A publication released the results of a study of the evolution of a certain mineral in the​ Earth's crust. Researchers estimate that the trace amount of this mineral x in reservoirs follows a uniform distribution ranging between 55 and 1010 parts per million

a. Find E(x) and interpret its value
b. Compute P(2.875 x35)
c. Computn Plx<4.125)

Final answer:

To find the probability of drawing two marbles without replacement, calculate the probability of each event separately and multiply them together.

Explanation:

To find the probability of drawing two marbles without replacement, we need to calculate the number of favorable outcomes and divide it by the total number of possible outcomes.

1. Both marbles are red:
First, we calculate the probability of drawing a red marble on the first draw (3 red marbles out of 6 total marbles). Then, we calculate the probability of drawing a red marble on the second draw, given that the first marble drawn was red (2 red marbles out of 5 total marbles remaining). The probability is (3/6) * (2/5) = 1/5.

2. Both marbles are yellow:
First, we calculate the probability of drawing a yellow marble on the first draw (2 yellow marbles out of 6 total marbles). Then, we calculate the probability of drawing a yellow marble on the second draw, given that the first marble drawn was yellow (1 yellow marble out of 5 total marbles remaining). The probability is (2/6) * (1/5) = 1/15.

Answer:

Null hypothesis:[tex]p=0.114[/tex]  

Alternative hypothesis:[tex]p \neq 0.114[/tex]  

[tex]z=\frac{0.152 -0.114}{\sqrt{\frac{0.114(1-0.114)}{850}}}=3.49[/tex]  

Since is a bilateral test the p value would be:  

[tex]p_v =2*P(z>3.49)=0.00049[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis.

[tex]0.152 - 1.96 \sqrt{\frac{0.152(1-0.152)}{850}}=0.128[/tex]

[tex]0.152 - 1.96 \sqrt{\frac{0.152(1-0.152)}{850}}=0.176[/tex]

And the 95% confidence interval would be given (0.128;0.176).

And support the conclusion obtained on the hypothesis test since the value of 0.114 is not in the confidence interval, so we have enough evidence to reject the null hypothesis.

Step-by-step explanation:

Data given and notation

n=850 represent the random sample taken

X=129 represent the number of housing units that are vacant.

[tex]\hat p=\frac{129}{850}=0.152[/tex] estimated proportion of housing units that are vacant.

[tex]p_o=0.114[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

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

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is equal is 0.114 or not:  

Null hypothesis:[tex]p=0.114[/tex]  

Alternative hypothesis:[tex]p \neq 0.114[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.152 -0.114}{\sqrt{\frac{0.114(1-0.114)}{850}}}=3.49[/tex]  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed is [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

[tex]p_v =2*P(z>3.49)=0.00049[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis.

Confidence interval

The confidence interval would be given by this formula

[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.

[tex]z_{\alpha/2}=1.96[/tex]

And replacing into the confidence interval formula we got:

[tex]0.152 - 1.96 \sqrt{\frac{0.152(1-0.152)}{850}}=0.128[/tex]

[tex]0.152 - 1.96 \sqrt{\frac{0.152(1-0.152)}{850}}=0.176[/tex]

And the 95% confidence interval would be given (0.128;0.176).

And support the conclusion obtained on the hypothesis test since the value of 0.114 is not in the confidence interval, so we have enough evidence to reject the null hypothesis.

We first set the null and alternative hypothesis and then conduct a z-test for proportions to calculate the z-score and subsequently the p-value. We use the p-value to decide whether to reject the null hypothesis. Finally, we construct a 95% confidence interval for the true proportion of vacant houses and check if this supports our test conclusion.

Firstly, define the proportion of vacant housing units in the country as p0 and in the randomly selected county as p. The null hypothesis (H0) states that the county isn't different, so H0: p = p0 = 0.114. The alternative hypothesis (Ha) would be that the county is different, so Ha: p ≠ 0.114.

Let's conduct a Test of Hypothesis using a z-test for proportions. The z-score is calculated as (p - p0) / sqrt((p0 * (1 - p0)) / n), where n represents the sample size. Substituting in your values, the z score will be calculated. This z-score can be used to find the p-value from a standard normal (Z) distribution table.

If the p-value is less than 0.05 (which is α, significance level), we reject the null hypothesis in favor of alternative hypothesis, else we do not reject the null hypothesis. Thus, our conclusion is formulated based on this p-value.

To construct a 95% confidence interval for the true proportion of vacant houses, we use the formula: p ± Z * sqrt((p * (1 - p)) / n). Here, Z will be the Z score corresponding to the desired confidence level, 95% (which is 1.96 for two-tailed test).

If the national proportion (0.114) doesn't lie within this interval, it supports our test conclusion of rejecting the null hypothesis and vice versa.

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Answer:

The variance of given sample is 33.64 square pounds.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 16

Standard deviation, s = 5.8 pounds

We have to find the variance of the given sample.

Variance is the square of the standard deviation.

[tex]\text{Variance} = (\text{Standard Deviation})^2\\= (5.8\text{ pounds})^2\\=33.64\text{ pound}^2[/tex]

Thus, the variance of given sample is 33.64 square pounds.

Answer: 29%

Step-by-step explanation:

I considered the question to be: for what proportion of cars was the number of occupants more than 1 standard deviation and greater than the mean.

The first step is to calculate the weighted mean of the number of occupants in the cars. The next step is to determine the standard deviation using the formula √[ Σ ( xi - μ )² / N ]. Subsequently, identify the number of occupants that are more than one standard deviation greater than the mean.

In this question, we're asked to calculate the proportion of cars that had more than one standard deviation above the mean number of occupants. First, we need to calculate the weighted mean (average) of the number of occupants in the cars. Based on the data, the mean can be calculated as:

Mean = (1*71 + 2*16 + 3*8 + 4*3 + 5*2) / 100 = 1.66

Next, we find the standard deviation. The standard deviation tells us how spread out the numbers are from the mean. Calculating standard deviation is a bit involved. The formula is √[ Σ ( xi - μ )² / N ]. Once you have these, look at the number of occupants that are more than one standard deviation greater than this mean.

Note: This is a statistical analysis question that requires knowledge of the concepts of mean and standard deviation.

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Answer:

Step-by-step explanation:

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: The distribution of severity of psoriasis cases at the end and prior are same.

Alternative hypothesis: The distribution of severity of psoriasis cases at the end and prior are different.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.

Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = k - 1 = 4 - 1

D.F = 3

(Ei) = n * pi

Category            observed Num      expected num      [(Or,c -Er,c)²/Er,c]

Remission             380                         20                           6480

Mild

symptoms               520                         770                       81.16883117

Moderate

symptoms                 95                         160                         24.40625

Severe

symptom                  5                             50                          40.5

Sum                          1000                       1000                       6628.075081

Χ2 = Σ [ (Oi - Ei)2 / Ei ]

Χ2 = 6628.08

Χ2Critical = 7.81

where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, Ei is the expected frequency count for level i, Oi is the observed frequency count for level i, and Χ2 is the chi-square test statistic.

The P-value is the probability that a chi-square statistic having 3 degrees of freedom is more extreme than 6628.08.

We use the Chi-Square Distribution Calculator to find P(Χ2 > 19.58) =less than 0.000001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we cannot accept the null hypothesis.

We reject H0, because 6628.08 is greater than 7.81. We have statistically significant evidence at alpha equals to 0.05 level to show that distribution of severity of psoriasis cases at the end of the clinical trial for the sample is different from the distribution of the severity of psoriasis cases prior to the administration of the drug suggesting the drug is effective.

The chi-square test is a statistical method that determines if there's a significant difference between observed and expected frequencies in different categories, such as symptom status in this clinical trial. Without post-treatment numbers, we can't run the exact test. However, if the test statistic exceeded the critical value, we could conclude that the drug significantly affected symptom statuses.

This question pertains to the use of a chi-squared test, which is a statistical method used to determine if there's a significant difference between observed frequencies and expected frequencies in one or more categories. For this case, the categories are the symptom statuses (remission, mild, moderate, and severe).

To conduct a chi-square test, you first need to know the observed frequencies (the initial percentages given in the question) and the expected frequencies (the percentages after treatment). As the question doesn't provide the numbers after treatment, I can't perform the exact chi-square test.

If the post-treatment numbers were provided, you would compare them to the pre-treatment numbers using the chi-squared formula, which involves summing the squared difference between observed and expected frequencies, divided by expected frequency, for all categories. The result is a chi-square test statistic, which you would then compare to a critical value associated with a chosen significance level (commonly 0.05) to determine if the treatment has a statistically significant effect.

To interpret a chi-square test statistic, if the calculated test statistic is larger than the critical value, it suggests that the drug made a significant difference in the distribution of symptom statuses. If not, we can't conclude the drug was effective.

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Answer:

- 21 x + 24 y + 15 z =120

Step-by-step explanation:

Given that

Po(3,-2,5), Qo (-3,-1,-5), and Ro (0,-4,4) ,These are the point in the space.

We know that equation of a plane is given as

[tex]\begin{vmatrix}x-x_1 & y-y_1 &z-z_1 \\ x_2-x_1 & y_2-y_1 &z_2-z_1 \\ x_3-x_1 &y_3-y_1 & z_3-z_1\end{vmatrix}=0\\[/tex]

[tex]\begin{vmatrix}x-0 & y+4 &z-4 \\ 3-0 & -2+4 &5-4 \\ -3-0 &-1+4 & -5-4\end{vmatrix}=0.[/tex]

[tex]\begin{vmatrix}x & y+4 &z-4 \\ 3 & 2 &1 \\ -3 &3 & -9\end{vmatrix}=0.[/tex]

Now by solving above determinate we get

x( -18 -3 ) -(y+4 ) ( -27 +3 ) + ( z- 4) (9+6) = 0

-21 x +24 y -24 x 4 + 15 z - 24 = 0

- 21 x + 24 y + 15 z -120 = 0

- 21 x + 24 y + 15 z =120

Therefore the equation of the plane will be

- 21 x + 24 y + 15 z =120

Answer:

C(X,Y)(a,b)=1−C(s(X),t(Y))(a,1−b).

Step-by-step explanation:

Let's introduce the cumulative distribution of (X,Y), X and Y :

F(X,Y)(x,y)=P(X≤x,Y≤y)

Likewise for (s(X),t(Y)), s(X) and t(Y) :

F(s(X),t(Y))(u,v)=P(s(X)≤u

Now, First establish the relationship between F(X,Y) and F(s(X),t(Y)) :

F(X,Y)(x,y)=P(X≤x,Y≤y)=P(s(X)≤s(x),t(Y)≥t(y))

The last step is obtained by applying the functions s and t since s preserves order and t reverses it.

This can be further transformed into

F(X,Y)(x,y)=1−P(s(X)≤s(x),t(Y)≤t(y))=1−F(s(X),t(Y))(s(x),t(y))

Since our random variables are continuous, we assume that the difference between t(Y)≤t(y) and t(Y)<t(y)) is just a set of zero measure.

Now, to transform this into a statement about copulas, note that

C(X,Y)(a,b)=F(X,Y)(F−1X(a), F−1Y(b))

Thus, plugging x=F−1X(a) and y=F−1Y(b) into our previous formula,

we get

F(X,Y)(F−1X(a),F−1Y(b))=1−F(s(X),t(Y))(s(F−1X(a)),t(F−1Y(b)))

The left hand side is the copula C(X,Y), the right hand side still needs some work.

Note that

Fs(X)(s(F−1X(a)))=P(s(X)≤s(F−1X(a)))=P(X≤F−1X(a))=FX(F−1X(a))=a

and likewise

Ft(Y)(s(F−1Y(b)))=P(t(Y)≤t(F−1Y(b)))=P(Y≥F−1Y(b))=1−FY(F−1Y(b))=1−b

Combining all results we obtain for the relationship between the copulas

C(X,Y)(a,b)=1−C(s(X),t(Y))(a,1−b).

Answer:

Step-by-step explanation:

Given that A is a square matrix and A is idempotent

[tex]A^2 = A[/tex]

Consider I-A

i) [tex](I-A)^2 = (I-A).(I-A)\\= I^2 -2A.I+A^2\\= I-2A+A\\=I-A[/tex]

It follows that I-A is also idempotent

ii) Consider the matrix 2A-I

[tex](2A-I).(2A-I)=\\4A^2-4AI+I^2\\= 4A-4A+I\\=I[/tex]

So it follows that 2A-I matrix is its own inverse.

Answer:

a) Equation is

[tex]y = 0.1135x+1.590[/tex]

b) Slope = 0.1135 represents the change in y for a unit change in x

i.e. When nitrate content is increasedby 1, absorbence is increased by 0.1135

Step-by-step explanation:

Nitrates Absorbence

x y

50 7

50 7.6

100 12.7

200 24

400 47

800 93

1200 138

1600 183

2000 231

2000 226

SUMMARY OUTPUT        

Regression Statistics        

Multiple R 0.999911043        

R Square 0.999822094        

Adjusted R Square 0.999799856        

Standard Error 1.2890282        

Observations 10        

Coefficients

Intercept 1.589782721

x 0.113500259

we get regression line as

y = 0.1135x+1.590

a) Equation is

[tex]y = 0.1135x+1.590[/tex]

b) Slope = 0.1135 represents the change in y for a unit change in x

i.e. When nitrate content is increasedby 1, absorbence is increased by 0.1135

Answer:

a) h'(t)= (6cos3t,-9sin3t,3[tex]\sqrt[]{5}[/tex]cos3t)

b) t=0.93994736+πn/3

c) Magnitude of h(t) is 3 which is a constant, so h(t) lies on a sphere

Step-by-step explanation:

Answer:

0.008 is the probability that a computer will take more than 42 seconds to boot up.                                  

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 30 seconds

Standard Deviation, σ = 5 second

We are given that the distribution of time taken for a computer to boot up is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

a) P(computer will take more than 42 seconds to boot up)

P(x > 42)

[tex]P( x > 42) = P( z > \displaystyle\frac{42 - 30}{5}) = P(z > 2.4)[/tex]

[tex]= 1 - P(z \leq 2.4)[/tex]

Calculation the value from standard normal z table, we have,  

[tex]P(x > 42) = 1 - 0.992 = 0.008[/tex]

0.008 is the probability that a computer will take more than 42 seconds to boot up.

Answer:

[tex] -3*(3y+2) + 9y = -6[/tex]

[tex] -9y -6 + 9y = -6[/tex]

[tex]-6=-6[/tex]

So then as we can see we can have infinite solutions.

[tex]S= [(x, \frac{x-2}{3}) , x \in R][/tex]

Step-by-step explanation:

Assuming the following system of equations:

[tex] 2x-6y =4[/tex]   (1)

[tex] -3x+9y =-6[/tex]   (2)

For this case we can use the substitution method in order to find the possible solutions for the system.

If we solve for x from equation (1) we got:

[tex] 2x = 6y +4[/tex]

[tex] x = 3y +2 [/tex]   (3)

Now we can replace equation (3) into equation (2) and we got:

[tex] -3*(3y+2) + 9y = -6[/tex]

[tex] -9y -6 + 9y = -6[/tex]

[tex]-6=-6[/tex]

So then as we can see we can have infinite solutions.

And the possible solutions are for a fixed value of x, we can solve y from equation (3) and we got:

[tex] y = \frac{x-2}{3}[/tex]

So the solution would be: [tex]S= [(x, \frac{x-2}{3}) , x \in R][/tex]

Answer:

ML = {bba, ab, a, bbaabb, ababb, aabb}

Step-by-step explanation:

By application of Union of a set.

M = {bba,ab, a}

L = {Λ,abb}

ML = {bba, ab, a, bbaabb, ababb, aabb}

Since the real track has a gauge of 3 feet but the model railroad track has a gauge of 3/4 inches, the scale factor must be 1/4, because you must follow the rule

"true measurement * scale factor = model measurement"

In fact, we have

[tex]3\cdot\dfrac{1}{4}=\dfrac{3}{4}[/tex]

This implies that the original driving wheels height, 44 inches, must be scaled down to

[tex]44\cdot\dfrac{1}{4}=11[/tex]

inches.

Answer:

[tex] ME = \frac{10.08}{2}= 5.04[/tex]

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=50.86[/tex] represent the sample mean  

[tex]\mu[/tex] population mean

s represent the sample standard deviation  

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:  

[tex] \bar X \pm ME[/tex]   (1)

Or equivalently:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (2)

Where the margin of error is given by:

[tex] ME=  t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]

For this case we have the confidence interval limits given (45.82, 55.90)

We can find the width of the interval like this:

[tex] Width =55.90-45.82= 10.08[/tex]

And now the margin of error would be the half of the width since we assume that the confidence interval is symmetrical.

[tex] ME = \frac{10.08}{2}= 5.04[/tex]

Answer:

You did the same on both exams.

Step-by-step explanation:

To compare both the scores, we need to compute the z scores of both the exams and then compare the values. The formula for z-score is:

Z = (X - μ)/σ

Where X = score obtained

           μ = mean score

           σ = standard deviation

For Exam 1:

Z = (95 - 79)/8

  = 16/8

Z = 2

For Exam 2:

Z = (90 - 60)/15

  = 30/15

Z = 2

The z-scores for both the tests are same hence the third option is correct i.e. you did the same on both exams.

Your performance on exam 1 and exam 2 can be compared using Z-scores, which measure how many standard deviations a score is from the mean. You scored 2 standard deviations above the mean on both exams, so you did the same on both exams.

In this question, your performance on exams is being compared relative to the mean of the class scores and their standard deviation. This is a concept in statistics known as Z-scores. The Z-score tells us how many standard deviations an observation (your score) is from the mean. The formula for Z-score is (observation - mean) / standard deviation.

For exam 1 your Z - score is (95-79) / 8 which equals 2. This means you scored 2 standard deviations above the mean on exam 1. For exam 2, your Z-score is (90-60) / 15 which equals 2. Again, this means you scored 2 standard deviations above the mean on exam 2. Because your Z-score for both exams is the same, you did the same on both exams.

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Answer:

(a) Probability that a randomly selected student is taking Spanish given that he or she is taking French = 0.5 .

(b) Probability that a randomly selected student is not taking French given that he or she is not taking Spanish = 0.6 .

Step-by-step explanation:

We are given that an elementary school is offering 2 language classes ;

 Spanish Language is denoted by S and French language is denoted by F.

Also we are given, P(S) = 0.5 {Probability of students taking Spanish language}

P(F) = 0.4 {Probability of students taking French language}

[tex]P(S\bigcup F)[/tex] = 0.7 {Probability of students taking Spanish or French Language}

We know that,  [tex]P(A\bigcup B)[/tex]  = [tex]P(A) + P(B) -[/tex] [tex]P(A\bigcap B)[/tex]

So, [tex]P(S\bigcap F)[/tex] = [tex]P(S) + P(F) - P(S\bigcup F)[/tex] = 0.5 + 0.4 - 0.7 = 0.2

[tex]P(S\bigcap F)[/tex] means Probability of students taking  both Spanish and French Language.

Also, P(S)' = 1 - P(S) = 1 - 0.5 = 0.5

         P(F)' = 1 - P(F) = 1 - 0.4 = 0.6

        [tex]P(S'\bigcap F')[/tex] = 1 -  [tex]P(S\bigcup F)[/tex] = 1 - 0.7 = 0.3

(a) Probability that a randomly selected student is taking Spanish given that he or she is taking French is given by P(S/F);

  P(S/F) = [tex]\frac{P(S\bigcap F)}{P(F)}[/tex] = [tex]\frac{0.2}{0.4}[/tex] = 0.5

(b) Probability that a randomly selected student is not taking French given that he or she is not taking Spanish is given by P(F'/S');

   P(F'/S') = [tex]\frac{P(S'\bigcap F')}{P(S')}[/tex] = [tex]\frac{1- P(S\bigcup F)}{1-P(S)}[/tex] = [tex]\frac{0.3}{0.5}[/tex] = 0.6 .

Note: 2. A pair of fair dice is rolled until a sum of either 5 or 7 appears  ; This question is incomplete please provide with complete detail.

Answer:

Option D

75.39

Step-by-step explanation:

When provided with the co-ordinates (x, y) and(a, b) then the distance between them is given by  [tex]\sqrt {(x-a)^{2}+(y-b)^{2}}[/tex]

Since  u = (25, $350) and v = (53, $420) then the Euclidean distance will be

[tex]\sqrt {(53-25)^{2}+(350-420)^{2}}=75.3923073\approx 75.39[/tex]

The dissimilarity between the two observations using Euclidean distance is approximately 75.39(Option d).

To calculate the dissimilarity between two observations using Euclidean distance, we need to find the distance between the corresponding elements in the two observations and then calculate the square root of the sum of their squared differences.

In this case, we have:

u = (25, $350) and v = (53, $420)

The distance between the ages is 53 - 25 = 28.

The distance between the amounts spent is $420 - $350 = $70.

Now we can use the formula for Euclidean distance:

Distance = sqrt((28)^2 + (70)^2) = sqrt(784 + 4900) = sqrt(5684) ≈ 75.39

Therefore, the dissimilarity between the two observations using Euclidean distance is approximately 75.39.

Learn more about Euclidean distance here:

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Answer:

0.0726 or 7.26%

Step-by-step explanation:

When choosing the first vial, there is a 16 in 58 chance that the vial does not have a hairline crack. When choosing the second vial, since on good vial was already picked, there is a 15 in 57 chance that the vial does not have a hairline crack. The probability that none of the 2 vials have hairline cracks is:

[tex]P = \frac{16}{58}*\frac{15}{57}\\P=0.0726[/tex]

There is a 0.0726 or 7.26% chance that none of the 2 vials have a hairline crack.

Answer:

a) Probability of picking the first two answers wrong & the third answer correctly in that order, P(WWC) = 0.128

b) All possible outcomes = WWC, WCW, CWW

P(WWC) = 0.128; P(WCW) = 0.128; P(CWW) = 0.128

c) Total Probability of picking only one correct answer and two incorrect answers from the first three attempts = 0.384

Step-by-step explanation:

P(Correct answer) = P(C) = number of correct options/total options available.

That is, P(C) = 1/5 = 0.2

P(Wrong answer) = P(W) = 1 - 0.2 = 0.8 or (number of wrong options)/(total options available) = 4/5 = 0.8

a) Using the multiple rule for independent events,

P(WWC) = P(W) × P(W) × P(C) = 0.8 × 0.8 × 0.2 = 0.128

Probability of picking the first two answers wrong & the third answer correctly in that order = 0.128

b) All possible outcomes of picking two wrong answers and one right answer in whichever order for the first 3 questions are (WWC, WCW, CWW)

P(WWC) = P(W) × P(W) × P(C) = 0.8 × 0.8 × 0.2 = 0.128

P(WCW) = P(W) × P(C) × P(W) = 0.8 × 0.2 × 0.8 = 0.128

P(CWW) = P(C) × P(W) × P(W) = 0.2 × 0.8 × 0.8 = 0.128

c) Using the addition rule for disjoint events,

Total Probability of picking only one correct answer and two incorrect answers from the first three attempts = P(WWC) + P(WCW) + P(CWW) = 0.128 + 0.128 + 0.128 = 0.384

QED!

The probability of getting exactly one correct answer when guessing on three multiple-choice questions with five possible answers each is [tex]\(\frac{3}{25}\)[/tex].

a. To find the probability that the first two guesses are wrong and the third is correct, denoted as P(WWC), we use the multiplication rule for independent events. Since there are five possible answers for each question and only one correct answer, the probability of guessing wrong on one question is [tex]\(\frac{4}{5}\)[/tex] and the probability of guessing correctly is [tex]\(\frac{1}{5}\)[/tex]. Therefore, the probability of WWC is:

[tex]\[ P(WWC) = P(W) \times P(W) \times P(C) = \left(\frac{4}{5}\right) \times \left(\frac{4}{5}\right) \times \left(\frac{1}{5}\right) = \frac{16}{125} \][/tex]

b. The different possible arrangements of two wrong answers and one correct answer (WWC) are:

1. WWC

2. WWC

3. CWW

4. CW

5. WCW

6. WCW

For each of these arrangements, the probability is the same as calculated in part (a), which is \(\frac{16}{125}\).

c. Since there are six different ways to arrange two wrong answers and one correct answer, and each of these arrangements has the same probability, we multiply the probability of one such arrangement by the number of arrangements to find the total probability of getting exactly one correct

[tex]\[ P(\text{exactly one correct answer}) = 6 \times P(WWC) = 6 \times \frac{16}{125} = \frac{96}{125} \][/tex]

However, we must note that we have counted each arrangement twice because the order of the wrong answers (W) does not matter. Therefore, we need to divide the total by 2 to get the correct probability:

[tex]\[ P(\text{exactly one correct answer}) = \frac{96}{2 \times 125} = \frac{48}{125} \][/tex]

Upon reviewing the calculations, it appears there was an error in the final step. We should not divide by 2 because the arrangements are distinct due to the position of the correct answer (C). The correct total probability is indeed [tex]\(\frac{96}{125}\)[/tex], but since we are looking for the probability of exactly one correct answer, we have to consider that there are three questions and the correct answer could be on any of them. Therefore, we have already accounted for all possible arrangements by multiplying by 6.

The correct probability of getting exactly one correct answer when guessing on three multiple-choice questions is:

[tex]\[ P(\text{exactly one correct answer}) = \frac{96}{125} \][/tex]

However, this is not the final answer. We need to consider that there are three different ways to get exactly one correct answer out of three questions (CWW, WCW, WWC). Since these are mutually exclusive events, we add their probabilities:

[tex]\[ P(\text{exactly one correct answer}) = P(CWW) + P(WCW) + P(WWC) \][/tex]

[tex]\[ P(\text{exactly one correct answer}) = \frac{16}{125} + \frac{16}{125} + \frac{16}{125} \][/tex]

[tex]\[ P(\text{exactly one correct answer}) = 3 \times \frac{16}{125} \][/tex]

[tex]\[ P(\text{exactly one correct answer}) = \frac{48}{125} \][/tex]

Thus, the final correct probability is [tex]\(\frac{48}{125}\)[/tex], which simplifies to [tex]\(\frac{3}{25}\)[/tex].

Answer:

linear: 5s

quadratic: 50s

log-linear: 0.75 s

cubic: 500s

Step-by-step explanation:

Let [tex]t_1,t_2[/tex] be the running time associated with the input of sizes [tex] s_1,s_2[/tex]

If the running time is linear

[tex]t_2 = t_1\frac{s_2}{s_1} = 0.5*\frac{1000}{100} = 0.5*10 = 5s[/tex]

If the running time is quadratic

[tex]t_2 = t_1\left(\frac{s_2}{s_1}\right)^2 = 0.5*\left(\frac{1000}{100}\right)^2 = 0.5*10^2 = 50s[/tex]

If the running time is log-linear

[tex]t_2 = t_1\frac{log(s_2)}{log(s_1)} = 0.5*\frac{log(1000)}{log(100)} = 0.5*1.5 = 0.75s[/tex]

If the running time is cubic:

[tex]t_2 = t_1\left(\frac{s_2}{s_1}\right)^3 = 0.5*\left(\frac{1000}{100}\right)^3 = 0.5*10^3 = 500s[/tex]

Answer: 13.04

Here are some consumer price indexes from the past 100+ years:

Year CPI

1909 9.1

1919 17.3

1929 17.1

1939 13.9

1949 23.8

1959 29.1

1969 36.7

1979 72.6

1989 118.3

1999 166.6

2009 214.5

2015 238.5

The admission price was $1.00 in 1909. How much would the Speedway have had to charge in 1989 to match the purchasing power of $1 in 1909? In other words, how much was that in 1989?

We are required to calculate the percentage increase in tax and determine who is right.

The percentage increase in tax is 25% and Linda is very correct

percentage increase = difference in tax /

percentage increase = difference in tax / old tax × 100

old tax = 2%

New tax = 2.5%

Difference = New tax - old tax

= 2.5% - 2%

= 0.5%

percentage increase = difference in tax /

percentage increase = difference in tax / old tax × 100

= 0.5% / 2% × 100

= 0.25 × 100

= 25%

Therefore, the percentage increase in tax is 25% and Linda is very correct

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Answer:

(√6/√2)ft

Step-by-step explanation:

cos 45 = m / √6 ft

m = cos 45 x √6 ft

m = (1 / √2) x √6 ft = (√6/√2)ft

Answer:

The right answer is option 3.

lim x → −3 f(x) = [infinity] means the values of f(x) can be made arbitrarily large by taking x sufficiently close to (but not equal to) −3.

Step-by-step explanation:

The limit of a function is a fundamental concept concerning the behavior of that function near a particular input.

A function f assigns an output f(x) to every input x. We say the function has a limit L at an input a: this means f(x) gets closer and closer to L as x moves closer and closer to a. More specifically, when f is applied to any input sufficiently close to a, the output value is forced arbitrarily close to L.

That is,

lim x → a f(x) = L

Hope this helps!

The limit lim x → −3 f(x) = [infinity] means that as x values get closer to -3 (without becoming -3), the value of the function f(x) goes towards infinity i.e., it grows without bound. This is akin to certain function behaviors near a value at which an asymptote is present. However, the second part about f(x) values getting close to 0 seems contradiction to the first statement.

The statement lim x → −3 f(x) = [infinity] is related to a concept in Calculus known as a limit. When we say that the limit of f(x) as x approaches -3 is infinity, we mean that as we make x values closer and closer to -3 (without letting x actually be -3), the value of the function f(x) becomes larger and larger without bound, i.e., approaches infinity.

This is similar to some function behaviors near an asymptote. For example, the function y = 1/x has a vertical asymptote at x = 0, where y approaches infinity as x approaches zero from either direction. Here, as x gets arbitrarily close to 0, the value of y = 1/x gets arbitrarily large, or 'approaches infinity'.

On the other hand, when the question states, 'The values of f(x) can be made arbitrarily close to 0 by taking x sufficiently close to (but not equal to) -3', it signifies the tendency of the function values to get closer and closer to 0 as x gets closer to -3. This indicates a certain limit behavior, but it seems to be contradictory with the first part where the limit was stated to be infinity. It is important to scrutinize the function's properties and behavior around x = -3 carefully.

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Answer:

300 million people

Step-by-step explanation:

If the participation rate is 60% and 200 million people 16 years or older are not in the labor force, it means that 200 million corresponds to 40% of people 16 years or older. Since 60% of people 16 years or older are in the labor force, the total number of people in the labor force is given by:

[tex]n=\frac{200}{0.4}-200\\ n= 300\ million\ people[/tex]

300 million people are in the labor force in this economy.

Answer:

There is a 25.52% probability of observating 4 our fewer succesful recommendations.

Step-by-step explanation:

For each recommendation, there are only two possible outcomes. Either it was a success, or it was a failure. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

In this problem we have that:

[tex]p = 0.553, n = 10[/tex]

If the claim is correct and the performance of recommendations is independent, what is the probability that you would have observed 4 or fewer successful:

This is

[tex]P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]

In which

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

[tex]P(X = 0) = C_{10,0}.(0.553)^{0}.(0.447)^{10} = 0.0003[/tex]

[tex]P(X = 1) = C_{10,1}.(0.553)^{1}.(0.447)^{9} = 0.0039[/tex]

[tex]P(X = 2) = C_{10,2}.(0.553)^{2}.(0.447)^{8} = 0.0219[/tex]

[tex]P(X = 3) = C_{10,3}.(0.553)^{3}.(0.447)^{7} = 0.0724[/tex]

[tex]P(X = 4) = C_{10,4}.(0.553)^{4}.(0.447)^{6} = 0.1567[/tex]

[tex]P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0003 + 0.0039 + 0.0219 + 0.0724 + 0.1567 = 0.2552[/tex]

There is a 25.52% probability of observating 4 our fewer succesful recommendations.

Answer:

Number of births to women below 22 years of age: 830

% of births occurred to women of age between 34 and 36: 6.35%

Step-by-step explanation:

There are two parts to this question:

To calculate the number of births, we look at the histogram below.

We see that each bar has a number on top, suggesting that particular for the age limit. Number of births to women below age of 22, we start adding all numbers below the mark of 22 on x-axis:

⇒ 39+134+29+363 = 830

To calculate the percentage, we divide the number of births in that particular interval by total number of births and then multiply by 100.

To calculate the total number of births, we add all the numbers on the top of the bars:

⇒ 39+ 134+294+363+391+425+460+474+432+343+250+163+98+49+17+4+1 Total births = 3937

[tex]\frac{250}{3937} \times 100\\ 0.0635 \times 100\\= 6.35 \%[/tex]

(a) The probability that a hand will contain exactly two wild cards is 0.076.

(b) The probability that a hand will contain two wild cards, two red cards, and three blue cards is 0.0007.

Let's solve these problems using the concept of combinations in probability.

Remember, [tex]C(n, k)[/tex] denotes the number of ways to choose k items from a set of n items, and is calculated as

[tex]C(n,k)=\frac{n!}{k!(n-k)!}[/tex]

where "!" denotes factorial. For example, [tex]5! = 5 \times 4 \times 3 \times 2 \times 1[/tex]

(a) The probability that a hand will contain exactly two wild cards:

The total number of ways to choose 7 cards out of 108 is [tex]C(108, 7)[/tex].

The number of ways to choose 2 wild cards out of 8 is [tex]C(8, 2)[/tex].

The number of ways to choose the remaining 5 cards out of the 100 non-wild cards is [tex]C(100, 5)[/tex].

So, the probability is [tex]\frac{C(8, 2) \times C(100, 5)}{C(108, 7)} \approx 0.076[/tex]

(b) The probability that a hand will contain two wild cards, two red cards, and three blue cards:

The number of ways to choose 2 wild cards out of 8 is [tex]C(8, 2)[/tex].

The number of ways to choose 2 red cards out of 25 is [tex]C(25, 2)[/tex].

The number of ways to choose 3 blue cards out of 25 is [tex]C(25, 3)[/tex].

So, the probability is [tex]\frac{C(8, 2) \times C(25, 2) \times C(25, 3)}{C(108, 7)} \approx 0.0007[/tex]

To Plotting lines , use the PFloor line tool and click your mouse for the first and second end-points, ensuring that the line intersects both the D1 and S1 lines and has a height greater than 50.

To plot the line described in the question, follow these steps:

Select the line tool on the tool palette labeled PFloor.

Click your mouse for the first end-point on the vertical axis.

Move your mouse to the right and click again for the second end-point.

The new line should intersect both the D1 and S1 lines and have a height greater than 50 on the vertical axis.

Learn more about Plotting lines here:

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