95 0.66

2(F) 101 7.15 6.31 0.63

Difference = mu (1) - mu (2)

Estimate for difference: 0.230000

95% lower bound for difference: -1.271079

T-Test of difference: T-value = 0.25 P-Value = 0.400 DF= 210

This output summarizes the information of the two samples and indicates the order the populations where studied.

It also informs you of the value of the statistic under the null hypothesis and the p-value.

Unfortunately, there is no information on the type of hypotheses that were tested, i.e. if they where two-tailed or one-tailed, in the latter case, there is no information if it was left-tailed or right-tailed). Likewise is not specified if the test was done for a specific value of the parameter. (for example μ₁ - μ₂ = 0 or μ₁ - μ₂ = θ₀)

For these reasons, the data provided by the output isn't enough to conclude any hypothesis.

From all the provided answers the one more likely to be correct is:

(iii) The data do not provide sufficient evidence to conclude that the mean depression score of male teens is larger than that of female teens.

I hope this helps!

Answer:

iv

Step-by-step explanation:

Since the p-value (0.4) is greater than the significance level (0.05), we can conclude that the result is not significant. This means that there is no enough statistical evidence to reject the null hypothesis H0. Therefore, we must accept it and conclude that the mean depression score for male and female teens do not differ.

Answer:

c

Step-by-step explanation:

Answer:

4

Step-by-step explanation:

Question- what is two plus two

Answer- 2+2=4

Answer:

0.0879 is the probability that out of 5 randomly selected consumers, three are comfortable with delivery by drones.        

Step-by-step explanation:

We are given the following information:

We treat drone deliveries as a success.

P(consumers comfortable having drones deliver) = 25% = 0.25

Then the number of consumers follows a binomial distribution, where

[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]

where n is the total number of observations, x is the number of success, p is the probability of success.

We have to evaluate:

P(Exactly 3 customers out of 5 are comfortable with delivery by drones)

Here,

[tex]n = 5\\x = 3\\p = 0.25\\q = 1 - p = 1-0.25=0.75[/tex]

Putting values, we get,

[tex]P(x =3)\\\\= \binom{5}{3}(0.25)^3(1-0.25)^2\\\\= 0.0879[/tex]

0.0879 is the probability that out of 5 randomly selected consumers, three are comfortable with delivery by drones.

Answer:

[tex]t=\frac{25.02-26}{\frac{4.83}{\sqrt{50}}}=-1.435[/tex]    

[tex]p_v =P(t_{(49)}<-1.435)=0.0788[/tex]  

If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its significant lower than 26 so then the specification is satisfied.

Step-by-step explanation:

Data given and notation  

[tex]\bar X=25.02[/tex] represent the sample mean

[tex]s=4.83[/tex] represent the sample standard deviation

[tex]n=50[/tex] sample size  

[tex]\mu_o =26[/tex] represent the value that we want to test

[tex]\alpha[/tex] represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is at least 26 mpg, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \geq 26[/tex]  

Alternative hypothesis:[tex]\mu < 26[/tex]  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

[tex]t=\frac{25.02-26}{\frac{4.83}{\sqrt{50}}}=-1.435[/tex]    

P-value

The first step is calculate the degrees of freedom, on this case:  

[tex]df=n-1=50-1=49[/tex]  

Since is a one sided test the p value would be:  

[tex]p_v =P(t_{(49)}<-1.435)=0.0788[/tex]  

Conclusion  

If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the height of men actually its significant lower than 26 so then the specification is satisfied.

Using the t-distribution, it is found that since the test statistic is more than the critical value for the left-tailed test, this is not strong evidence that they have failed to attain their fuel economy goal.

At the null hypothesis, it is tested if the fleet average is of at least 26 miles per gallon, that is:

[tex]H_0: \mu \geq 26[/tex]

At the alternative hypothesis, it is tested if it is less, that is:

[tex]H_1: \mu < 26[/tex]

We have the standard deviation for the sample, hence, the t-distribution is used to solve this question.

The test statistic is given by:

[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]

The parameters are:

For this problem, the values of the parameters are: [tex]\overline{X} = 25.02, \mu = 26, s = 4.83, n = 50[/tex].

Hence, the value of the test statistic is:

[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]

[tex]t = \frac{25.02 - 26}{\frac{4.83}{\sqrt{50}}}[/tex]

[tex]t = -1.43[/tex]

The critical value for a left-tailed test, as we are testing if the mean is less than a value, with a significance level of 0.05 and 50 - 1 = 49 df is of [tex]t^{\ast} = -1.677[/tex]

Since the test statistic is more than the critical value for the left-tailed test, this is not strong evidence that they have failed to attain their fuel economy goal.

To learn more about the t-distribution, you can take a look at https://brainly.com/question/13873630

Answer:

r = w/t

Step-by-step explanation:

To solve this problem, we need to isolate the variable r, or get it alone by itself on the left side of the equation. To do this, we must get rid of the t. Since the r and t are multiplied together, we must divide by t to get the r alone. This is modeled below:

rt = w

rt/t = w/t

r = w/t

Hope this helps!

Answer:

-0.03

Step-by-step explanation:

Answer:

-.03 hope this helps

Answer:

Check the explanation

Step-by-step explanation:

Ans=

A: For m = 5: P(³≥1) = 1 – P(³=0) = 1 – 0.9973^5 = 0.0134

M = 10: 1 – 0.9973^10 = 0.0267

M = 20: 1 – 0.9973^20 = 0.0526

M = 30: 1 – 0.9973^30 = 0.0779

M = 50: 1 – 0.9973^50 = 0.126

18)

Ans=

Going by the question and the explanation above, we derived sample values of the mean as well as standard deviation in calculating our probability, since that is the necessary value in determining the probability of an out-of-bounds point being plotted. Furthermore, we would know that that value for the possibility would likely be a poor es²ma²on, cas²ng doubt on anycalcula²ons we made using those values

Answer:

Step-by-step explanation:

Hello!

The variable of interest is

X: the amount of coffee a person drinks per day. (ounces)

This variable has a normal distribution with μ= 15 ounces and σ=4.5 ounces.

10 randomly selected persons where surveyed.

a) X~N(μ;σ²) ⇒ X~N(15;20.25)

b) The sample means has the same distribution as the original variable except that the population variance is affected by the sample size:

X[bar]~N(μ;σ²/n) ⇒ X[bar]~N(15;20.25/10) ⇒ X[bar]~N(15;2.025)

c)

You need to find the probability that one random person drinks between 14.5 and 16.5 ounces of coffee per day, symbolically:

P(14.5 ≤ X ≤ 16.5)

For this part, since you need to calculate the probability that one person drinks between 14.5 and 16.5 ounces of coffee daily out of the whole population of people that drink coffee, you have to work using the distribution of the variable:

X~N(15;20.25)

To reach the correspondent probability you have to first standardize both bonds of the interval using Z=(X-μ)/δ~N(0;1):

P(14.5 ≤ X ≤ 16.5)

P(X ≤ 16.5) - P(X ≤ 14.5)

P(Z≤ (16.5-15)/4.5) - P(Z≤ (14.5-15)/4.5)

P(Z≤ 0.33) - P(Z≤ -0.11)= 0.62930 - 0.45620= 0.1731

d)

In this item, you need to calculate the probability that one out of the ten people surveyed drinks on average between 14.5 and 16.5 ounces of coffee.

P(14.5 ≤ X[bar] ≤ 16.5)

To calculate the correspondent probability you have to work using the distribution od the sample mean: X[bar]~N(15;2.025) and the standard normal: Z=(X[bar]-μ)/(δ/√n)~N(0;1)

P(X[bar] ≤ 16.5) - P(X[bar] ≤ 14.5)

P(Z≤ (16.5-15)/1.42) - P(Z≤ (14.5-15)/1.42)

P(Z≤ 1.06) - P(Z≤ -0.35)= 0.85543 - 0.36317= 0.49226

e)

Yes. Without the assumption that the variable had a normal distribution, you wouldn't be able to use the standard normal distribution to calculate the asked probabilities.

There is a possibility to apply the Central Limit Theorem to approximate the sampling distribution to normal if the sample was large enough (n ≥30) which is not the case.

f)

You need to find the IQR for the average consumption of coffee for a sample of 10 persons.

For this, you have to work using the distribution of the sample mean X[bar]~N(15;2.025)

The 1st quartile is the value that divides the bottom 25% of the distribution from the top 75%, symbolically:

P(X[bar]≤x[bar]₀)= 0.25

First, you have to determine the value of Z that accumulates 0.25 of probability:

P(Z≤z₀)= 0.25

z₀= -0.674

Now using the formula Z=(X[bar]-μ)/(δ/√n), you have to "reverse" the standardization, i.e. clear the value of X[bar]:

z₀=(x[bar]₀-μ)/(δ/√n)

-0.674=(x[bar]₀-15)/1.42

-0.674*1.42= (x[bar]₀-15)

x[bar]₀= (-0.674*1.42)+15

1st Quartile x[bar]₀= 14.0429 ounces

The third quartile is the value that divides the bottom 75% of the distribution from the top 25% of it:

P(X[bar]≤x[bar]₀)= 0.75

As before, the first step is to determine the corresponding value of Z:

P(Z≤z₀)= 0.75

z₀= 0.674

z₀=(x[bar]₀-μ)/(δ/√n)

0.674= (x[bar]₀-15)/1.42

0.674*1.42= x[bar]₀-15

x[bar]₀= (0.674*1.42)+15

3rd Quartile x[bar]₀= 15.9571

The IQR is the difference between Q3 and Q1:

IQR= 15.9571 - 14.0429= 1.9142

I hope this helps!

Answer:

each friend would get 1/6 when you divide 1/2 by 3

Step-by-step explanation:

Each friend receives 1/6 of the pie.

The student asked how much pie each of three friends would get if they share 1/2 of a blueberry pie. To find the answer, you simply divide the half of the pie by three, since there are three friends sharing it. That is, 1/2 ÷ 3.

You can think of this division as splitting the 1/2 of the pie into three equal parts. Each friend would get 1/6 of the pie, because 1/2 ÷ 3 = 1/6. This intuitively makes sense because if a pie is cut into six equal slices, three slices (which is half the pie) can be shared by three friends, giving each one slice.

Answer:

attached

Step-by-step explanation:

attached

Answer:

[tex]t=\frac{15.1-12.8}{\frac{6.8}{\sqrt{31}}}=1.883[/tex]  

[tex]p_v =P(t_{30}>1.883)=0.0347[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12.8 at 5% of signficance and the claim makes sense.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X=15.1[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

s=6.8 represent the sample standard deviation

n=31 represent the sample size  

Solution to the problem

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if  the pilots claim that they are not slowing down, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 12.8[/tex]  

Alternative hypothesis:[tex]\mu > 12.8[/tex]  

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

[tex]t=\frac{15.1-12.8}{\frac{6.8}{\sqrt{31}}}=1.883[/tex]  

P-value  

The degrees of freedom are given by:

[tex] df = n-1=31-1=30[/tex]

Since is a right tailed test the p value would be:  

[tex]p_v =P(t_{30}>1.883)=0.0347[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12.8 at 5% of signficance and the claim makes sense.

Answer:

28x7=196 (Pls give Brainliest)

Answer:

[tex] \huge \pi \: units [/tex]

Step-by-step explanation:

[tex]l = \frac{48 \degree}{360 \degree} \times 2\pi \: r \\ \\ = \frac{2}{24} \times 2 \times \pi \times 6 \\ \\ = \frac{24}{24} \times \pi \\ \\ = \pi \: units \\ [/tex]

Answer:

It was hypothesized that more people would choose the number 7 as their 'lucky' number than any other number.

Step-by-step explanation:

Given that one variable chi-square is used to test whether a single categorical variable follows a hypothesized population distribution. The Chi Square statistic compares the tallies or counts of categorical responses between two (or more) independent groups

The null hypothesis (H0) for the test is that all proportions are equal.

The alternate hypothesis (H1) is given condition in the question.

A. It was hypothesized that more people would choose the number 7 as their 'lucky' number than any other number.

This is suited for testing with a one-variable chi-square test because we are testing if the proportion of people who choose number 7 is greater than the proportion of any other numbers. So, we are therefore comparing more than 2 proportions.

B. People who choose the number 7 as their 'lucky' number are significantly more superstitious than people who choose the number 13 as their 'lucky' number.

This is not suited for testing with a one-variable chi-square test. A z test is more preferable in this instance because we are testing just two proportions.

C. Choice of 'lucky' number is directly related to measures superstition.

This is not suited for testing with a one-variable chi-square test because chi square test is not used for showing relationship between variables.

D. All of these. Since option A is correct, this option can not be correct.

Let A be some subset of a universal set U. The "complement of A" is the set of elements in U that do not belong to A.

For example, if U is the set of all integers {..., -2, -1, 0, 1, 2, ...} and A is the set of all positive integers {1, 2, 3, ...}, then the complement of A is the set {..., -2, -1, 0}.

Notice that the union of A and its complement make up the universal set U.

In this case,

U = {1, 2, 3, 6, 10, 13, 14, 16, 17}

The set {3, 10, 16} is a subset of U, since all three of its elements belong to U.

Then the complement of this set is all the elements of U that aren't in this set:

{1, 2, 6, 13, 14, 17}

The complement of a set A with respect to a set B includes elements in B that are not in A. The complement of the set {3,10,16} within {1,2,3,6,10,13,14,16,17} is {1,2,6,13,14,17}. The process involves removing elements in A from B.

In mathematics, specifically in set theory, the complement of a set A, with respect to a set B, refers to the elements in set B that are not in set A. Let's use this definition to find the complement of {3,10,16} in the universal set U={1,2,3,6,10,13,14,16,17}.

First, list all the elements in U. Next, remove those elements which appear in the set {3,10,16}. The remaining elements constitute the complement of {3,10,16} given U. With this procedure, we find that the complement of {3,10,16} with respect to U is {1,2,6,13,14,17}. This method can be applied to any sets within a given universal set to find their complements.

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Answer: 2925 feet a day

Step-by-step explanation:

you gotta multiply 975 x 3 and you gonna 2925.

Answer:

The probability that the number of correct answers is 4 is 0.2461.

Step-by-step explanation:

Let X = number of correct answers.

The probability that an answer is correct is,P (X) = p = 0.50.

The total number of questions is, n = 9.

The event of an answer being correct is independent of the other answers.

The success of each trial is defined as a correct answer with equal probability of success for each trial, i.e. 0.50.

The random variable X follows a Binomial distribution with parameter n = 9 and p = 0.50.

The probability mass function of X is:

[tex]P(X=x)={9\choose x}\times0.50^{x}\times (1-0.50)^{9-x};\ x=0,1,2,3...[/tex]

Compute the value of P (X = 4) as follows:

[tex]P(X=4)={9\choose 4}\times(0.50)^{4}\times (1-0.50)^{9-4}[/tex]

                [tex]=126\times 0.0625\times 0.03125\\=0.24609375\\\approx 0.2461[/tex]

Thus, the probability that the number of correct answers is 4 is 0.2461.

Answer:

Step-by-step explanation:

The mean of the set of data given is

Mean = (29.7 + 29.4 + 31.7 + 29.0 + 29.1 + 30.5 + 29.1 + 29.8)/8 = 29.788 = $29788

Standard deviation = √(summation(x - mean)/n

n = 8

Summation(x - mean) = (29700 - 29788)^2 + (29400 - 29788)^2 + (31700 - 29788)^2 + (29000 - 29788)^2 + (29100 - 29788)^2 + (30500 - 29788)^2 + (29100 - 29788)^2 + (29800 - 29788)^2 = 5888752

Standard deviation = √(5888752/8) = 857.96

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ < 30000

For the alternative hypothesis,

µ > 30000

It is a right tailed test

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 8,

Degrees of freedom, df = n - 1 = 8 - 1 = 7

t = (x - µ)/(s/√n)

Where

x = sample mean = 29788

µ = population mean = 30000

s = samples standard deviation = 857.96

t = (29788 - 30000)/(857.96/√8) = - 0.7

We would determine the p value using the t test calculator. It becomes

p = 0.253

Since alpha, 0.05 < than the p value, 0.253, then we would fail to reject the null hypothesis. Therefore, At a 5% level of significance, the sample data did not show evidence that the mean wedding cost is not less than $30,000 as advertised